日期:2014-05-18  浏览次数:20597 次

【100分,确实有些疑难】关于按照某个字段部分内容来进行Group By的问题。
我要做个电子邮件发送统计
表的数据如下:
Mail_From | Mail_To
----------------------------------------------------
liming |aa@sina.com;bb@sina.com;mmm@yahoo.com.cn
liming |aa@sina.com
liming |bb@sina.com
liming |baby@hotmail.com;mmm@yahoo.com.cn
liming |mmm@yahoo.com.cn
liming |aa@sina.com;mmm@yahoo.com.cn

我想统计出来的结果是这样
 Mail_To |Num
-------------------------------
mmm@yahoo.com.cn |4
aa@sina.com |3
bb@sina.com |2
baby@hotmail.com |1

也就是统计liming都给哪些人发了邮件,按次数排倒序,最麻烦的是收件人是用";"分割开的,搞了好久没有搞出来,大家谁有什么好办法?谢谢大家100分不够再加!

------解决方案--------------------
SQL code
----------------------------
-- Author  :SQL77(只为思齐老)
-- Date    :2010-03-09 15:17:54
-- Version:
--      Microsoft SQL Server  2000 - 8.00.194 (Intel X86) 
--    Aug  6 2000 00:57:48 
--    Copyright (c) 1988-2000 Microsoft Corporation
--    Desktop Engine on Windows NT 5.1 (Build 2600: Service Pack 3)
--
----------------------------
--> 测试数据:#TB
if object_id('tempdb.dbo.#TB') is not null drop table #TB
go 
create table #TB([Mail_From] varchar(6),[Mail_To] varchar(40))
insert #TB
select 'liming','aa@sina.com;bb@sina.com;mmm@yahoo.com.cn' union all
select 'liming','aa@sina.com' union all
select 'liming','bb@sina.com' union all
select 'liming','baby@hotmail.com;mmm@yahoo.com.cn' union all
select 'liming','mmm@yahoo.com.cn' union all
select 'liming','aa@sina.com;mmm@yahoo.com.cn'
--------------开始查询--------------------------
SELECT  Mail_From, Mail_TO,COUNT(1) C FROM 
(
select Mail_From,
SUBSTRING( Mail_To,NUMBER,CHARINDEX(';',Mail_To+';',NUMBER)-NUMBER)Mail_To
FROM  #TB,MASTER..SPT_VALUES 
WHERE TYPE='P' AND SUBSTRING(';'+Mail_To,NUMBER,1)=';'
)T GROUP BY 
Mail_From, Mail_TO ORDER BY 3 DESC
----------------结果----------------------------

(所影响的行数为 6 行)

Mail_From Mail_TO                                  C           
--------- ---------------------------------------- ----------- 
liming    mmm@yahoo.com.cn                         4
liming    aa@sina.com                              3
liming    bb@sina.com                              2
liming    baby@hotmail.com                         1

(所影响的行数为 4 行)


/* 
*/

------解决方案--------------------
SQL code
--> 测试数据:[tb]
if object_id('[tb]') is not null drop table [tb]
create table [tb]([Mail_From] varchar(6),[Mail_To] varchar(40))
go
insert [tb]
select 'liming','aa@sina.com;bb@sina.com;mmm@yahoo.com.cn' union all
select 'liming','aa@sina.com' union all
select 'liming','bb@sina.com' union all
select 'liming','baby@hotmail.com;mmm@yahoo.com.cn' union all
select 'liming','mmm@yahoo.com.cn' union all
select 'liming','aa@sina.com;mmm@yahoo.com.cn'

select Mail_To=substring(t.Mail_To,r.number,charindex(';',t.Mail_To+';',r.number)-r.number),
count(1) as Num
from tb t,master..spt_values r
where r.number<=len(t.Mail_To) and r.type='p' 
and substring(';'+t.Mail_To,r.number,1)=';'
group by substring(t.Mail_To,r.number,charindex(';',t.Mail_To+';',r.number)-r.number)

Mail_To                                  Num         
---------------------------------------- ----------- 
aa@sina.com                              3
baby@hotmail.com                         1
bb@sina.com                              2
mmm@yahoo.com.cn                         4

(所影响的行数为 4 行)

------解决方案--------------------
SQL code
--参考
拆分表:

--> --> (Roy)生成測試數據
 
if not object_id('Tab') is null
    drop table Tab
Go
Create table Tab([Col1] int,[COl2] nvarchar(5))
Insert Tab
select 1,N'a,b,c' union all
select 2,N'd,e' union all
select 3,N'f'
Go

--SQL2000用辅助表:
if object_id('Tempdb..#Num') is not null
    drop table #Num
go
select top 100 ID=Identity(int,1,1) into #Num from sysc