http://zonghl8006.blog.163.com/blog/static/4528311520083995931317/

? over(Partition by...) 一個(gè)超級(jí)牛皮的ORACLE特有函數(shù)。

天天都用ORACLE，用了快2年了。最近才接觸到這個(gè)功能強(qiáng)大而靈活的函數(shù)。真實(shí)慚愧啊！

oracle的分析函數(shù)over 及開(kāi)窗函數(shù)
一：分析函數(shù)over
Oracle從8.1.6開(kāi)始提供分析函數(shù)，分析函數(shù)用于計(jì)算基于組的某種聚合值，它和聚合函數(shù)的不同之處是
對(duì)于每個(gè)組返回多行，而聚合函數(shù)對(duì)于每個(gè)組只返回一行。?
下面通過(guò)幾個(gè)例子來(lái)說(shuō)明其應(yīng)用。???????????????????????????????????????
1：統(tǒng)計(jì)某商店的營(yíng)業(yè)額。????????
???? date?????? sale
???? 1?????????? 20
???? 2?????????? 15
???? 3?????????? 14
???? 4?????????? 18
???? 5?????????? 30
??? 規(guī)則：按天統(tǒng)計(jì)：每天都統(tǒng)計(jì)前面幾天的總額
??? 得到的結(jié)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:統(tǒng)計(jì)各班成績(jī)第一名的同學(xué)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??????????????
???
??? 通過(guò)：???
??? --
??? select * from???????????????????????????????????????????????????????????????????????
??? (????????????????????????????????????????????????????????????????????????????
??? select name,class,s,rank()over(partition by class order by s desc) mm from t2
??? )????????????????????????????????????????????????????????????????????????????
??? where mm=1?
??? --
??? 得到結(jié)果：
??? NAME?? CLASS S?????????????????????? MM????????????????????????????????????????????????????????????????????????????????????????
??? ----- ----- ---------------------- ----------------------?
??? dss??? 1????? 95????????????????????? 1??????????????????????
??? gds??? 2????? 92????????????????????? 1??????????????????????
??? gf???? 3????? 99????????????????????? 1??????????????????????
??? ddd??? 3????? 99????????????????????? 1??????????
???
??? 注意：
??? 1.在求第一名成績(jī)的時(shí)候，不能用row_number()，因?yàn)槿绻嘤袃蓚€(gè)并列第一，row_number()只返回一個(gè)結(jié)果?????????
??? 2.rank()和dense_rank()的區(qū)別是：
????? --rank()是跳躍排序，有兩個(gè)第二名時(shí)接下來(lái)就是第四名
????? --dense_rank()l是連續(xù)排序，有兩個(gè)第二名時(shí)仍然跟著第三名
?????
?????
3.分類(lèi)統(tǒng)計(jì) (并顯示信息)
??? 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????????????????
?? 得到結(jié)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 則只能得到
?? A?? SUM(C)????????????????????????????
?? -- ----------------------?
?? h?? 3??????????????????????
?? m?? 4??????????????????????
?? n?? 6??????????????????????
?? x?? 9??????????????????????
?? 無(wú)法得到B列值???????
??
＝＝＝＝＝
select * from test

數(shù)據(jù):
A B C?
1 1 1?
1 2 2?
1 3 3?
2 2 5?
3 4 6

---將B欄位值相同的對(duì)應(yīng)的C 欄位值加總
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

?

---如果不需要已某個(gè)欄位的值分割,那就要用 null

eg: 就是將C的欄位值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

?

求個(gè)人工資占部門(mén)工資的百分比

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：開(kāi)窗函數(shù)???????????
????? 開(kāi)窗函數(shù)指定了分析函數(shù)工作的數(shù)據(jù)窗口大小，這個(gè)數(shù)據(jù)窗口大小可能會(huì)隨著行的變化而變化，舉例如下：?
1：?????
?? over（order by salary） 按照salary排序進(jìn)行累計(jì)，order by是個(gè)默認(rèn)的開(kāi)窗函數(shù)
?? over（partition by deptno）按照部門(mén)分區(qū)
2：
? over（order by salary range between 5 preceding and 5 following）
?? 每行對(duì)應(yīng)的數(shù)據(jù)窗口是之前行幅度值不超過(guò)5，之后行幅度值不超過(guò)5
?? 例如:對(duì)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）
?? 得出的結(jié)果是
??????????? AA?????????????????????? SUM
??????????? ---------------------- -------------------------------------------------------?
??????????? 1?????????????????????? 10??????????????????????????????????????????????????????
??????????? 2?????????????????????? 14??????????????????????????????????????????????????????
??????????? 2?????????????????????? 14??????????????????????????????????????????????????????
??????????? 2?????????????????????? 14??????????????????????????????????????????????????????
??????????? 3?????????????????????? 18??????????????????????????????????????????????????????
??????????? 4?????????????????????? 18??????????????????????????????????????????????????????
??????????? 5?????????????????????? 22??????????????????????????????????????????????????????
??????????? 6?????????????????????? 18????????????????????????????????????????????????????????????????
??????????? 7?????????????????????? 22????????????????????????????????????????????????????????????????
??????????? 9?????????????????????? 9?????????????????????????????????????????????????????????????????
?????????????
?? 就是說(shuō)，對(duì)于aa=5的一行 ，sum為?? 5-1<=aa<=5+2 的和
?? 對(duì)于aa=2來(lái)說(shuō) ，sum=1+2+2+2+3+4=14???? ；
?? 又如 對(duì)于aa=9 ，9-1<=aa<=9+2 只有9一個(gè)數(shù)，所以sum=9??? ；
??????????????
3：其它：
???? over（order by salary rows between 2 preceding and 4 following）
????????? 每行對(duì)應(yīng)的數(shù)據(jù)窗口是之前2行，之后4行?
4：下面三條語(yǔ)句等效：???????????
???? over（order by salary rows between unbounded preceding and unbounded following）
????????? 每行對(duì)應(yīng)的數(shù)據(jù)窗口是從第一行到最后一行，等效：
???? over（order by salary range between unbounded preceding and unbounded following）
?????????? 等效
???? over(partition by null)

?

常用的分析函數(shù)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

*******************************************************************

關(guān)于partition by

這些都是分析函數(shù)，好像是8.0以后才有的 row_number()和rownum差不多，功能更強(qiáng)一點(diǎn)（可以在各個(gè)分組內(nèi)從1開(kāi)時(shí)排序） rank()是跳躍排序，有兩個(gè)第二名時(shí)接下來(lái)就是第四名（同樣是在各個(gè)分組內(nèi)） dense_rank()l是連續(xù)排序，有兩個(gè)第二名時(shí)仍然跟著第三名。相比之下row_number是沒(méi)有重復(fù)值的 lag（arg1,arg2,arg3): arg1是從其他行返回的表達(dá)式 arg2是希望檢索的當(dāng)前行分區(qū)的偏移量。是一個(gè)正的偏移量，時(shí)一個(gè)往回檢索以前的行的數(shù)目。 arg3是在arg2表示的數(shù)目超出了分組的范圍時(shí)返回的值。

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;--每行記錄后都有總計(jì)值? select deptno, sum(sal) from emp group by deptno;
7. 求每個(gè)部門(mén)的平均工資以及每個(gè)人與所在部門(mén)的工資差額

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;

轉(zhuǎn)載于:https://www.cnblogs.com/hailei/p/4494066.html

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