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1:统计某商店的营业额。 date sale 1 20 2 15 3 14 4 18 5 30 规则:按天统计:每天都统计前面几天的总额 得到的结果: DATE SALE SUM ----- -------- ------ 1 20 20 --1天 2 15 35 --1天+2天 3 14 49 --1天+2天+3天 4 18 67 . 5 30 97 . 2:统计各班成绩第一名的同学信息 NAME CLASS S ----- ----- ---------------------- fda 1 80 ffd 1 78 dss 1 95 cfe 2 74 gds 2 92 gf 3 99 ddd 3 99 adf 3 45 asdf 3 55 3dd 3 78 通过: -- select * from ( select name ,class,s,rank()over(partition by class order by s desc ) mm from t2 ) where mm=1 -- 得到结果: NAME CLASS S MM ----- ----- ---------------------- ---------------------- dss 1 95 1 gds 2 92 1 gf 3 99 1 ddd 3 99 1 注意: 1.在求第一名成绩的时候,不能用row_number(),因为如果同班有两个并列第一,row_number()只返回一个结果 2.rank()和dense_rank()的区别是: --rank()是跳跃排序,有两个第二名时接下来就是第四名 --dense_rank()l是连续排序,有两个第二名时仍然跟着第三名 3.分类统计 (并显示信息) A B C -- -- ---------------------- m a 2 n a 3 m a 2 n b 2 n b 1 x b 3 x b 2 x b 4 h b 3 select a,c, sum (c)over(partition by a) from t2 得到结果: A B C SUM (C)OVER(PARTITIONBYA) -- -- ------- ------------------------ h b 3 3 m a 2 4 m a 2 4 n a 3 6 n b 2 6 n b 1 6 x b 3 9 x b 2 9 x b 4 9 如果用 sum , group by 则只能得到 A SUM (C) -- ---------------------- h 3 m 4 n 6 x 9 无法得到B列值 ===== select * from test 数据: A B C 1 1 1 1 2 2 1 3 3 2 2 5 3 4 6 ---将B栏位值相同的对应的C 栏位值加总 select a,b,c, SUM (C) OVER (PARTITION BY B) C_Sum from test A B C C_SUM 1 1 1 1 1 2 2 7 2 2 5 7 1 3 3 3 3 4 6 6 ---如果不需要已某个栏位的值分割,那就要用 null eg: 就是将C的栏位值summary 放在每行后面 select a,b,c, SUM (C) OVER (PARTITION BY null ) C_Sum from test A B C C_SUM 1 1 1 17 1 2 2 17 1 3 3 17 2 2 5 17 3 4 6 17 求个人工资占部门工资的百分比 SQL> select * from salary; NAME DEPT SAL ---------- ---- ----- a 10 2000 b 10 3000 c 10 5000 d 20 4000 SQL> select name ,dept,sal,sal*100/ sum (sal) over(partition by dept) percent from salary; NAME DEPT SAL PERCENT ---------- ---- ----- ---------- a 10 2000 20 b 10 3000 30 c 10 5000 50 d 20 4000 100 二:开窗函数 开窗函数指定了分析函数工作的数据窗口大小,这个数据窗口大小可能会随着行的变化而变化,举例如下: 1: over( order by salary) 按照salary排序进行累计, order by 是个默认的开窗函数 over(partition by deptno)按照部门分区 2: over( order by salary range between 5 preceding and 5 following) 每行对应的数据窗口是之前行幅度值不超过5,之后行幅度值不超过5 例如:对于以下列 aa 1 2 2 2 3 4 5 6 7 9 sum (aa)over( order by aa range between 2 preceding and 2 following) 得出的结果是 AA SUM ---------------------- ------------------------------------------------------- 1 10 2 14 2 14 2 14 3 18 4 18 5 22 6 18 7 22 9 9 就是说,对于aa=5的一行 , sum 为 5-1<=aa<=5+2 的和 对于aa=2来说 , sum =1+2+2+2+3+4=14 ; 又如 对于aa=9 ,9-1<=aa<=9+2 只有9一个数,所以 sum =9 ; 3:其它: over( order by salary rows between 2 preceding and 4 following) 每行对应的数据窗口是之前2行,之后4行 4:下面三条语句等效: over( order by salary rows between unbounded preceding and unbounded following) 每行对应的数据窗口是从第一行到最后一行,等效: over( order by salary range between unbounded preceding and unbounded following) 等效 over(partition by null ) 常用的分析函数如下所列: row_number() over(partition by ... order by ...) rank() over(partition by ... order by ...) dense_rank() over(partition by ... order by ...) count () over(partition by ... order by ...) max () over(partition by ... order by ...) min () over(partition by ... order by ...) sum () over(partition by ... order by ...) avg () over(partition by ... order by ...) first_value() over(partition by ... order by ...) last_value() over(partition by ... order by ...) lag() over(partition by ... order by ...) lead() over(partition by ... order by ...) 示例 SQL> select type,qty from test; TYPE QTY ---------- ---------- 1 6 2 9 SQL> select type,qty,to_char(row_number() over(partition by type order by qty))|| ‘/‘ ||to_char( count (*) over(partition by type)) as cnt2 from test; TYPE QTY CNT2 ---------- ---------- ------------ 3 1/2 1 6 2/2 2 5 1/3 7 2/3 2 9 3/3 SQL> select * from test; ---------- ------------------------------------------------- 1 11111 2 22222 3 33333 4 44444 SQL> select t.id,mc,to_char(b.rn)|| ‘/‘ ||t.id)e 2 from test t, ( select rownum rn from ( select max (to_number(id)) mid from test) connect by rownum <=mid ))L 4 where b.rn<=to_number(t.id) order by id ID MC TO_CHAR(B.RN)|| ‘/‘ ||T.ID --------- -------------------------------------------------- --------------------------------------------------- 1 11111 1/1 2 22222 1/2 2 22222 2/2 3 33333 1/3 3 33333 2/3 3 33333 3/3 44444 1/4 44444 2/4 4 44444 3/4CNOUG4 44444 4/4 10 rows selected ******************************************************************* |
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1. select deptno,row_number() over(partition by deptno order by sal) from emp order by deptno; 2. select deptno,rank() over (partition by deptno order by sal) from emp order by deptno; 3. select deptno,dense_rank() over(partition by deptno order by sal) from emp order by deptno; 4. select deptno,ename,sal,lag(ename,1, null ) over(partition by deptno order by ename) from emp ord er by deptno; 5. select deptno,ename,sal,lag(ename,2, ‘example‘ ) over(partition by deptno order by ename) from em p order by deptno; 6. select deptno, sal, sum (sal) over(partition by deptno) from emp; --每行记录后都有总计值 select deptno, sum(sal) from emp group by deptno; 7. 求每个部门的平均工资以及每个人与所在部门的工资差额 select deptno,ename,sal , round( avg (sal) over(partition by deptno)) as dept_avg_sal, round(sal- avg (sal) over(partition by deptno)) as dept_sal_diff from emp; |
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原文地址:http://www.cnblogs.com/Unrmk-LingXing/p/4607858.html