OVER(PARTITION BY)函数介绍

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

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?? over（partition by deptno order by salary）

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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

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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

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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

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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;

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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

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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。

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在first_value和last_value中ignore nulls的使用数据如下：

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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：

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--lag() over()函数用法（取出前n行数据）
lag(expresstion,<offset>,<default>)
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,<offset>,<default>)
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