日韩性视频-久久久蜜桃-www中文字幕-在线中文字幕av-亚洲欧美一区二区三区四区-撸久久-香蕉视频一区-久久无码精品丰满人妻-国产高潮av-激情福利社-日韩av网址大全-国产精品久久999-日本五十路在线-性欧美在线-久久99精品波多结衣一区-男女午夜免费视频-黑人极品ⅴideos精品欧美棵-人人妻人人澡人人爽精品欧美一区-日韩一区在线看-欧美a级在线免费观看

歡迎訪問 生活随笔!

生活随笔

當前位置: 首頁 > 编程资源 > 编程问答 >内容正文

编程问答

row_number() OVER(PARTITION BY)函数

發布時間:2024/4/15 编程问答 40 豆豆
生活随笔 收集整理的這篇文章主要介紹了 row_number() OVER(PARTITION BY)函数 小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

OVER(PARTITION?BY)函數介紹

開窗函數??????????
?????Oracle從8.1.6開始提供分析函數,分析函數用于計算基于組的某種聚合值,它和聚合函數的不同之處是:對于每個組返回多行,而聚合函數對于每個組只返回一行。

??????開窗函數指定了分析函數工作的數據窗口大小,這個數據窗口大小可能會隨著行的變化而變化,舉例如下:
1:over后的寫法:????
???over(order by salary) 按照salary排序進行累計,order by是個默認的開窗函數
???over(partition by deptno)按照部門分區

?

???over(partition by deptno order by salary)

?

2:開窗的窗口范圍
over(order by salary?range?between 5 preceding and 5 following):窗口范圍為當前行數據幅度減5加5后的范圍內的。

舉例:

?

--sum(s)over(order by s range between 2 preceding and 2 following)?表示加2或2的范圍內的求和

?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減2即43到47,但是s在這個范圍內只有45
asdf???????3????????55????????55
cfe????????2????????74????????74
3dd????????3????????78????????158 --78在76到80范圍內有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):窗口范圍為當前行前后各移動5行。

舉例:

?

--sum(s)over(order by s rows between 2 preceding and 2 following)表示在上下兩行之間的范圍內
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、與over函數結合的幾個函數介紹

row_number()over()、rank()over()和dense_rank()over()函數的使用

下面以班級成績表t2來說明其應用

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;
得到的結果是:
dss????????1????????95????????1
ffd????????1????????95????????1
gds????????2????????92????????1
gf?????????3????????99????????1
ddd????????3????????99????????1?

注意:
????1.在求第一名成績的時候,不能用row_number(),因為如果同班有兩個并列第一,row_number()只返回一個結果;
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有兩名但是只顯示一個
2????????92????????1
3????????99????????1 --99有兩名但也只顯示一個

????2.rank()和dense_rank()可以將所有的都查找出來:
如上可以看到采用rank可以將并列第一名的都查找出來;
?????rank()和dense_rank()區別:
?????--rank()是跳躍排序,有兩個第二名時接下來就是第四名;
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是連續排序,有兩個第二名時仍然跟著第三名
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 --連續排序(仍為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 --根據班級進行分數求和
dss????????1????????95????????190??--由于兩個95都是第一名,所以累加時是兩個第一名的相加
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 ?



--找出這三條電路每條電路的第一條記錄類型和最后一條記錄類型

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時不使用rows BETWEEN unbounded preceding AND unbounded following的結果

?

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;

如下圖可以看到,如果不使用

rows BETWEEN unbounded preceding AND unbounded following,取出的last_value由于與res_type進行進行排列,因此取出的電路的最后一行記錄的類型就不是按照電路的范圍提取了,而是以res_type為范圍進行提取了。

?

?

?

?

?

在first_value和last_value中ignore nulls的使用

數據如下:

?

?

取出該電路的第一條記錄,加上ignore nulls后,如果第一條是判斷的那個字段是空的,則默認取下一條,結果如下所示:

?

?

--lag() over()函數用法(取出前n行數據)
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()函數用法(取出后N行數據)

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)函數用法 Ratio_to_report() 括號中就是分子,over() 括號中就是分母
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?--分母缺省就是整個占比
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;--分組后的占比


轉載于:https://blog.51cto.com/wilsonking/1600004

與50位技術專家面對面20年技術見證,附贈技術全景圖

總結

以上是生活随笔為你收集整理的row_number() OVER(PARTITION BY)函数的全部內容,希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯,歡迎將生活随笔推薦給好友。