oracle比较decode(case when) 、4种去重---和去重中的统计函数

oracle比较decode/case when 、4种去重和去重中的统计函数一：decode 和case whendecode (expression, search_1, result_1, search_2, result_2, ...., search_n, result_n, default) decode函数比较表达式和搜索字，如果匹配，返回结果；如果不匹配，返回default值；如果未定义default值，则返回空值。

select username,decode(lock_date,null,'unlocked','locked') status from t;----------------如果lock_date是null就返回unlocked 如果不是null就返回lockedselect username,decode(lock_date,null,'unlocked') status from t;----------------如果lock_date是null就返回unlocked 否则是空(因为没有定义)例如有个学生表...... 行转列-------create table score2(name varchar2(10),Language number(3),Math number(3),English number(3));insert into score2 values('Zhang',80,67,95);insert into score2 values('Li',79,84,62);insert into score2(name,Language) values('Chen',88);commit;Select * from score2; 显示成报表的格式.......select name,sum(decode(subject,'Language', grade,0)) "Language",sum(decode(subject,'Math', grade,0)) "Math",sum(decode(subject,'English', grade,0)) "English"from scoregroup by name;NAME Language Math English-------------------- ---------- ---------- ----------Zhang 80 92 76Wang 73 0 0Li 81 95 50第二列，如果subject='Language',那么就显示成绩,否则显示为0第三列，如果subject='Math',那么就显示成绩,否则显示为0第四列，如果subject='English',那么就显示成绩，否则显示为0Case whenCase [selector] ---selector可以不设置When id=1 THEN ‘id是1’;ELSE ‘没有id值’;END CASE; ---->如果ID=1 就显示id是1 否则显示没有id值~~~~~上面用case when显示.................select ,sum( casewhen t.subject = 'Language' then t.gradeelse 0end ) dd,sum( casewhen t.subject = 'Math' then t.gradeelse 0end ) Math,sum( casewhen t.subject = 'English' then t.gradeelse 0end ) Englishfrom score t group by ;二：去重的4中方法create table test (c1 int ,c2 varchar2(10));insert into test values (1,'Smith');insert into test values (1,'Smith');insert into test values (2,'John');insert into test values(1,'Jack');insert into test values (2,'Jack');insert into test values (2,'John');insert into test values (2,'John');commit;一种方法：distinct 把之前的表去重显示并创建,drop table old_table;create table tmp_test as select distinct * from test1; ---创建临时表drop table test1;alter table tmp_test rename to test1;第二种rowiddelete from testwhere rowid <> ( select min(rowid)from test bwhere b.c1 = test.c1and b.c2 = test.c2 )第三种方法:分组，rowiddelete from test t where t.rowid not in (select min(rowid) from test group by c1,c2 ); commit;Rowid为伪列是物理地址OOOOOO: 数据库对象号FFF: 表空间相关的数据文件号(相对文件号)BBBBBB: 数据块号RRR: 在块中的行号第四种方法，分析函数dense_rank()drop table test;create table test (c1 int ,c2 varchar2(10));insert into test values (1,'Smith');insert into test values (1,'Smith');insert into test values (2,'John');insert into test values(1,'Jack');insert into test values (2,'Jack');insert into test values (2,'John');insert into test values (2,'John');commit;select c1,c2,rowid rd,row_number() over(partition by c1,c2 order by c1) rn from test;不重复的只有1重复的就会出现2，3，4第一次出现，不重复的时候，rn为1相同的记录，重复出现，第二次，就记录为2第三次，3怎么找不出重复的行select b.c1,b.c2 from宣城在线:/(select c1,c2,rowid rd,row_number() over(partition by c1,c2 order by c1) rn from test) bwhere b.rn = 1;不等于号用了不走索引效率很低分页row_number要求emp表中的5-10的记录？select * from(select a.*,row_number() over(order by empno desc) rk from emp a) whererk<=10 and rk>=5;select rn,empno,ename from (select rownum rn,empno,ename from emp) where rn>=5 and rn<=10;select *from (select deptno,ename,sal,dense_rank() over(partition by deptno order by sal desc) rkfrom emp)where rk<=3; --->发现部门为20的scott 和ford都是3k，所以并列第一有2.3名select *from (select deptno,ename,sal,rank() over(partition by deptno order by sal desc) rk from emp )where rk<=3; --->发现部门为20的scott 和ford都是3k，所以并列第一没2有3名Partition by == group by分析函数里用的是partition by普通函数用的是group byselect *from(select deptno,ename,sal,row_number() over(partition by deptno order by sal desc) rnfrom emp)where rn<=3; --->发现部门为20的scott 和ford按照默认显示出现1和2 名次那么看下这些区别：select *from(select deptno,ename,sal,row_number() over(partition by deptno order by sal desc) rn, rank() over(partition by deptno order by sal desc) rk,dense_rank() over(partition by deptno order by sal desc) drk from emp) where drk<=3;/blog/1546668Oracle统计函数。