Oracle OCP SQL 3 Single-Row Functions
ORACLE_分析函数大全
ORACLE_分析函数大全Oracle分析函数是一种高级SQL函数，它可以在查询中实现一系列复杂的分析操作。
这些函数可以帮助我们在数据库中执行各种数据分析和报表生成任务。
本文将介绍Oracle数据库中的一些常用分析函数。
1.ROW_NUMBER函数：该函数为查询结果中的每一行分配一个唯一的数字。
可以用它对结果进行排序或分组。
例如，可以使用ROW_NUMBER函数在结果集中为每个员工计算唯一的编号。
2.RANK和DENSE_RANK函数：这两个函数用于计算结果集中每个行的排名。
RANK函数返回相同值的行具有相同的排名，并且下一个排名值将被跳过。
DENSE_RANK函数类似，但是下一个排名值不会被跳过。
G和LEAD函数：LAG函数返回结果集中指定列的前一个（上一个）行的值，而LEAD函数返回后一个（下一个）行的值。
这些函数通常用于计算增长率或发现趋势。
4.FIRST和LAST函数：这两个函数用于返回结果集中分组的第一个和最后一个行的值。
可以与GROUPBY子句一起使用。
5.CUME_DIST函数：该函数用于计算给定值的累积分布。
它返回值的累积分布在结果集中的位置（百分比）。
6.PERCENT_RANK函数：该函数用于计算结果集中每个行的百分位数排名。
它返回值的百分位数排名（0到1之间的小数）。
7. NTILE函数：该函数用于将结果集分成指定数量的桶（Bucket），并为每个行分配一个桶号。
通常用于将数据分组为更小的块。
8.LISTAGG函数：该函数将指定列的值连接成一个字符串，并使用指定的分隔符分隔每个值。
可以用它将多个值合并在一起形成一个字符串。
9.AVG、SUM、COUNT和MAX/MIN函数：这些是常见的聚合函数，可以在分析函数中使用。
它们用于计算结果集中的平均值、总和、计数和最大/最小值。
以上只是Oracle数据库中的一些常用分析函数。
还有其他一些分析函数，如PERCENTILE_CONT、PERCENTILE_DISC等可以用于更高级的分析计算。
oraclefunction用法
oraclefunction⽤法函数调⽤限制1、SQL语句中只能调⽤存储函数(服务器端),⽽不能调⽤客户端的函数2、SQL只能调⽤带有输⼊参数,不能带有输出,输⼊输出函数3、SQL不能使⽤PL/SQL的特有数据类型(boolean,table,record等)4、SQL语句中调⽤的函数不能包含INSERT,UPDATE和DELETE语句1.function函数的语法如下：1create or replace function function_name (2 argu1 [mode1] datatype1, --定义参数变量3 argu2 [mode2] datatype2 --定义参数变量4 ) return datatype --定义返回的数据类型5is67begin89end;执⾏：var v1 varchar2(100)exec :v1:=function_name2.不带任何参数的定义create or replace function get_userreturn varchar2isResult varchar2(50); --定义变量beginselect username into Result from user_users;return(Result); --返回值end get_user;3.带有in参数的create or replace function get_sal(empname in varchar2) return numberisResult number;beginselect sal into Result from emp where ename=empname;return(Result);end;执⾏：SQL>var sal numberSQL>exec :sal:=get_sal('scott');4.带out参数的create or replace function get_info(e_name varchar2,job out varchar2) return numberIsResult number;beginselect sal,job into Result,job from emp where ename=e_name;return(Result);end;执⾏：SQL>var job varchar2(20)SQL>var dname varchar2(20)SQL>exec :dname:=get_info('SCOTT',:job)5.带in out参数的6.函数调⽤举例create or replace function f_sys_getseqid(v_seqname IN VARCHAR2,v_provincecode IN VARCHAR2--省编码) return Varchar2ISiv_date VARCHAR2(8);iv_seqname VARCHAR2(50);iv_sqlstr VARCHAR2(200);iv_seq VARCHAR2(8);iv_seqid VARCHAR2(16);BEGINiv_seqname :=LOWER(TRIM(v_seqname));iv_sqlstr :='SELECT '||iv_seqname||'.nextval FROM DUAL';EXECUTE IMMEDIATE iv_sqlstr INTO iv_seq;--执⾏动态的sql语句，执⾏相似的⼀组语句IF v_seqname ='SEQ_FUNCROLE_ID'THENiv_seqid:='ESS'|| LPAD(iv_seq,5,'0');ELSESELECT substrb(v_provincecode,1,2)||TO_CHAR(SYSDATE,'yymmdd') INTO iv_date FROM DUAL;iv_seqid:= iv_date || LPAD(iv_seq,8,'0');END IF;RETURN iv_seqid;EXCEPTIONWHEN OTHERS THENRETURN NULL;END;调⽤⽅式如下：SELECT TO_NUMBER(F_SYS_GETSEQID('SEQ_TERMTRADE_ID', V_PROVINCE_CODE)) INTO V_BATCH_ID FROM DUAL; EXECUTE IMMEDIATE的说明：执⾏动态的sql语句。
ocp 考试题型
ocp 考试题型OCP（Oracle Certified Professional）认证是针对Oracle数据库专业人员的一种全球性认证。
凭借OCP认证，可以证明自己在设计、开发和管理Oracle数据库方面具备专业的技能和知识。
在考试中，存在不同类型的题型，以下将对OCP考试题型进行详细的介绍。
1. 单选题单选题是OCP考试中最常见的题型之一。
在单选题中，为了测试考生对特定概念、技术或应用的理解和掌握程度，会提供一个问题和多个备选答案，考生需要选择出唯一正确的答案。
举例：Question: Oracle数据库中，以下哪个命令用于创建一个新的表？A. CREATE INDEXB. CREATE TRIGGERC. CREATE PROCEDURED. CREATE TABLE在这个问题中，正确答案是D，因为在Oracle数据库中，创建新表需要使用CREATE TABLE命令。
2. 多选题多选题是OCP考试中的另一个常见题型。
与单选题类似，多选题也会提供一个问题和多个备选答案，但考生需要选择所有正确的答案。
举例：Question: 以下哪些语句可以修改Oracle数据库中已有表的列？A. ALTER TABLEB. UPDATE TABLEC. MODIFY COLUMND. CHANGE COLUMN在这个问题中，正确答案是A和C，因为ALTER TABLE和MODIFY COLUMN可以用于修改已有表的列。
3. 填空题填空题是OCP考试中用于测试考生对特定概念或语法的掌握程度的一种题型。
考生需要根据问题的要求，在空白处填写正确的答案。
举例：Question: 使用Oracle SQL语句查询表中的所有数据，可使用_____命令。
在这个问题中，正确答案是SELECT。
4. 情景题情景题是对考生在实际工作场景中解决问题的能力的测试。
会给出一个具体的情景描述，考生需要根据情境选择合适的解决方案。
举例：Question: 在一个高负载的Oracle数据库环境中，发现部分查询性能下降。
oracle 的function方法
oracle 的function方法Oracle的Function方法是Oracle数据库中一种非常重要的功能，它允许用户在数据库中定义自己的函数，以实现特定的业务需求。
本文将介绍Oracle的Function方法的定义、使用场景以及一些常见的示例。
我们来了解一下Oracle的Function方法的定义。
Function方法是一种存储过程，它接收输入参数并返回一个值。
与存储过程不同的是，Function方法必须返回一个值，并且可以直接在SQL语句中调用。
它可以用于计算、转换数据、执行复杂的业务逻辑等。
接下来，我们来看一些使用Oracle的Function方法的场景。
首先，Function方法可以用于计算某个数据列的总和、平均值、最大值、最小值等统计信息。
例如，我们可以定义一个名为get_total_sales 的Function方法，用于计算某个销售表中所有销售额的总和。
Function方法还可以用于数据转换。
例如，我们可以定义一个名为convert_currency的Function方法，用于将某个货币金额转换为其他货币的金额。
这在跨国企业的财务报表中非常常见。
Function方法还可以用于执行复杂的业务逻辑。
例如，我们可以定义一个名为check_stock的Function方法，用于检查某个产品的库存是否充足。
如果库存不足，则返回一个提示信息；如果库存充足，则返回一个成功信息。
下面，我们来看一些具体的Oracle的Function方法的示例。
首先，我们定义一个名为get_total_sales的Function方法，用于计算某个销售表中所有销售额的总和。
该方法的定义如下：CREATE OR REPLACE FUNCTION get_total_salesRETURN NUMBERIStotal_sales NUMBER := 0;BEGINSELECT SUM(sales_amount)INTO total_salesFROM sales_table;RETURN total_sales;END;在上述示例中，我们使用了SUM函数来计算销售表中所有销售额的总和，并将结果保存在total_sales变量中。
oracle ocp考试内容
oracle ocp考试内容
OracleOCP考试是一种面向数据库管理员和开发人员的专业认证考试，它涵盖了多个方面的数据库技术，包括数据库设计、安装和升级、备份和恢复、性能调优等。
以下是Oracle OCP考试的具体内容： 1. 数据库设计：包括数据库的范式化、数据建模、数据库对象的创建和管理等；
2. 数据库安装和升级：包括操作系统的选择和安装、数据库软件的选择、安装和升级、数据库实例的创建和配置等；
3. 数据库备份和恢复：包括备份和恢复的类型、备份和恢复的策略、备份和恢复的命令和工具等；
4. 性能调优：包括性能优化的方法、性能调优的工具和技术、SQL语句的调优等；
5. 安全管理：包括用户和权限管理、安全策略的制定和实施、数据加密和解密等；
6. 高可用性：包括故障转移、数据复制、容错管理等。
以上是Oracle OCP考试的主要内容，考生需要通过在线测试和实验来完成考试，获得Oracle OCP认证。
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OCP认证详细介绍
在全球近乎半数的企业与组织在使用Oracle的数据库产品，很多技术Oracle都保持着行业内独一无二的地位。
Oracle技术岗位是世界上市场上炙手可热的紧缺型技术人才种类，也是IT从业人员获得高薪的保障之一。
在中国oracle的技术广泛应用于各行各业，其中电信、电力、金融、政府及大型制造业都需要oracle技术人才，未来5年有多达10万相关技术人员的缺口，在Oracle 技术需求这么旺盛的时期，尤其数据库技术人才，具备Oracle OCP认证牌照的技术人才在就业方面具有相当大的竞争力。
Oracle认证中企业最看重，参加人员最多的是DBA认证。
Oracle认证目前分为三个层次OCA、OCP、OCM。
OCA即Oracle Certified Associate，OCP即Oracle Certified Professional，OCM即Oracle Certified Master。
其中OCP包括了数据库管理方向的DBA、数据库开发方向的DEV及Oracle应用产品方向的Applications专家。
DBA认证中最受欢迎的是Oracle OCP认证。
要成为OCP，必须先获得OCA证书，然后才能参加OCP要求的其他考试。
参加OCP认证的学员必须至少在Oracle大学或者其授权培训中心学习一门课程才能获得OCP证书。
这些考试也必须在授权的国际认证考试中心或者Oracle授权培训中心进行。
通过Oracle认证后，便成为了Oracle专家，Oracle专家能跟上如今日趋复杂的系统环境需求。
最佳的Oracle DBA们都在幕后工作，他们小心地维护着系统，使得系统能每天都平稳地运转，并且防止意外灾难的发生，譬如数据库崩溃或成小时地宕机。
这项艰巨的任务需要对Oracle数据库的结构和运行方式有着广泛深入的了解，并且有丰富的实战经验。
最佳的DBA能在取得最佳运行状态及防止他们公司停止运行的突发事件中找到平衡。
正是因为Oracle证书获取的苛刻条件，所以Oracle认证专家证书成了衡量人们具备Oracle相关知识与技能的最高标志，Oracle认证就业前景良好。
ocp课程大纲
ocp课程大纲摘要：一、OCP课程简介二、OCP课程大纲概述1.课程模块划分2.课程内容简介三、OCP课程实战应用1.实战案例介绍2.实践技能提升四、OCP课程学习建议1.学习方法与策略2.考试准备与技巧五、OCP课程的价值与意义1.对个人职业发展的影响2.对企业运维能力的提升正文：一、OCP课程简介OCP（Oracle Certified Professional）课程是Oracle 官方认证课程，旨在培养具备Oracle 数据库管理和开发能力的专业人才。
通过学习OCP课程，学员可以熟练掌握Oracle数据库的核心技术，提升工作效率，增强职业竞争力。
二、OCP课程大纲概述1.课程模块划分OCP课程大纲分为五个模块，分别是：数据库管理基础、数据库性能优化、数据库安全、SQL语言和PL/SQL编程。
2.课程内容简介（1）数据库管理基础：介绍Oracle数据库的基本概念、架构和常用管理工具。
（2）数据库性能优化：讲解数据库性能诊断、优化策略和调整技巧。
（3）数据库安全：阐述数据库安全策略、访问控制、加密技术等。
（4）SQL语言：深入解析SQL语句的编写原则和技巧，包括数据查询、数据插入、数据更新等。
（5）PL/SQL编程：介绍PL/SQL编程基础、流程控制、异常处理等。
三、OCP课程实战应用1.实战案例介绍OCP课程紧密结合实际工作场景，通过多个实战案例让学员深入了解Oracle数据库的运维和管理。
例如，数据库性能诊断与优化、数据备份与恢复、SQL语句优化等。
2.实践技能提升课程结束后，学员将具备独立管理和优化Oracle数据库的能力，提升工作效率，为个人和企业创造价值。
四、OCP课程学习建议1.学习方法与策略（1）认真阅读教材，理解课程内容。
（2）动手实践，跟随课程案例操作，巩固理论知识。
（3）定期复习，总结学习笔记，加深印象。
2.考试准备与技巧（1）参加OCP认证考试前，确保掌握课程全部内容。
（2）多做模拟试题，熟悉考试题型和答题技巧。
oracle常用sql函数
oracle常用sql函数Oracle是一款非常优秀的数据库管理系统，它提供了很多强大的功能与工具，其中最基本而实用的功能就是SQL函数了。
下面我们将重点介绍一些常用的Oracle SQL函数，帮助读者更好地了解和掌握这些常用的函数，从而更加有效地处理和管理数据。
一、数值型函数1. ABS()函数：用于计算数值的绝对值，比如ABS(-10)会返回10。
2. CEIL()函数：用于向上取整，比如CEIL(3.14)会返回4。
3. FLOOR()函数：用于向下取整，比如FLOOR(3.14)会返回3。
4. MOD()函数：用于求模，比如MOD(12,5)会返回2。
5. ROUND()函数：用于四舍五入，比如ROUND(3.49)会返回3，而ROUND(3.51)会返回4。
二、字符型函数1. CONCAT()函数：用于连接两个字符串，比如CONCAT('hello','world')会返回helloworld。
2. SUBSTR()函数：用于截取字符串，比如SUBSTR('hello',2,3)会返回ell。
3. UPPER()函数：用于将字符串转换成大写，比如UPPER('hello')会返回HELLO。
4. LOWER()函数：用于将字符串转换成小写，比如LOWER('HELLO')会返回hello。
5. TRIM()函数：用于去掉字符串首尾的空格，比如TRIM(' hello ')会返回hello。
三、日期型函数1. CURRENT_DATE函数：用于获取当前日期，比如SELECT CURRENT_DATE FROM DUAL会返回当前日期。
2. SYSDATE函数：用于获取系统当前时间，比如SELECTSYSDATE FROM DUAL会返回系统当前时间。
3. TO_CHAR()函数：用于将日期转换成字符型，比如TO_CHAR(SYSDATE,'yyyy-mm-dd')会返回当前日期的年月日格式。
OCP考试047认证题库1z1-047
Oracle 1Z1-047Oracle® Database SQL ExpertQ&A DEMOEnglish: BIG5: GB: TestInside,help you pass any IT exam!1. Which statements are true? (Choose all that apply.)A. The data dictionary is created and maintained by the database administrator.B. The data dictionary views can consist of joins of dictionary base tables and user-defined tables.C. The usernames of all the users including the database administrators are stored in the data dictionary.D. The USER_CONS_COLUMNS view should be queried to find the names of the columns to which a constraint applies.E. Both USER_OBJECTS and CAT views provide the same information about all the objects that are owned by the user.F. Views with the same name but different prefixes, such as DBA, ALL and USER, use the same base tables from the data dictionaryAnswer: CDF2. View the Exhibit and examine the structure of the MARKS_DETAILS and MARKStables.Which is the best method to load data from the MARKS_DETAILStable to the MARKStable?A. Pivoting INSERTB. Unconditional INSERTC. Conditional ALL INSERTD. Conditional FIRST INSERTAnswer: A3. Which three statements are true regarding single-row functions? (Choose three.)A. They can accept only one argument.B. They can be nested up to only two levels.C. They can return multiple values of more than one data type.D. They can be used in SELECT, WHERE, and ORDER BY clauses.E. They can modify the data type of the argument that is referenced.F. They can accept a column name, expression, variable name, or a user-supplied constant as arguments. Answer: DEF4. Which two statements are true regarding operators used with subqueries? (Choose two.)A. The NOT IN operator is equivalent to IS NULL.B. The <ANY operator means less than the maximum.C. =ANY and =ALL operators have the same functionality.D. The IN operator cannot be used in single-row subqueries.E. The NOT operator can be used with IN, ANY and ALL operators.Answer: BE5. Which two statements are true regarding views? (Choose two.)A. A simple view in which column aliases have been used cannot be updated.B. A subquery used in a complex view definition cannot contain group functions or joins.C. Rows cannot be deleted through a view if the view definition contains the DISTINCT keyword.D. Rows added through a view are deleted from the table automatically when the view is dropped.E. The OR REPLACE option is used to change the definition of an existing view without dropping and re-creating it.F. The WITH CHECK OPTION constraint can be used in a view definition to restrict the columns displayed through the view.Answer: CE6. Which two statements are true about sequences created in a single instance database? (Choose two.)A. The numbers generated by a sequence can be used only for one table.B. DELETE <sequencename> would remove a sequence from the database.C. CURRV AL is used to refer to the last sequence number that has been generated.D. When the MAXV ALUE limit for a sequence is reached, you can increase the MAXV ALUE limit by using the ALTER SEQUENCE statement.E. When a database instance shuts down abnormally, the sequence numbers that have been cached but not used would be available once again when the database instance is restarted.Answer: CD7. Which statements are correct regarding indexes? (Choose all that apply.)A. When a table is dropped, the corresponding indexes are automatically dropped.B. For each DML operation performed, the corresponding indexes are automatically updated.C. Indexes should be created on columns that are frequently referenced as part of an expression.D. A non-deferrable PRIMARY KEY or UNIQUE KEY constraint in a table automatically creates a unique index. Answer: ABD8. The ORDERS table belongs to the user OE. OE has granted the SELECT privilege on the ORDERS table to the user HR.Which statement would create a synonym ORD so that HR can execute the following query successfully? SELECT * FROM ord;A. CREATE SYNONYM ord FOR orders; This command is issued by OE.B. CREATE PUBLIC SYNONYM ord FOR orders; This command is issued by OE.C. CREATE SYNONYM ord FOR oe.orders; This command is issued by the database administrator.D. CREATE PUBLIC SYNONYM ord FOR oe.orders; This command is issued by the database administrator. Answer: D9. Which statement is true regarding synonyms?A. Synonyms can be created for tables but not views.B. Synonyms are used to reference only those tables that are owned by another user.C. A public synonym and a private synonym can exist with the same name for the same table.D. The DROP SYNONYM statement removes the synonym, and the status of the table on which the synonym hasbeen created becomes invalid.Answer: C10. ORD is a private synonym for the OE. ORDERS table.The user OE issues the following command:DROP SYNONYM ord;Which statement is true regarding the above SQL statement?A. Only the synonym would be dropped.B. The synonym would be dropped and the corresponding table would become invalid.C. The synonym would be dropped and the packages referring to the synonym would be dropped.D. The synonym would be dropped and any PUBLIC synonym with the same name becomes invalid. Answer: A。
oracle ocp 中文题库
Oracle OCP 中文题库一、引言Oracle OCP（Oracle Certified Professional）认证是全球范围内认可的数据库专业人士认证体系之一，该认证体系包括了数据库管理、应用开发等多个方向。
作为一名希望取得Oracle OCP认证的考生，熟悉并掌握题库中的内容是非常重要的。
本文将针对Oracle OCP中文题库进行详细的介绍和讲解，希望能够对广大考生有所帮助。
二、Oracle OCP中文题库的组成1. 数据库概念部分：包括数据库的概念、体系结构、原理等内容。
2. SQL部分：主要涉及SQL语言的基本语法、查询、数据操作等内容。
3. 数据库管理部分：包括数据库的安装、配置、备份恢复、性能优化等内容。
4. 安全性部分：主要介绍数据库的安全管理和权限控制等内容。
5. 高级应用开发部分：包括存储过程、触发器、PL/SQL等高级内容。
三、Oracle OCP中文题库的重要性1. 作为备考工具：Oracle OCP中文题库是备考的重要工具，通过不断地练习题库中的题目，可以使考生更加熟练地掌握数据库的相关知识和技能。
2. 考核知识点：题库中的题目涵盖了Oracle数据库的各个知识点，考生可以通过题库了解考试的重点和难点。
3. 提高应试技巧：通过练习题库，考生可以提高自己的答题速度和技巧，为顺利通过考试打下基础。
四、如何有效使用Oracle OCP中文题库1. 制定学习计划：考生应根据自身的实际情况，合理制定学习计划，安排每天的学习时间，保证有足够的时间来练习题库。
2. 分模块练习：将题库中的题目按照不同的模块进行分类，有条不紊地进行练习，逐渐提高各个模块的掌握程度。
3. 定期自测：在练习题库的过程中，考生应定期进行自测，了解自己的学习情况和水平，及时调整学习计划。
4. 结合其他学习资源：除了题库，考生还可以结合其他学习资源，比如教材、考试指南、网络课程等，全面提升自己的学习效果。
五、总结Oracle OCP中文题库是备考Oracle OCP认证考试的重要辅助工具，考生应合理利用题库，制定科学的学习计划，通过不断的练习和自我调整，提高自己的学习和应试技巧，从而顺利通过考试取得认证资格。
Oracle OCP11g考试题库
Which statement is true regarding the COALESCE function?A. It can have a maximum of five expressions in a listB. It returns the highest NOT NULL value in the list for all rowsC. It requires that all expressions in the list must be of the same data typeD. It requires that at least one of the expressions in the list must have a NOT NULL valueAnswer: C2.View the Exhibit and examine the structure of the PROMOTIONS table.Which SQL statements are valid? (Choose all that apply.)A. SELECT promo_id, DECODE (NVL(promo_cost,0), promo_cost,promo_cost * 0.25, 100) "Discount"FROM promotions;B. SELECT promo_id, DECODE (promo_cost, 10000,DECODE (promo_category, 'G1', promo_cost *.25, NULL), NULL) "Catcost"FROM promotions;C SELECT promo_id, DECODE(NULLIF(promo_cost, 10000), NULL,promo_cost*.25, 'N/A') "Catcost"FROM promotions;D. SELECTpromo_id,DECODE(promo_cost, >10000, 'High',<10000, 'Low')"Range"FROM promotions;Answer: A, BView the Exhibit and examine the structure of ORDERS and CUSTOMERS tables. There is only one customer with the cust _last_name column having value Roberts. Which INSERT statement should be used to add a row into the ORDERS table for the customer whose CUST LAST NAME is Roberts and CREDIT LIMIT is 600?A. INSERT INTO orders V ALUES (1, '10-mar-2007', 'direct',(SELECT customer_idFROM customersWHERE cust last name= 'Roberts' ANDcredit_limit=600), 1000);B. INSERT INTO orders (order_id,order_date,order_mode,(SELECT customer_idFROM customersWHERE cust last name= 'Roberts' ANDcredit_limit=600),order_total)V ALUES(1, '10-mar-2007', 'direct', &&customer_id, 1000);C. INSERT INTO(SELECT o.order_id, o.order_date,o.order_mode,c.customer_id,o.order_totalFROM orders o, customers cWHERE o.customer_id=c.customer_idAND c.cust_last_name='Roberts' ANDc.credit_limit=600 |V ALUES (1,'10-mar-2007', 'direct',(SELECT customer_idFROM customersWHERE cust_last_name= 'Roberts' AND credit_limit=600 ), 1000);D. INSERT INTO orders (order_id,order_date,order_mode,(SELECT customer_idFROM customersWHERE cust_last_name= 'Roberts' ANDcredit_limit=600),order_total)V ALUES(1,'10-mar-2007', 'direct', &customer_id, 1000);Answer: A4.View the Exhibit and examine the structure of the CUSTOMERS table.Evaluate the following SQL statementSQL> SELECT cust_city, COUNT(cust_last_name)FROM customersWHERE cust_credit_limit > 1000GROUP BY cust_cityHA VING A VG(cust_credit_limit) BETWEEN 5000 AND 6000;Which statement is true regarding the outcome of the above query?A. It executes successfullyB. It returns an error because the BETWEEN operator cannot be used in the HA VING clauseC. It returns an error because WHERE and HA VING clauses cannot be used in the same SELECT statementD. It returns an error because WHERE and HA VING clauses cannot be used to apply conditions on the same columnAnswer: AView the Exhibit and examine the structure of the PROMOTIONS table. Examine the following two SQL statements:Which statement is true regarding the above two SQL statements?A. statement 1 gives an error, statement 2 executes successfullyB. statement 2 gives an error, statement 1 executes successfullyC. statement 1 and statement 2 execute successfully and give the same outputD. statement 1 and statement 2 execute successfully and give a different output Answer: DYou created an ORDERS table with the following description:You inserted some rows in the table. After some time, you want to alter the table by creating the PRIMARY KEY constraint on the ORD_ID column. Which statement is true in this scenario?A. You cannot have two constraints on one columnB. You cannot add a primary key constraint if data exists in the columnC. The primary key constraint can be created only at the time of table creationD. You can add the primary key constraint even if data exists, provided that there are no duplicate valuesAnswer: D7.When does a transaction complete? (Choose all that apply.)A. when a DELETE statement is executedB. when a ROLLBACK command is executedC. when a PL/SQL anonymous block is executedD. when a data definition language (DDL) statement is executedE. when a TRUNCATE statement is executed after the pending transact ionAnswer: B, D, E8.You need to display the first names of all customers from the CUSTOMERS table that contain the character 'e' and have the character 'a' in the second last positionWhich query would give the required output?A. SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, 'e') <>0 ANDSUBSTR(cust_first_name, -2, l) ='a';B. SELECT cust first nameFROM customersWHERE INSTR(cust_first_name, 'e') <>" ANDSUBSTR(cust_first_name, -2, l)='a';C. SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, 'e') IS NOT NULL ANDSUBSTR(cust_first_name, l, -2)='a';D. SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, 'e')<>0 ANDSUBSTR(cust_first_name, LENGTH(cust_first_name),-2)='a';Answer: A9.The ORDERS table belongs to the user OE. OE has granted the SELECT privilege on the ORDERS table to the user HR.Which statement would create a synonym ORD so that HR can execute the following query successfully?SELECT * FROM ord;A. CREATE SYNONYM ord FOR orders; This command is issued by OEB. CREATE PUBLIC SYNONYM ord FOR orders; This command is issued by OEC. CREATE SYNONYM ord FOR oe.orders; This command is issued by the database administratorD. CREATE PUBLIC SYNONYM ord FOR oe.orders; This command is issued by the database administratorAnswer: D10.View the Exhibit and examine the structure of the PRODUCTS tableYou need to generate a report in the following format:Which two queries would give the required output? (Choose two.)A. SELECT prod_name || q"'s category is ' || prod_category CATEGORIESFROM products;B. SELECT prod_name || q'['s]'category is' || prod_category CATEGORIESFROM products;C. SELECT prod_name || q'\'s\' category is '|| prod_category CATEGORIESFROM products;D. SELECT prod_name ||q'<'s>'|| 'category is ' || prod_category CATEGORIESFROM products;Answer: C, D11.Which statement is true regarding the INTERSECT operator?A. It ignores NULL valuesB. Reversing the order of the intersected tables alters the resultC. The names of columns in all SELECT statements must be identicalD. The number of columns and data types must be identical for all SELECT statements in the queryAnswer: D12.Which two statements are true regarding the USING and ON clauses in table joins? (Choose two.)A. Both USING and ON clauses can be used for equijoins and nonequijoinsB. A maximum of one pair of columns can be joined between two tables using the ON clauseC. The ON clause can be used to join tables on columns that have different names but compatible data typesD. The WHERE clause can be used to apply additional conditions in SELECT statements containing the ON or the USING clauseAnswer: C, D13.Examine the structure of the PROGRAMS tableWhich two SQL statements would execute successfully? (Choose two.)A. SELECT NVL(ADD_MONTHS(END_DATE,l),SYSDATE)FROM programs;B. SELECT TO_DATE(NVL(SYSDATE-END_DATE,SYSDATE))FROM programs;C. SELECT NVL(MONTHS_BETWEEN(start_date,end_date),'Ongoing')FROM programs;D. SELECTNVL(TO_CHAR(MONTHS_BETWEEN(start_date,end_date)),'Ongoing') FROM programs;Answer: A, D14.Where can subqueries be used? (Choose all that apply.)A. field names in the SELECT statementB. the FROM clause in the SELECT statementC. the HA VING clause in the SELECT statementD. the GROUP BY clause in the SELECT statementE. the WHERE clause in only the SELECT statementF. the WHERE clause in SELECT as well as all DML statementsAnswer: A, B, C, F15.View the Exhibits and examine the structures of the PRODUCTS, SAL ES, and CUSTOMERS tablesYou need to generate a report that gives details of the customer's last name, name of the product, and the quantity sold for all customers in 'Tokyo'Which two queries give the required result? (Choose two.)A. SELECT c.cust_last_name,p.prod_name,s.quantity_soldFROM sales s JOIN products pUSING(prod_id)JOIN customers cB. SELECT c.cust_last_name, p.prod_name, s.quantity_soldFROM products p JOIN sales s JOIN customers cON(p.prod_id=s.prod_id)ON(s.cust_id=c.cust_id)WHERE c.cust_city='Tokyo';C. SELECT c.cust_last_name, p.prod_name, s.quantity_soldFROM products p JOIN sales sON(p.prod_id=s.prod_id)JOIN customers cON(s.cust_id=c.cust_id)AND c.cust_city='Tokyo';D. SELECT c.cust_id, c.cust_last_name, p.prod_id, p.prod_name, s.quantity_soldFROM products p JOIN sales sUSING(prod_id)JOIN customers cUSING(cust_id)WHERE c.cust_city='Tokyo';Answer: A, C16.View the Exhibit and evaluate the structure and data in the CUST_STATUS table You issue the following SQL statementSQL> SELECT custno, NVL2(NULLIF(amt_spent,credit_limit), 0, 1000) "BONUS"FROM cust_status;Which statement is true regarding the execution of the above query?A. It produces an error because the AMT_SPENT column contains a null valueB. It displays a bonus of 1000forall customers whose AMT_SPENT is less than CREDIT_LIMITC. It displays a bonus of 1000 for all customers whose AMT_SPENT equals CREDIT_LIMIT, or AMT_SPENT is nullD. It produces an error because the TO_NUMBER function must be used to convert the result of the NULLIF function before it can be used by the NVL2 functionAnswer: C17.Examine the structure and data in the PRICE LIST tableYou plan to give a discount of 25% on the product price and need to display the discount amount in the same format as the PROD_PRICE.Which SQL statement would give the required result?A. SELECT TO_CHAR(prod_price* .25, '$99,999.99')FROM PRICE_LIST;B. SELECT TO_CHAR(TO_NUMBER(prod_price)* .25, '$99,999.00')FROM PRICE_LIST;C. SELECT TO_CHAR(TO_NUMBER(prod_price, '$99,999.99')* .25,'$99,999.00')FROM PRICE_LIST;D. SELECT TO_NUMBER(TO_NUMBER(prod_price,$99,999.99')* .25,'$99,999.00')FROM PRICE_LIST;Answer: C18.You need to generate a list of all customer last names with their credit limits from the CUSTOMERS table. Those customers who do not have a credit limit should appear last in the list.Which two queries would achieve the required result? (Choose two.)A. SELECT cust_last_name, cust_credit_limitFROM customersORDER BY cust_credit_limit DESC;B. SELECT cust_last_name, cust_credit_limitFROM customersORDER BY cust_credit_limit;C. SELECT cust_last_name, cust_credit_limitFROM customersORDER BY cust_credit_limit NULLS LAST;D. SELECT cust_last_name, cust_credit_limitFROM customersORDER BY cust_last_name, cust_credit_limit NULLSLAST;Answer: B, C19.Which two statements are true regarding the COUNT function? (Choose two.)A. The COUNT function can be used only for CHAR, V ARCHAR2, and NUMBER data typesB. COUNT (*) returns the number of rows including duplicate rows and rows containing NULL value in any of the columnsC. COUNT (cust_id) returns the number of rows including rows with duplicate customer IDs and NULL value in the CUST_ID columnD. COUNT(DISTINCT inv_amt) returns the number of rows excluding rows containing duplicates and NULL values in the INV_AMT columnE. A SELECT statement using the COUNT function with a DISTINCT keyword cannot have a WHERE clauseAnswer: B, D20.Which two statements are true regarding single row functions? (Choose two.)A. They accept only a single argumentB. They can be nested only to two levelsC. Arguments can only be column values or constantsD. They always return a single result row for every row of a queried tableE. They can return a data type value different from the one that is referencedAnswer: D, E21.View the Exhibit and examine the data in the COSTS table.You need to generate a report that displays the IDs of all products in the COSTS table whose unit price is at least 25% more than the unit cost. The details should be displayed in the descending order of 25% of the unit cost.You issue the following query:SQL>SELECT prod_idFROM costsWHERE unit_price >= unit_cost * 1.25ORDER BY unit_cost * 0.25 DESC;Which statement is true regarding the above query?A. It executes and produces the required resultB. It produces an error because an expression cannot be used in the ORDER By clauseC. It produces an error because the DESC option cannot be used with an expression in the ORDER BY clauseD. It produces an error because the expression in the ORDER BY clause should also be specified in the SELECT clauseAnswer: A22.View the Exhibit and examine the structure of the CUSTOMERS tableWhich statement would display the highest credit limit available in each income level in each city in the CUSTOMERS table?A. SELECT cust_city, cust_income_level, MAX(cust_credit_limit)FROM customersGROUP BY cust_city, cust_income_level, cust_credit_limit;B. SELECT cust_city, cust_income_level, MAX(cust_credit_limit)FROM customersGROUP BY cust_city, cust_income_level;C. SELECT cust_city, cust_income_level, MAX(cust_credit_limit)FROM customersGROUP BY cust_credit_limit, cust_income_level, cust_city;D. SELECT cust_city, cust_income_level, MAX(cust_credit_limit)FROM customersGROUP BY cust_city, cust_income_level, MAX(cust_credit_limit);Answer: B23.Which two statements are true regarding subqueries? (Choose two.)A. A subquery can retrieve zero or more rowsB. Only two subqueries can be placed atone levelC. A subquery can be used only in SQL query statementsD. A subquery can appear on either side of a comparison operatorE. There is no limit on the number of subquery levels in the WHERE clause of a SELECT statementAnswer: A, D24.View the Exhibit and examine the structure of the PROMOTIONS table.Evaluate the following SQL statement:The above query generates an error on execution.Which clause in the above SQL statement causes the error?A. WHEREB. SELECTC. GROUP BYD. ORDER BYAnswer: C25.You need to create a table for a banking application One of the columns in the table has the following requirements:1) You want a column in the table to store the duration of the credit period2) The data in the column should be stored in a format such that it can be easily addedand subtracted with DATE data type without using conversion functions3) The maximum period of the credit provision in the application is 30days4) The interest has to be calculated for the number of days an individual has taken a credit for.Which data type would you use for such a column in the table?A. DATEB. NUMBERC. TIMESTAMPD. INTERV AL DAY TO SECONDE. INTERV AL YEAR TO MONTHAnswer: D26.View the Exhibit to examine the description for the SALES tableWhich views can have all DML operations performed on it? (Choose all that apply.)A. CREATE VIEW v3AS SELECT * FROM SALESWHERE cust_id = 2034WITH CHECK OPTION;B. CREATE VIEW vlAS SELECT * FROM SALESWHERE time_id <= SYSDATE - 2*365WITH CHECK OPTION;C. CREATE VIEW v2AS SELECT prod_id, cust_id, time_id FROM SALESWHERE time id <= SYSDATE -2*365WITH CHECK OPTION;D. CREATE VIEW v4AS SELECT prod_id, cust_id, SUM(quantity_sold) FROM SALESWHERE time id <= SYSDATE -2*365GROUP BY prod_id, cust_idWITH CHECK OPTION;Answer: A, B27.Which is the valid CREA TE TABLE statement?A. CREATE TABLE emp9$# (emp_no NUMBER(4));B. CREATE TABLE 9emp$# (emp_no NUMBER(4));C. CREATE TABLE emp*123 (emp_no NUMBER(4));D. CREATE TABLE emp9$# (emp_no NUMBER (4), date DATE);Answer: A28.View the Exhibit and examine the data in the PRODUCTS table.You need to display product names from the PRODUCTS table that belong to the'Software/Other' category with minimum prices as either $2000 or $4000 and no unit of measure.You issue the following query:SQL>SELECT prod_name, prod_category, prod_min_priceFROM productsWHERE prod_category LIKE '%Other%' AND (prod_min_price = 2000 OR prod_min_price = 4000) AND prod_unit_of_measure <> ";Which statement is true regarding the above query?A. It executes successfully but returns no resultB. It executes successfully and returns the required resultC. It generates an error because the condition specified forPROD_UNIT_OF_MEASURE is not validD. It generates an error because the condition specified for the PROD_CATEGORY Column is not validAnswer: AView the Exhibit to examine the description for the SALES and PRODUCTS tables. You want to create a SALE _PROD view by executing the following SQL statement:Which statement is true regarding the execution of the above statement?A. The view will be created and you can perform DML operations on the viewB. The view will be created but no DML operations will be allowed on the viewC. The view will not be created because the join statements are not allowed for creating a viewD. The view will not be created because the GROUP BY clause is not allowed for creating a viewAnswer: BWhich three statements are true regarding the data types in Oracle Database 10g/11g? (Choose three.)A. Only one LONG column can be used per tableB. A TIMESTAMP data type column stores only time values with fractional secondsC. The BLOB data type column is used to store binary data in an operating system fileD. The minimum column width that can be specified for a V ARCHAR2 data type column is oneE. The value for a CHAR data type column is blank-padded to the maximum defined column widthAnswer: A, D, E31.View the Exhibit and examine the structure of the PRODUCTS table.Evaluate the following query:What would be the outcome of executing the above SQL statement?A. It produces an errorB. It shows the names of all products in the tableC. It shows the names of products whose list price is the second highest in the tableD. It shows the names of all products whose list price is less than the maximum list priceAnswer: CView the Exhibit and examine the structure of ORD and ORD_ITEMS tables.The ORD_NO column is PRIMARY KEY in the ORD table and the ORD NO and ITEM_NO columns are composite PRIMARY KEY in the ORD ITEMS table. Which two CREATE INDEX statements are valid? (Choose two.)A. CREATE INDEX ord_idxlON ord(ord_no);B. CREATE INDEX ord_idx2ON ord_items(ord_no);C. CREATE INDEX ord_idx3ON ord_items(item_no);D. CREATE INDEX ord_idx4ON ord,ord_items(ord_no, ord_date,qty);Answer: B, C33.Which statement is true regarding subqueries?A. The LIKE operator cannot be used with single-row subqueriesB. The NOT IN operator is equivalent to IS NULL with single-row subqueriesC. =ANY and =ALL operators have the same functionality in multiple-row subqueriesD. The NOT operator can be used with IN, ANY, and ALL operators in multiple-row subqueriesAnswer: DYou are currently located in Singapore and have connected to a remote database in Chicago.You issue the following command:SQL> SELECT ROUND (SYSDATE-promo_begin_date, 0)FROM promotionsWHERE (SYSDATE-promo_begin_date)/365 >2;PROMOTIONS is the public synonym for the public database link for the PROMOTIONS table.What is the outcome?A. an error because the ROUND function specified is invalidB. an error because the WHERE condition specified is invalidC. number of days since the promo started based on the current Chicago date and timeD. number of days since the promo started based on the current Singapore date and timeAnswer: C35.View the Exhibit and examine the structure of the PRODUCTS table.Using the PRODUCTS table, you issue the following query to generate the names, current list price, and discounted list price for all those products whose list price falls below $10 after a discount of 25% is applied on it.SQL>SELECT prod_name, prod_list_price,prod_list_price-(prod_list_price*.25) "DISCOUNTED_PRICE"FROM productsWHERE discounted_price < 10;The query generates an error.What is the reason for the error?A. The parenthesis should be added to enclose the entire expressionB. The double quotation marks should be removed from the column aliasC. The column alias should be replaced with the expression in the WH ERE clauseD. The column alias should be put in uppercase and enclosed within double quotation marks in the WHERE clauseAnswer: C36.View the Exhibit for the structure of the STUDENT and FACULTY tables.You need to display the faculty name followed by the number of students handled by the faculty at the base location.Examine the following two SQL statements:Which statement is true regarding the outcome?A. Only statement 1 executes successfully and gives the required resultB. Only statement 2 executes successfully and gives the required resultC. Both statements 1 and 2 execute successfully and give different resultsD. Both statements 1 and 2 execute successfully and give the same required resultAnswer: DView the Exhibit and examine the structure of CUSTOMERS and SALES tables. Evaluate the following SQL statement:Which statement is true regarding the execution of the above UPD ATE statement? A. It would not execute because two tables cannot be used in a single UPDATE statementB. It would not execute because the SELECT statement cannot be used in place of the table nameC. It would execute and restrict modifications to only the columns specified in the SELECT statementD. It would not execute because a subquery cannot be used in the WHERE clause of an UPDATE statementAnswer: CEvaluate the following query:SQL> SELECT TRUNC(ROUND(156.00,-1),-1)FROM DUAL;What would be the outcome?A. 16B. 100C. 160D. 200E. 150Answer: C39.View the Exhibits and examine the structures of the PROMOTIONS and SALES tables.Evaluate the following SQL statement:Which statement is true regarding the output of the above query?A. It gives the details of promos for which there have been salesB. It gives the details of promos for which there have been no salesC. It gives details of all promos irrespective of whether they have resulted in a sale or notD. It gives details of product 105 that have been sold irrespective of whether they had a promo or notAnswer: C40.View the Exhibit and examine the description of SALES and PROMOTIONS tables You want to delete rows from the SALES table, where the PROMO_NAME column in the PROMOTIONS table has either blowout sale or everyday low price as values Which DELETE statements are valid? (Choose all that apply.)A. DELETEFROM salesWHERE promo_id = (SELECT promo_idFROM promotionsWHERE promo_name = 'blowout sale' ) AND promo_id = (SELECT promo_idFROM promotionsWHERE promo_name = 'everyday low price' );B. DELETEFROM salesWHERE promo_id = (SELECT promo_idFROM promotionsWHERE promo_name = 'blowout sale' ) OR promo_id = (SELECT promo_idFROM promotionsWHERE promo_name = 'everyday low price' );C. DELETEFROM salesWHERE promo_id IN (SELECT promo_idFROM promotionsWHERE promo_name = 'blowout sale' )OR promo_name = 'everyday low price' );D. DELETEFROM salesWHERE promo_id IN (SELECT promo_idFROM promotionsWHERE promo_name IN ('blowout sale', 'everyday lowprice' ));Answer: B, C, D41.View the Exhibit and examine the structure of the PROMOTIONS table.You have to generate a report that displays the promo name and start date for all promos that started after the last promo in the' INTERNET' category.Which query would give you the required output?A. SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date > ALL (SELECT MAX(promo_begin_date)FROM promotions )ANDpromo_category = 'INTERNET');B. SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date IN (SELECT promo_begin_date)FROM promotionsWHERE promo_category = 'INTERNET');C. SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date > ALL (SELECT promo_begin_dateFROM promotionsWHERE promo_category = 'INTERNET');D. SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date > ANY (SELECT promo_begin_dateFROM promotionsWHERE promo_category = 'INTERNET'); Answer: C42.View the Exhibit and examine the data in the CUSTOMERS tableEvaluate the following query:SQL> SELECT cust_name AS "NAME", cust_credit_limit/2 AS MIDPOINT,MIDPOINT+100AS "MAX LOWER LIMIT" FROM customers;The above query produces an error on executionWhat is the reason for the error?A. An alias cannot be used in an expressionB. The alias NAME should not be enclosed within double quotation marksC. The MIDPOINT+100 expression gives an error because CUST_CREDIT _LIMIT contains NULL valuesD. The alias MIDPOINT should be enclosed within double quotation marks for the CUST_CREDIT_LIMIT/2 expressionAnswer: A43.View the Exhibit and examine the structure of the ORD tableEvaluate the following SQL statements that are executed in a user session in the specified order.What would be the outcome of the above statements?A. All the statements would execute successfully and the ORD_NO column would contain the value 2 for the CUST ID 101B. The CREATE SEQUENCE command would not execute because the minimum value and maximum value for the sequence have not been specifiedC. The CREATE SEQUENCE command would not execute because the starting value of the sequence and the increment value have not been specifiedD. All the statements would execute successfully and the ORD_NO column would have the value 20 for the CUST_ID 101 because the default CACHE value is 20Answer: A44.Using the CUSTOMERS table, you need to generate a report that shows 50% of each credit amount in each income level. The report should NOT show any repeated credit amounts in each income level.Which query would give the required result?A. SELECT cust_income_level, DISTINCT cust_credit_limit * 0.50AS "50% Credit Limit"FROM customers;B. SELECT DISTINCT cust_income_level, DISTINCT cust_credit_limit * 0.50AS "50% Credit Limit"FROM customers;C. SELECT DISTINCT cust_income level || ' ' || cust_credit_limit * 0.50AS "50% Credit Limit"FROM customers;D. S ELECT cust income level || ' ' || cust_credit_limit * 0.50 AS "50% Credit Limit"ROM customers;Answer: C45.View the Exhibit and examine the structure of the CUSTOMERS tableEvaluate the query statement:What would be the outcome of the above statement?A. It executes successfullyB. It produces an error because the condition on CUST_LAST NAME is invalidC. It executes successfully only if the CUST_CREDIT_LIMIT column does not contain any null valuesD. It produces an error because the AND operator cannot be used to combine multiple BETWEEN clausesAnswer: A。
Oracle考试报名 OCA OCP OCM
OCP认证考试流程在竞争日益激烈的IT界，越来越多的学生意识到就业压力的巨大，因此，“OCP认证”作为数据库认证的权威认证受到了广大师生的重视与青睐。
OAI成员学校的学生可享受 Oracle 认证考试 40% 的折扣优惠，同时参加OCP认证的流程也有所不同，请OAI成员特别关注。
下文以Oracle9i DBA为例，详细介绍OAI成员学校学生获得OCP证书的流程及注意事项。
一、登陆Oracle University官方网站，查看通过OCP 认证必须通过的考试。
具体步骤如下：1、登陆官网，点击屏幕左下方的“认证”，图示如下：2、进入“认证”页面后点击屏幕中部“数据库”链接，图示如下：3、进入“数据库”页面后，选择你想获得的认证，此处以获得Oracle 9i DBA OCP为例。
图示如下：4、在Oracle9i Database Administrator Certified Professional页面查看要参加的考试及考试号。
注意：参加OCP认证必须已经获得OCA认证，因此，若你还未通过OCA认证，则必须参加获得OCA认证所要求的考试。
在Oracle9i DBA的OCP认证过程中，必须通过的考试为图中圈出的所有考试。
图示如下：二、确定对应考试科目及考试号后，进行Oracle考试网上报名。
下文以” Oracle9i Database: Performance Tuning”（IZ0-033）考试为例，具体报名步骤如下(以下步骤为一门考试的报名步骤，若有多门考试，需重复以下步骤)：1、访问官方网站：/Index.asp，选择考试所在”Country”，这里选”China”，”State”可以不选择，图示如下：2、选择Oracle考试，图示如下，阅读考试协议，点击Next。
3、选择考试科目，图示如下4、选择考点，找到自己需要考试的考点。
预订考试时，需用prometric帐户登录。
如果从未参加过Oracle考试，在第一次注册报名的时候，系统会自动分配一个考试号，例如，SR255xxxx。
【CUUG内部资料】OCP最新考试题库-1Z0-062(5)
AUTHID CURRENT_USER AS BEGIN
OCP最新考试题库 1015267481 INSERT INTO departments VALUES (v_deptno, v_dname, v_mgr, v_loc);
END; 2、 调用该存储过程： CALL create_dept(44, 'Information Technology', 200, 1700);
】 E. Execute FLASHBACK TABLE OCA.EXAM_RESULTS TO BEFORE DROP RENAME TO 料 EXAM_RESULTS; connected as the OCP user. 部资 Correct Answer: C 【CUUG内 -1Z0-062 Section: (none)
能够执行这个存储过程，所以如果是 definer's rights 的话，执行就会出错，因为该模式下只 能对 HR.DEPARTMENTS 的表进行操作；可能的情况下是这些用户自己有跟 HR 用户相同的
OCP最新考试题库 1015267481 表名字也叫 DEPARTMENTS，所以当执行这个存储过程的时候，修改的是自己的表，可以实
Explanation（验证过） （解析：因为是删除了用户，而且数据库版本是 11g，没有表恢复功能，但是从恢复工作量
oracle ocp 中文题库
oracle ocp 中文题库摘要：1.Oracle OCP 中文题库概述2.Oracle OCP 083 题库的内容3.如何获取Oracle OCP 题库4.Oracle OCP 的重要性正文：一、Oracle OCP 中文题库概述Oracle OCP（Oracle Certified Professional）中文题库是一个包含了大量Oracle 数据库相关问题的题库，旨在帮助准备参加Oracle OCP 认证考试的人更好地理解和掌握Oracle 数据库的知识和技能。
该题库涵盖了Oracle 数据库的各个方面，如数据库管理、SQL 语句、备份与恢复、性能优化等。
二、Oracle OCP 083 题库的内容Oracle OCP 083 题库是Oracle OCP 中文题库中的一个部分，主要涉及Oracle 数据库19c 及更高版本的相关知识。
题库中包含了关于闪回日志（Flashback Logs）在Oracle 数据库中的使用、回收空间（Reclaiming Space）等方面的问题。
每天提供十道题目，可以帮助考生逐步掌握Oracle 数据库的知识点。
三、如何获取Oracle OCP 题库考生可以通过私聊的方式获取Oracle OCP 083 题库的全部题目。
同时，在准备考试的过程中，建议考生参加Oracle 官方提供的培训课程，结合实际操作来更好地理解和应用所学知识。
四、Oracle OCP 的重要性Oracle OCP 认证是Oracle 公司对其数据库专业人才的一种官方认证，拥有该认证证书的人员可以证明自己具备较高的Oracle 数据库管理与优化能力。
对于数据库管理员和开发人员来说，获得Oracle OCP 认证可以提高个人职业素质，增加就业机会，提升薪资水平。
ocp证书的类型 -回复
ocp证书的类型-回复OCP证书的类型导语：随着信息技术的迅猛发展，越来越多的人开始关注和追求计算机领域的证书。
OCP（Oracle Certified Professional）证书作为计算机行业中备受瞩目的证书之一，涵盖了多个不同的子领域。
在本篇文章中，我们将一步一步地回答以下问题，包括OCP证书的类型，各个类型的特点以及如何获得这些证书。
第一部分：OCP证书的概述OCP证书是由Oracle公司授权的一系列认证证书，旨在证明持有人在Oracle相关领域具备一定的技术能力和知识。
这些证书被全球范围内的企业和组织广泛认可，持有人可以通过这些证书证明自己具备在Oracle技术方面的专业能力，提高就业竞争力和职业发展机会。
第二部分：OCP证书的类型OCP证书类型众多，下面我们将介绍其中几个常见的类型：1. OCP数据库管理员证书（OCP DBA）：这是最受欢迎的OCP证书之一，主要关注数据库管理方面的技能和知识。
持有这个证书的人可以展示他们在Oracle数据库管理方面的专业能力，包括安装、配置、备份与恢复、性能调优等方面。
2. OCP开发者证书（OCP Developer）：这个证书主要关注Oracle开发方面的技能和知识，包括Java开发、PL/SQL编程等。
持有这个证书的人可以证明他们在Oracle开发方面有着深入的理解和技术能力。
3. OCP业务智能证书（OCP Business Intelligence）：这个证书主要关注Oracle业务智能方面的技能和知识，包括数据分析、报告和仪表板设计等。
持有这个证书的人可以证明他们在利用Oracle工具和技术进行商业智能分析方面具备专业能力。
4. OCP Java软件工程师证书（OCP Java SE）：这个证书主要关注Java 软件开发方面的技能和知识。
持有这个证书的人可以证明他们在Java语言和相关技术方面具备专业能力。
5. OCP企业架构师证书（OCP Enterprise Architect）：这个证书主要关注企业级系统架构设计和实施方面的技能和知识。
oracle function 语法
Oracle Function 语法Oracle Function 是一种可重用的程序单元，它接受输入参数并返回一个值。
Function 可以在 SQL 查询中使用，也可以在 PL/SQL 块中调用。
本文将详细介绍Oracle Function 的语法，包括创建和使用 Function 的步骤，以及一些常用的用法和示例。
创建 Function在 Oracle 中，创建 Function 需要使用CREATE FUNCTION语句。
下面是创建Function 的基本语法：CREATE [OR REPLACE] FUNCTION function_name[ (parameter_name [IN | OUT | IN OUT] data_type [, ...]) ]RETURN return_data_type[DETERMINISTIC]{IS | AS}[PRAGMA AUTONOMOUS_TRANSACTION;]BEGIN-- Function 的逻辑代码RETURN return_value;END;•CREATE FUNCTION：创建 Function 的关键字。
•OR REPLACE：可选项，表示如果 Function 已经存在，则替换它。
•function_name：Function 的名称，遵循标识符的命名规则。
•parameter_name：可选项，Function 的输入参数名称。
•IN | OUT | IN OUT：可选项，指定参数的传递方式。
默认为IN，表示参数是输入参数。
•data_type：参数的数据类型。
•RETURN return_data_type：指定 Function 的返回值数据类型。
•DETERMINISTIC：可选项，表示 Function 的结果只依赖于输入参数，没有其他影响。
•IS | AS：Function 的逻辑代码开始的关键字。
•PRAGMA AUTONOMOUS_TRANSACTION;：可选项，表示 Function 可以在独立的事务中执行。
ocp证书的类型 -回复
ocp证书的类型-回复OCP证书的类型OCP（Oracle Certified Professional）证书是由Oracle公司提供的IT 技术认证。
它分为多个不同的类型，每种类型都代表着不同的专业领域和技能要求。
在本文中，我们将一步一步地回答有关OCP证书类型的问题，以便更好地了解其内容。
第一步：初识OCP证书OCP证书是Oracle公司为IT专业人士提供的一种技术认证，旨在验证他们在Oracle相关领域的专业知识和技能。
这些认证证明了考试者在特定领域的专业能力，并为他们在求职市场上提供了竞争优势。
第二步：OCP证书的主要类型OCP证书分为多个主要类型，包括数据库管理员（DBA）、Java开发人员和应用开发人员等。
下面我们将详细介绍每种类型。
1. OCP数据库管理员（DBA）证书：这是最常见和受欢迎的OCP证书类型。
它涵盖了Oracle数据库的管理和维护等方面的知识和技能，适用于那些希望在数据库管理和维护领域工作的专业人士。
2. OCP Java开发人员证书：这是Oracle针对Java开发人员提供的证书类型。
它涵盖了Java编程语言和相关技术的知识和技能，在Java开发领域有很高的认可度。
3. OCP应用开发人员证书：这种类型的证书适用于那些希望证明自己在应用开发领域的专业能力的人士。
它包括了Oracle数据库和相关工具的应用开发知识和技能。
第三步：OCP证书的级别OCP证书分为不同的级别，每个级别对应着不同的技能水平和专业能力。
以下是一些常见的OCP证书级别：1. OCA（Oracle Certified Associate）：这是OCP证书体系中的最低级别。
它是向那些刚开始学习Oracle技术的人士提供的，并验证了他们对特定领域的基本知识和技能。
2. OCP（Oracle Certified Professional）：这是OCP证书体系中的中级别证书。
它要求考生通过更高级别的考试来验证他们在特定领域的专业能力。
oracle ocp 报考流程
oracle ocp 报考流程下载温馨提示:该文档是我店铺精心编制而成，希望大家下载以后，能够帮助大家解决实际的问题。
文档下载后可定制随意修改，请根据实际需要进行相应的调整和使用，谢谢!并且，本店铺为大家提供各种各样类型的实用资料，如教育随笔、日记赏析、句子摘抄、古诗大全、经典美文、话题作文、工作总结、词语解析、文案摘录、其他资料等等，如想了解不同资料格式和写法，敬请关注!Download tips: This document is carefully compiled by theeditor.I hope that after you download them,they can help yousolve practical problems. The document can be customized andmodified after downloading,please adjust and use it according toactual needs, thank you!In addition, our shop provides you with various types ofpractical materials,such as educational essays, diaryappreciation,sentence excerpts,ancient poems,classic articles,topic composition,work summary,word parsing,copy excerpts,other materials and so on,want to know different data formats andwriting methods,please pay attention!Oracle OCP 认证考试报考全面指南Oracle Certified Professional (OCP) 是Oracle公司提供的高级数据库管理员认证，对于希望提升自己在Oracle数据库管理领域专业技能的专业人士来说，是一项重要的资格认证。
OCP认证完全的指南
OCP认证完全指南1.1 概述OCP 是Oracle 公司推出的一项认证计划，全称为：Oracle Certified Professional。
由于Oracle 数据库在全球高端数据库领域占有的份额以与Oracle 数据库本身的复杂性致使这份认证的含金量颇高。
但是需要先纠正一个概念，很多认证机构将OCP 吹嘘成高薪的充分条件，似乎只要通过了OCP 就会拿到一份有优厚待遇的工作，其实并不是这样，特别是在目前中国IT 行业的大背景下，很少国企业会雇佣专门的数据库管理员来管理数据库，往往需要同时兼作数据库设计、数据库管理和数据库开发，而考取多门的OCP 认证在精力和财力上对于个人来说都是难以接受的，所以有些人在取得了一门OCP 认证之后反而会有一种失落感，好像是付出了努力，却并不能立刻感觉到在自己的职业生涯里有明显的变化。
端正思想吧，因为考取OCP 并不是一项艰难的任务，所以也不要把自己的全部理想押在OCP 上面。
虽然OCP 不是万金油，但是如果对于数据库相关的工作感兴趣，那么考取OCP 仍然是当今数一数二的选择。
理由在于，参加OCP 考试会强迫自己系统而深入地学习Oracle，而同时得到的OCP 证书以与Oracle 的知识又在任何时候都会是自己能力的一个砝码。
1.2 OCP 考试分类（Certification Path）下面列出OCP 认证的详细分类，对于类似于Oracle8 OCP 这样已经取消的OCP 认证，本文不再进行介绍。
（1）Database Administrator。
Oracle8i Certified Professional，下文简称为8iOCP。
Oracle9i Upgrade from Oracle8i OCP，下文简称为8ito9i。
Oracle9i Certified Associate，下文简称为9iOCA。
Oracle9i Certified Professional，下文简称为9iOCP。
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Two Types of SQL Functions
Functions
Single-row functions
Multiple-row functions
3-4
Copyright © Oracle Corporation, 2001. All rights reserved.
Single-Row Functions
•
•
Oracle database stores dates in an internal numeric format: century, year, month, day, hours, minutes, seconds. The default date display format is DD-MON-RR.
• • •
Describe various types of functions available in SQL Use character, number, and date functions in SELECT statements Describe the use of conversion functions
3-9
Copyright © Oracle Corporation, 2001. All rights reserved.
Using Case Manipulation Functions
Display the employee number, name, and department number for employee Higgins:
Character-manipulation functions
CONCAT SUBSTR LENGTH INSTR LPAD | RPAD TRIM REPLACE
3-7
Copyright © Oracle Corporation, 2001. All rights reserved.
Case Manipulation Functions
3-19
Copyright © Oracle Corporation, 2001. All rights reserved.
Using Arithmetic Operators with Dates
SELECT last_name, (SYSDATE-hire_date)/7 AS WEEKS FROM employees WHERE department_id = 90;
3-10
Copyright © Oracle Corporation, 2001. All rights reserved.
Character-Manipulation Functions
These functions manipulate character strings: Function CONCAT('Hello', 'World') SUBSTR('HelloWorld',1,5) LENGTH('HelloWorld') INSTR('HelloWorld', 'W') LPAD(salary,10,'*') RPAD(salary, 10, '*') TRIM('H' FROM 'HelloWorld') Result HelloWorld Hello 10 6 *****24000 24000***** elloWorld
TRUNC: Truncates value to specified decimal MOD: Returns remainder of division
MOD(1600, 300)
3-13
Copyright © Oracle Corporation, 2001. All rights reserved.
SELECT last_name, salary, MOD(salary, 5000) FROM employees WHERE job_id = 'SA_REP';
3-16
Copyright © Oracle Corporation, 2001. All rights reserved.
Working with Dates
1
2
3
3-15
Copyright © Oracle Corporation, 2001. All rights reserved.
Using the MOD Function
Calculate the remainder of a salary after it is divided by 5000 for all employees whose job title is sales representative.
expression
function_name [(arg1, arg2,...)]
3-5
Copyright © Oracle Corporation, 2001. All rights reserved.
Single-Row Functions
Character
General Single-row functions
–
–
Allows you to store 21st century dates in the 20th century by specifying only the last two digits of the year. Allows you to store 20th century dates in the 21st century in the same way.
SELECT employee_id, last_name, department_id FROM employees WHERE last_name = 'higgins'; no rows selected SELECT employee_id, last_name, department_id FROM employees WHERE LOWER(last_name) = 'higgins';
These functions convert case for character strings. Function Result
LOWER('SQL Course')
UPPER('SQL Course')
sql course
SQL COURSE
INITCAP('SQL Course') Sql Course
SELECT last_name, hire_date FROM employees WHERE last_name like 'G%';
3-17
Copyright © Oracle Corporation, 2001. All rights reserved.
Working with Dates
2
3
1
2
3
3-12
Copyright © Oracle Corporation, 2001. All rights reserved.
Number Functions
•
• •
ROUND: Rounds value to specified decimal
ROUND(45.926, 2) TRUNC(45.926, 2) 45.93 45.92 100
Single row functions:
• Manipulate data items • Accept arguments and return one value • Act on each row returned • Return one result per row • May modify the data type • Can be nested • Accept arguments which can be a column or an
3-20
Copyright © Oracle Corporation, 2001. All rights reserved.
Date Functions
Function MONTHS_BETWEEN Description Number of months between two dates
ADD_MONTHS
NEXT_DAY LAST_DAY ROUND TRUNC
Add calendar months to date
Next day of the date specified Last day of the month Round date Truncate date
3-21
Copyright © Oracle Corporation, 2001. All rights reserved.
3-2
Copyright © Oracle Corporation, 2001. All rights reserved.
SQL Functions
Input
arg 1 arg 2
Function
Output
Function performs action
Result value
arg n
3-3
Copyright © Oracle Corporation, 2001. All rights reserved.
Single-Row Functions
Copyright © Oracle Corporation, 2001. All rights reserved.
Objectives
After completing this lesson, you should be able to do the following: