oracle分析函数系列之sum(col1) over(partition by col2 order by col3):实现分组汇总或递增汇总教程
oracle分析函数系列之sum(col1) over(partition by col2 order by col3):实现分组汇总或递增汇总
rfb0204421 分享于 2012-06-18
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语法:sum(col1) over(partition by col2 order by col3 )
准备数据:
DEPT\_ID ENAME SAL
1 1000 A 2500
2 1000 B 3500
3 1000 C 1500
4 1000 D 2000
5 2000 E 2500
6 2000 F 2000
7 2000 G 3500
主要有四种情况:
**sum(sal) over (partition by deptno order by ename) 按部门“连续”求总和
sum(sal) over (partition by deptno) 按部门求总和
sum(sal) over (order by deptno，ename) 不按部门“连续”求总和
sum(sal) over () 不按部门，求所有员工总和，效果等同于sum(sal)。**
1.有partition by有order by : 在partition by分组下,按照不同的order by col3实现递增汇总..
SQL>>select DEPT\_ID,ENAME,SAL,sum(SAL) over(partition by dept\_id order by ENAME) AS TOTAL from dept\_sal
结果:按照部门分组,按名字排序实现递增汇总.
DEPT\_ID ENAME SAL TOTAL
1 1000 A 3500 3500
2 1000 B 3500 7000
3 1000 C 1500 8500
4 1000 D 2000 10500
5 2000 E 2500 2500
6 2000 F 2000 4500
7 2000 G 3500 8000
如果col3重复会只加总一次(当然在本例中这种写法毫无意义):
SQL>> select DEPT\_ID,ENAME,SAL,sum(SAL) over(partition by dept\_id order by SAL) AS TOTAL from dept\_sal
DEPT\_ID ENAME SAL TOTAL
11000 C 1500 1500
2 1000 D 2000 3500
3 1000 A 3500 10500
4 1000 B 3500 10500
5 2000 F 2000 2000
6 2000 E 2500 4500
7 2000 G 3500 8000
2.有partition by无order by: 实现分组内所有数据的汇总
SQL>>select DEPT\_ID,ENAME,SAL,sum(SAL) over(partition by dept\_id) AS TOTAL from dept\_sal
DEPT\_ID ENAME SAL TOTAL
1 1000 A 3500 10500
2 1000 B 3500 10500
3 1000 C 1500 10500
4 1000 D 2000 10500
5 2000 E 2500 8000
6 2000 F 2000 8000
7 2000 G 3500 8000
3.无partition by有order by : 直接按order by 字段实现递增汇总
SQL>>select DEPT\_ID,ENAME,SAL,sum(SAL) over(order by ENAME) AS TOTAL from dept\_sal
**DEPT\_ID ENAME SAL TOTAL
1 1000 A 3500 3500
2 1000 B 3500 7000
3 1000 C 1500 8500
4 1000 D 2000 10500
5 2000 E 2500 13000
6 2000 F 2000 15000
7 2000 G 3500 18500**
如果order by 的值相同,会进行汇总,但汇总后显示的值会是一样的,如下:
SQL>>select DEPT\_ID,ENAME,SAL,sum(SAL) over(order by DEPT\_ID) AS TOTAL from dept\_sal**
DEPT\_ID ENAME SAL TOTAL
1 1000 A 3500 10500
2 1000 B 3500 10500
3 1000 C 1500 10500
4 1000 D 2000 10500
5 2000 E 2500 18500
6 2000 F 2000 18500
7 2000 G 3500 18500
4.无partition by无order by: 所有数据相加.
SQL>>select DEPT\_ID,ENAME,SAL,sum(SAL) over() AS TOTAL from dept\_sal**
DEPT\_ID ENAME SAL TOTAL
1 1000 A 3500 18500
2 1000 B 3500 18500
3 1000 C 1500 18500
4 1000 D 2000 18500
5 2000 E 2500 18500
6 2000 F 2000 18500
7 2000 G 3500 18500
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分析sum(x) over(partition by y ORDER BY z) [分析函数]
Oracle 作者：SmartWilson 时间：2015-05-14 10:34:21 1672 0
[本文来源于http://www.cnblogs.com/luhe/p/4155612.html
sum(x) over( partition by y ORDER BY z ) 分析](http://www.cnblogs.com/luhe/p/4155612.html)
之前用过row\_number()，rank()等排序与over( partition by ... ORDER BY ...)，这两个比较好理解: 先分组，然后在组内排名。
今天突然碰到sum(...) over( partition by ... ORDER BY ... )，居然搞不清除怎么执行的，所以查了些资料，做了下实操。
1. 从最简单的开始
sum(...) over( )，对所有行求和
sum(...) over( order by ... )，和 = 第一行 到 与当前行同序号行的最后一行的所有值求和，文字不太好理解，请看下图的算法解析。
with aa as ( SELECT 1 a,1 b, 3 c FROM dual union SELECT 2 a,2 b, 3 c FROM dual union SELECT 3 a,3 b, 3 c FROM dual union SELECT 4 a,4 b, 3 c FROM dual union SELECT 5 a,5 b, 3 c FROM dual union SELECT 6 a,5 b, 3 c FROM dual union SELECT 7 a,2 b, 3 c FROM dual union SELECT 8 a,2 b, 8 c FROM dual union SELECT 9 a,3 b, 3 c FROM dual
) SELECT a,b,c, sum(c) over(order by b) sum1,--有排序，求和当前行所在顺序号的C列所有值 sum(c) over() sum2 FROM aa;--无排序，求和 C列所有值
2. 与 partition by 结合
sum(...) over( partition by... )，同组内所行求和
sum(...) over( partition by... order by ... )，同第1点中的排序求和原理，只是范围限制在组内
with aa as ( SELECT 1 a,1 b, 3 c FROM dual union SELECT 2 a,2 b, 3 c FROM dual union SELECT 3 a,3 b, 3 c FROM dual union SELECT 4 a,4 b, 3 c FROM dual union SELECT 5 a,5 b, 3 c FROM dual union SELECT 6 a,5 b, 3 c FROM dual union SELECT 7 a,2 b, 3 c FROM dual union SELECT 7 a,2 b, 8 c FROM dual union SELECT 9 a,3 b, 3 c FROM dual
) SELECT a,b,c,sum(c) over( partition by b ) partition_sum, sum(c) over( partition by b order by a desc) partition_order_sum FROM aa;
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OVER(PARTITION BY) 函数
0.2 2018.08.26 11:44 字数 553 阅读 5287评论 0喜欢 1
最近在项目中遇到了对每一个类型进行求和并且求该类型所占的比例的需求。
一开始使用的是自表的连接，后来发现这样做太复杂，更改后的SQL的解决方法是：
<pre class="has">SELECT T.CHANNEL AS PATTERN,
COUNT(T.TRANSACTIONKEY) AS T_COUNT,
SUM(T.AMT) AS T_AMT,
ROUND(100 * SUM(T.AMT) / SUM(SUM(T.AMT)) OVER(PARTITION BY 1), 2) AS AMT_PERCENT,
ROUND(100 * COUNT(T.TRANSACTIONKEY) / SUM(COUNT(T.TRANSACTIONKEY)) OVER(PARTITION BY 1),2) AS COUNT_PERCENTFROM XX(表名) T
WHERE T.PARTY_ID = '100579050'
GROUP BY T.CHANNEL
1. over函数的写法：
`over(partition by class order by sroce)` 按照`sroce`排序进行累计，`order by`是个默认的开窗函数，按照`class`分区。
2. 开窗的窗口范围：
`over(order by sroce range between 5 preceding and 5 following)`：窗口范围为当前行数据幅度减5加5后的范围内的。
`over(order by sroce rows between 5 preceding and 5 following)`：窗口范围为当前行前后各移动5行。
3. 与`over()`函数结合的函数的介绍
(1). 查询每个班的第一名的成绩：如下
``` SELECT * FROM (select t.name,t.class,t.sroce,rank() over(partition by t.class order by t.sroce desc) mm from T2_TEMP t) where mm = 1; ``` `````` 得到的结果是: dss 1 95 1 ffd 1 95 1 gds 2 92 1 gf 3 99 1 ddd 3 99 1 ``` 注意：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。 `````` SELECT * FROM (select t.name,t.class,t.sroce,row_number() over(partition by t.class order by t.sroce desc) mm from T2_TEMP t) where mm = 1; ``` 结果为： `````` dss 1 95 1 gfs 2 92 1 ddd 3 99 1 ``` 可以看出，本来第一名是两个人的并列，结果只显示了一个。 (2). `rank()`和`dense_rank()`可以将所有的都查找出来，`rank`可以将并列第一名的都查找出来；`rank()`和`dense_rank()`区别：`rank()`是跳跃排序，有两个第二名时接下来就是第四名。 求班级成绩排名： `````` select t.name,t.class,t.sroce,rank() over(partition by t.class order by t.sroce desc) mm from T2_TEMP t; ``` 查询结果： `````` dss 1 95 1 ffd 1 95 1 fda 1 80 3 gds 2 92 1 cfe 2 74 2 gf 3 99 1 ddd 3 99 1 3dd 3 78 3 asdf 3 55 4 adf 3 45 5 ``` `dense_rank()`是连续排序，有两个第二名时仍然跟着第三名 `````` select t.name,t.class,t.sroce,dense_rank() over(partition by t.class order by t.sroce desc) mm from T2_TEMP t; ``` 查询结果： `````` dss 1 95 1 ffd 1 95 1 fda 1 80 2 gds 2 92 1 cfe 2 74 2 gf 3 99 1 ddd 3 99 1 3dd 3 78 2 asdf 3 55 3 adf 3 45 4 ``` (3). `sum() over()`的使用 根据班级进行分数求和 `````` select t.name,t.class,t.sroce,sum(t.sroce) over(partition by t.class order by t.sroce desc) mm from T2_TEMP t; ``` `````` dss 1 95 190 --由于两个95都是第一名，所以累加时是两个第一名的相加 ffd 1 95 190 fda 1 80 270 --第一名加上第二名的 gds 2 92 92 cfe 2 74 166 gf 3 99 198 ddd 3 99 198 3dd 3 78 276 asdf 3 55 331 adf 3 45 376 ``` (4). `first_value () over()`和`last_value() over()`的使用 `````` select t.name,t.class,t.sroce,first_value(t.sroce) over(partition by t.class order by t.sroce desc) mm from T2_TEMP t; select t.name,t.class,t.sroce,last_value(t.sroce) over(partition by t.class order by t.sroce desc) mm from T2_TEMP t; ``` 分别求出第一个和最后一个成绩。 (5). `sum() over()`的使用 `````` select t.name,t.class,t.sroce,sum(t.sroce) over(partition by t.class order by t.sroce desc) mm from T2_TEMP t; ``` 求出班级的总分。 下面还有很多用法，就不一一列举了，简单介绍一下，和上面用法类似： `````` count() over(partition by ... order by ...)--求分组后的总数。 max() over(partition by ... order by ...)--求分组后的最大值。 min() over(partition by ... order by ...)--求分组后的最小值。 avg() over(partition by ... order by ...)--求分组后的平均值。 lag() over(partition by ... order by ...)--取出前n行数据。 lead() over(partition by ... order by ...)--取出后n行数据。 ratio_to_report() over(partition by ... order by ...)--Ratio_to_report() 括号中就是分子，over() 括号中就是分母。 percent_rank() over(partition by ... order by ...)-- ``` (6). `over partition by`与`group by`的区别： `group by`是对检索结果的保留行进行单纯分组，一般和聚合函数一起使用例如`max`、`min`、`sum`、`avg`、`count`等一块用。`partition by`虽然也具有分组功能，但同时也具有其他的高级功能。 \----------------------------------------------------------------------------------------------------------------------------------------------------------------- OVER(PARTITION BY)函数介绍 开窗函数 Oracle从8.1.6开始提供分析函数，分析函数用于计算基于组的某种聚合值，它和聚合函数的不同之处是：对于每个组返回多行，而聚合函数对于每个组只返回一行。 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后的写法： 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 over（partition by deptno order by salary） 2：开窗的窗口范围： over（order by salary range between 5 preceding and 5 following）：窗口范围为当前行数据幅度减5加5后的范围内的。 举例： \--sum(s)over(order by s range between 2 preceding and 2 following) 表示加2或2的范围内的求和 select name,class,s, sum(s)over(order by s range between 2 preceding and 2 following) mm from t2 adf 3 45 45 --45加2减2即43到47，但是s在这个范围内只有45 asdf 3 55 55 cfe 2 74 74 3dd 3 78 158 --78在76到80范围内有78，80，求和得158 fda 1 80 158 gds 2 92 92 ffd 1 95 190 dss 1 95 190 ddd 3 99 198 gf 3 99 198 over（order by salary rows between 5 preceding and 5 following）：窗口范围为当前行前后各移动5行。 举例： \--sum(s)over(order by s rows between 2 preceding and 2 following)表示在上下两行之间的范围内 select name,class,s, sum(s)over(order by s rows between 2 preceding and 2 following) mm from t2 adf 3 45 174 （45+55+74=174） asdf 3 55 252 （45+55+74+78=252） cfe 2 74 332 （74+55+45+78+80=332） 3dd 3 78 379 （78+74+55+80+92=379） fda 1 80 419 gds 2 92 440 ffd 1 95 461 dss 1 95 480 ddd 3 99 388 gf 3 99 293 over（order by salary range between unbounded preceding and unbounded following）或者 over（order by salary rows between unbounded preceding and unbounded following）：窗口不做限制 3、与over函数结合的几个函数介绍 **row\_number()over()、rank()over()和dense\_rank()over()函数的使用** 下面以班级成绩表t2来说明其应用 t2表信息如下： cfe 2 74 dss 1 95 ffd 1 95 fda 1 80 gds 2 92 gf 3 99 ddd 3 99 adf 3 45 asdf 3 55 3dd 3 78 select \* from ( select name,class,s,rank()over(partition by class order by s desc) mm from t2 ) where mm=1； 得到的结果是: dss 1 95 1 ffd 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; select \* from ( select name,class,s,row\_number()over(partition by class order by s desc) mm from t2 ) where mm=1； 1 95 1 --95有两名但是只显示一个 2 92 1 3 99 1 --99有两名但也只显示一个 2.rank()和dense\_rank()可以将所有的都查找出来： 如上可以看到采用rank可以将并列第一名的都查找出来； rank()和dense\_rank()区别： --rank()是跳跃排序，有两个第二名时接下来就是第四名； select name,class,s,rank()over(partition by class order by s desc) mm from t2 dss 1 95 1 ffd 1 95 1 fda 1 80 3 --直接就跳到了第三 gds 2 92 1 cfe 2 74 2 gf 3 99 1 ddd 3 99 1 3dd 3 78 3 asdf 3 55 4 adf 3 45 5 --dense\_rank()l是连续排序，有两个第二名时仍然跟着第三名 select name,class,s,dense\_rank()over(partition by class order by s desc) mm from t2 dss 1 95 1 ffd 1 95 1 fda 1 80 2 --连续排序（仍为2） gds 2 92 1 cfe 2 74 2 gf 3 99 1 ddd 3 99 1 3dd 3 78 2 asdf 3 55 3 adf 3 45 4 **--sum()over（）的使用** select name,class,s, sum(s)over(partition by class order by s desc) mm from t2 --根据班级进行分数求和 dss 1 95 190 --由于两个95都是第一名，所以累加时是两个第一名的相加 ffd 1 95 190 fda 1 80 270 --第一名加上第二名的 gds 2 92 92 cfe 2 74 166 gf 3 99 198 ddd 3 99 198 3dd 3 78 276 asdf 3 55 331 adf 3 45 376 **first\_value() over()和last\_value() over()的使用**  \--找出这三条电路每条电路的第一条记录类型和最后一条记录类型 SELECT opr\_id,res\_type, first\_value(res\_type) over(PARTITION BY opr\_id ORDER BY res\_type) low, last\_value(res\_type) over(PARTITION BY opr\_id ORDER BY res\_type **rows BETWEEN unbounded preceding AND unbounded following**) high FROM rm\_circuit\_route WHERE opr\_id IN ('000100190000000000021311','000100190000000000021355','000100190000000000021339') ORDER BY opr\_id;  注：**rows BETWEEN unbounded preceding AND unbounded following** 的使用 \--取last\_value时不使用**rows BETWEEN unbounded preceding AND unbounded following**的结果** SELECT opr\_id,res\_type, first\_value(res\_type) over(PARTITION BY opr\_id ORDER BY res\_type) low, last\_value(res\_type) over(PARTITION BY opr\_id ORDER BY res\_type) high FROM rm\_circuit\_route WHERE opr\_id IN ('000100190000000000021311','000100190000000000021355','000100190000000000021339') ORDER BY opr\_id; 如下图可以看到，如果不使用 ****rows BETWEEN unbounded preceding AND unbounded following，**取出的last\_value由于与res\_type进行进行排列，因此取出的电路的最后一行记录的类型就不是按照电路的范围提取了，而是以res\_type为范围进行提取了。**  ***在first\_value和last\_value中ignore nulls的使用*** 数据如下：  取出该电路的第一条记录，加上***ignore nulls****后，如果第一条是判断的那个字段是空的，则默认取下一条，结果如下所示：*   **--lag() over()函数用法**（取出前n行数据） **lag(expresstion,, )** with a as (select 1 id,'a' name from dual union select 2 id,'b' name from dual union select 3 id,'c' name from dual union select 4 id,'d' name from dual union select 5 id,'e' name from dual ) select id,name,lag(id,1,'')over(order by name) from a; **--lead() over()函数用法**（取出后N行数据） **lead(expresstion, , )** with a as (select 1 id,'a' name from dual union select 2 id,'b' name from dual union select 3 id,'c' name from dual union select 4 id,'d' name from dual union select 5 id,'e' name from dual ) select id,name,lead(id,1,'')over(order by name) from a; **--ratio\_to\_report(a)函数用法 Ratio\_to\_report() 括号中就是分子，over() 括号中就是分母** with a as (select 1 a from dual union all select 1 a from dual union all select 1 a from dual union all select 2 a from dual union all select 3 a from dual union all select 4 a from dual union all select 4 a from dual union all select 5 a from dual ) select a, ratio\_to\_report(a)over(partition by a) b from a order by a; with a as (select 1 a from dual union all select 1 a from dual union all select 1 a from dual union all select 2 a from dual union all select 3 a from dual union all select 4 a from dual union all select 4 a from dual union all select 5 a from dual ) select a, ratio\_to\_report(a)over() b from a --分母缺省就是整个占比 order by a; with a as (select 1 a from dual union all select 1 a from dual union all select 1 a from dual union all select 2 a from dual union all select 3 a from dual union all select 4 a from dual union all select 4 a from dual union all select 5 a from dual ) select a, ratio\_to\_report(a)over() b from a group by a order by a;--分组后的占比 **percent\_rank用法** 计算方法：所在组排名序号-1除以该组所有的行数-1，如下所示自己计算的pr1与通过percent\_rank函数得到的值是一样的： SELECT a.deptno, a.ename, a.sal, a.r, b.n, (a.r-1)/(n-1) pr1, percent\_rank() over(PARTITION BY a.deptno ORDER BY a.sal) pr2 FROM (SELECT deptno, ename, sal, rank() over(PARTITION BY deptno ORDER BY sal) r --计算出在组中的排名序号 FROM emp ORDER BY deptno, sal) a, (SELECT deptno, COUNT(1) n FROM emp GROUP BY deptno) b --按部门计算每个部门的所有成员数 WHERE a.deptno = b.deptno;  **cume\_dist函数** 计算方法：所在组排名序号除以该组所有的行数，但是如果存在并列情况，则需加上并列的个数-1， 如下所示自己计算的pr1与通过percent\_rank函数得到的值是一样的： SELECT a.deptno, a.ename, a.sal, a.r, b.n, c.rn, (a.r + c.rn - 1) / n pr1, cume\_dist() over(PARTITION BY a.deptno ORDER BY a.sal) pr2 FROM (SELECT deptno, ename, sal, rank() over(PARTITION BY deptno ORDER BY sal) r FROM emp ORDER BY deptno, sal) a, (SELECT deptno, COUNT(1) n FROM emp GROUP BY deptno) b, (SELECT deptno, r, COUNT(1) rn,sal FROM (SELECT deptno,sal, rank() over(PARTITION BY deptno ORDER BY sal) r FROM emp) GROUP BY deptno, r,sal ORDER BY deptno) c --c表就是为了得到每个部门员工工资的一样的个数 WHERE a.deptno = b.deptno AND a.deptno = c.deptno(+) AND a.sal = c.sal;  **percentile\_cont函数** 含义：输入一个百分比（该百分比就是按照percent\_rank函数计算的值），返回该百分比位置的平均值 如下，输入百分比为0.7，因为0.7介于0.6和0.8之间，因此返回的结果就是0.6对应的sal的1500加上0.8对应的sal的1600平均 SELECT ename, sal, deptno, percentile\_cont(0.7) within GROUP(ORDER BY sal) over(PARTITION BY deptno) "Percentile\_Cont", percent\_rank() over(PARTITION BY deptno ORDER BY sal) "Percent\_Rank" FROM emp WHERE deptno IN (30, 60);  若输入的百分比为0.6，则直接0.6对应的sal值，即1500 SELECT ename, sal, deptno, percentile\_cont(0.6) within GROUP(ORDER BY sal) over(PARTITION BY deptno) "Percentile\_Cont", percent\_rank() over(PARTITION BY deptno ORDER BY sal) "Percent\_Rank" FROM emp WHERE deptno IN (30, 60);  **PERCENTILE\_DISC函数** 功能描述：返回一个与输入的分布百分比值相对应的数据值，分布百分比的计算方法见函数CUME\_DIST，如果没有正好对应的数据值，就取大于该分布值的下一个值。 注意：本函数与PERCENTILE\_CONT的区别在找不到对应的分布值时返回的替代值的计算方法不同 SAMPLE：下例中0.7的分布值在部门30中没有对应的Cume\_Dist值，所以就取下一个分布值0.83333333所对应的SALARY来替代 SELECT ename, sal, deptno, percentile\_disc(0.7) within GROUP(ORDER BY sal) over(PARTITION BY deptno) "Percentile\_Disc", cume\_dist() over(PARTITION BY deptno ORDER BY sal) "Cume\_Dist" FROM emp WHERE deptno IN (30, 60);  详细参考http://www.cnblogs.com/lanzi/archive/2010/10/26/1861338.html \--select max(t.check\_date),t.user\_id from attendance t group by t.user\_id; insert into attendance\_day (id, user\_id, check\_day, gps\_x, gpx\_y) select seq\_attendance.nextval, t.user\_id, t.check\_date, t.gps\_x, t.gpx\_y \--from attendance t from (select m.\*, row\_number() over(partition by m.user\_id order by m.check\_date desc) rn from attendance m) t where rn = 1;