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ORACLE分组查询和统计等

时间:2018-05-24 12:44:13      阅读:203      评论:0      收藏:0      [点我收藏+]

标签:区别   eve   12月   where   条件   总数   问题   star   查询   

select flow_id,rw from (select t.flow_id ,rownum as rw from apex_030200.wwv_flow_list_templates t)  where rw >= 5


1.rownum只能用<如果使用>加别名

2.子查询引用只能在查询出的结果中引用,比如子查询没有查出flow_id,外层不能用,另外外层不能引用内层的t

3.薪水前三名,内层查出薪水 order desc的虚表外层使用rownum<3

4.merge可以实现存在数据就update不存在你就insert

merge into product a

using (select  1717 product_id, ‘002‘ req_no from dual b)

on (a.product_id = b.product.id and a.req_no = b.req.no)

when matched then

update set  product name = ‘‘.....................

when not matched then

insert () values ()

5.start with connect by 可以查询出一组树的数据,注意最后connect by的条件(父节点=子节点 向上查询 反之向下查询)

可以order排序,可以加入两棵树(or),也可以加入where条件

select * from emp

where......

start with empnc = 7369 or empnc = 7204(注意不能用and )

connect by prior mgr = empno

order by ...

6 份额(查询某数据占有总数据的百分比)

select t.empno,t.ename,t.sal,
100*round(sal/sum(sal) over(),5)
 from emp t

7 连续求和(同名分为同组 累加)

select t.empno,t.ename,t.sal,
sum(sal) over(order by sal)
 from emp t

8.带条件的连续求和(分部门连续求和)

select t.empno,t.ename,t.sal,t.deptno,
sum(sal) over(partition by t.deptno order by sal)
 from emp t

9.分部门总和(取出orderby)

select t.empno,t.ename,t.sal,t.deptno,
sum(sal) over(partition by t.deptno)
 from emp t

10工资的分组查询份额(总数的百分比)带上部门分组

select t.empno,t.ename,t.sal,t.deptno,
100*round(sal/sum(sal) over (partition by t.deptno),4)
 from emp t

注意这里查询的是“”分组“,因此这里查询的是变成一组为一个100%,查询的是一个部门中员工在本部分所占用的薪水比例

11分组查询出单一条件并分级(查询某一个部门的薪水的级别)注意rank()和row_number()的区别 rank是跳跃性并列(1.1.3.3.5) row_number(1.2.3.4.5)

select t.*,ROW_NUMBER() over(partition by t.deptno order by t.sal desc) rank from emp t

12“总”。。。。。。这个字眼一般使用group by(区分于over(partition by order by))

按部门分组查询部门的总薪水

select sum(t.sal),t.deptno from emp t group by t.deptno

13 总的基础上再次分组 group by + rollup

select sum(t.sal),t.deptno from emp t group by rollup (t.deptno)汇总后将总和进行求和

select sum(t.sal),t.job,t.deptno from emp t group by rollup (t.deptno,t.job)注意多个rollup其实只有第一个参数有效

14cube 连接在order by后面(代替ROLLUP) rollup升级版 全部分组

15grouping实现不用java代码就可以对oracle 查询出的null字段进行赋值 0 本身结果 1合计结果

select sum(t.sal),t.deptno,
(case
       when((grouping(t.job)=1 and grouping(t.deptno)=0)) then ‘部门小计‘
       when((grouping(t.job)=1 and grouping(t.deptno)=1)) then ‘部门总计‘
else t.job end) as
job from emp t group by rollup (t.deptno,t.job)

16分组后的字段累加(比如按照员工名称,根据月份分组,实现自1月份到12月份工资累加,即二月份是 1月 + 2月 。。)
select t.empno, t.ename, t.sal, sum(sal) over (partition by t.ename order by t.sal desc) from emp t

17分组最高值 最低值 平均值 使用 max() over(partition by order by)代替sum() 还可以用min() avg()

18select * from v$transaction 查看事务

19多层分组函数和子查询之间的冲突问题

 

select   a.CREATOR, a.count_Sum ,b.full_name,b.dept_code,b.area_code from (    
select temp.c CREATOR,count(temp.c) as count_Sum from (  
select   
         t.ca,  
         c.lv,instr(t.ca, ‘,‘, 1, c.lv) + 1,  
         substr(t.ca,  
                instr(t.ca, ‘,‘, 1, c.lv) + 1,  
                instr(t.ca, ‘,‘, 1, c.lv + 1) -  
                (instr(t.ca, ‘,‘, 1, c.lv) + 1)) AS c  
    from (select   
                 ‘,‘ || checker || ‘,‘ AS ca,  
                 checker,  
                 LENGTH(checker),  
                 length(checker || ‘,‘),  
                 REPLACE(checker, ‘,‘),  
                 length(REPLACE(checker, ‘,‘)),   
                 nvl(length(REPLACE(checker, ‘,‘)), 0),   
                 length(checker || ‘,‘) - nvl(length(REPLACE(checker, ‘,‘)), 0) AS cnt   
            FROM wm_time_info a where a.check_result !=1) t,  
            (select LEVEL lv from dual CONNECT BY LEVEL <= 100) c  
   where c.lv <= t.cnt  ) temp group by temp.c) a ,base_user b where a.CREATOR = b.user_code  

  外层查询和内层分组冲突

--select   a.CREATOR, a.count_Sum ,b.full_name,b.dept_code,b.area_code from (    
select o.CREATOR,count(o.CREATOR) as count_Sum,FULLNAME,DEPTCODE,AREACODE from (  
select temp.c CREATOR,b.full_name FULLNAME,b.dept_code DEPTCODE,b.area_code AREACODE,work_date from  
(  
select   
         work_date,  
         t.ca,  
         c.lv,  
         instr(t.ca, ‘,‘, 1, c.lv) + 1,  
         substr(t.ca,  
                instr(t.ca, ‘,‘, 1, c.lv) + 1,  
                instr(t.ca, ‘,‘, 1, c.lv + 1) -  
                (instr(t.ca, ‘,‘, 1, c.lv) + 1)) AS c  
    from (  
    ---  
    select   
                 work_date,  
                 ‘,‘ || checker || ‘,‘ AS ca,  
                 checker,  
                 LENGTH(checker),  
                 length(checker || ‘,‘),  
                 REPLACE(checker, ‘,‘),  
                 length(REPLACE(checker, ‘,‘)),   
                 nvl(length(REPLACE(checker, ‘,‘)), 0),   
                 length(checker || ‘,‘) - nvl(length(REPLACE(checker, ‘,‘)), 0) AS cnt   
            FROM wm_time_info a where a.check_result !=1  
     ---         
            ) t,  
            (select LEVEL lv from dual CONNECT BY LEVEL <= 100) c   
   where c.lv <= t.cnt   
   ) temp ,base_user b where temp.c = b.user_code ) o  
   --where work_date >=‘2016-01-10‘   
   group by o.CREATOR,FULLNAME,DEPTCODE,AREACODE  
   --) a ,base_user b where a.CREATOR = b.user_code  

  20 注意 本条select中的分组和子查询都不可以作为函数的参数传入

ORACLE分组查询和统计等

标签:区别   eve   12月   where   条件   总数   问题   star   查询   

原文地址:https://www.cnblogs.com/guohu/p/9082239.html

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