【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的有关问题

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-- 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 syscolumns a,syscolumns b
Select 
    a.Col1,COl2=substring(a.Col2,b.ID,charindex(',',a.Col2+',',b.ID)-b.ID) 
from 
    Tab a,#Num b
where
    charindex(',',','+a.Col2,b.ID)=b.ID --也可用 substring(','+a.COl2,b.ID,1)=','
--2000不使用辅助表
Select
    a.Col1,COl2=substring(a.Col2,b.number,charindex(',',a.Col2+',',b.number)-b.number) 
from 
    Tab a join master..spt_values  b 
    ON B.type='p' AND B.number BETWEEN 1 AND LEN(A.col2)
where
     substring(','+a.COl2,b.number,1)=','


SQL2005用Xml:

select 
    a.COl1,b.Col2
from 
    (select Col1,COl2=convert(xml,'<root><v>'+replace(COl2,',','</v><v>')+'</v></root>') from Tab)a
outer apply
    (select Col2=C.v.value('.','nvarchar(100)') from a.COl2.nodes('/root/v')C(v))b




SQL05用CTE:

;with roy as 
(select Col1,COl2=cast(left(Col2,charindex(',',Col2+',')-1) as nvarchar(100)),Split=cast(stuff(COl2+',',1,charindex(',',Col2+','),'') as nvarchar(100)) from Tab
union all
select Col1,COl2=cast(left(Split,charindex(',',Split)-1) as nvarchar(100)),Split= cast(stuff(Split,1,charindex(',',Split),'') as nvarchar(100)) from Roy where split>''
)
select COl1,COl2 from roy order by COl1 option (MAXRECURSION 0)

生成结果:
/*
Col1        COl2
----------- -----
1           a
1           b
1           c
2           d
2           e
3           f
*/