/*
标题：简单数据拆分(version 2.0)
作者：爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间：2010-05-07
地点：重庆航天职业学院
描述：

有表tb, 如下:
id          value
----------- -----------
1           aa,bb
2           aaa,bbb,ccc
欲按id,分拆value列, 分拆后结果如下:
id          value
----------- --------
1           aa
1           bb
2           aaa
2           bbb
2           ccc
*/

--1. 旧的解决方法(sql server 2000)

create table tb(id int,value varchar(30))
insert into tb values(1,'aa,bb')
insert into tb values(2,'aaa,bbb,ccc')
go

--方法1.使用临时表完成
SELECT TOP 8000 id = IDENTITY(int, 1, 1) INTO # FROM syscolumns a, syscolumns b 

SELECT A.id, value = SUBSTRING(A.[value], B.id, CHARINDEX(',', A.[value] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[value], B.id, 1) = ','

DROP TABLE #

--方法2.如果数据量小，可不使用临时表
select a.id , value = substring(a.value , b.number , charindex(',' , a.value + ',' , b.number) - b.number) 
from tb a join master..spt_values  b 
on b.type='p' and b.number between 1 and len(a.value)
where substring(',' + a.value , b.number , 1) = ','

--2. 新的解决方法(sql server 2005)
create table tb(id int,value varchar(30))
insert into tb values(1,'aa,bb')
insert into tb values(2,'aaa,bbb,ccc')
go

--方法1.使用xml完成
SELECT A.id, B.value FROM
(
  SELECT id, [value] = CONVERT(xml,'<root><v>' + REPLACE([value], ',', '</v><v>') + '</v></root>') FROM tb
) A OUTER APPLY
(
  SELECT value = N.v.value('.', 'varchar(100)') FROM A.[value].nodes('/root/v') N(v)
) B

--方法2.使用CTE完成
;with tt as 
(select id,[value]=cast(left([value],charindex(',',[value]+',')-1) as nvarchar(100)),Split=cast(stuff([value]+',',1,charindex(',',[value]+','),'') as nvarchar(100)) from tb
union all
select id,[value]=cast(left(Split,charindex(',',Split)-1) as nvarchar(100)),Split= cast(stuff(Split,1,charindex(',',Split),'') as nvarchar(100)) from tt where split>''
)
select id,[value] from tt order by id option (MAXRECURSION 0)


DROP TABLE tb

/*
id          value
----------- ------------------------------
1           aa
1           bb
2           aaa
2           bbb
2           ccc

(5 行受影响)
*/

------解决方案--------------------

SQL code
create table tb(id int,name varchar(100))
insert into tb values('1', '张三，是好人<br>，李四，是好人<br>，')
insert into tb values('2', '张三，是好人<br>，李四，是好人<br>，赵六，是好人<br>，')
insert into tb values('3','赵六，是好人<br>，李四，是好人<br>，')
insert into tb values('4', '张三，是好人<br>，赵六，是好人<br>，')
insert into tb values('5', '张三，是好人<br>，王五，是好人<br>，')
insert into tb values('6', '王五，是好人<br>，李四，是好人<br>，')
insert into tb values('7', '张三，是好人<br>，王五，是好人<br>，赵六，是好人<br>，李四，是好人<br>，')
insert into tb values('8', '张三，是好人<br>，李四，是好人<br>，')
go

select left(name ,charindex('，',name)-1) name , count(1) [count] from
(
select a.id , name = substring(a.name , b.number , charindex('<br>，' , a.name + '<br>，' , b.number) - b.number) 
from tb a join master..spt_values  b 
on b.type='p' and b.number between 1 and len(a.name)
where substring('<br>，' + a.name , b.number , 5) = '<br>，'
) t
group by left(name ,charindex('，',name)-1)

drop table tb

/*
name