SQLServer 簡單數據拆分
阿新 • • 發佈:2017-09-25
imp arc des osi border 解決方法 arch form lis
--1. 舊的解決方法(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.使用臨時表完成 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.如果數據量小,可不使用臨時表
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. 新的解決方法(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,‘‘ + REPLACE([value], ‘,‘, ‘‘) + ‘‘) 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
--1. 舊的解決方法(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.使用臨時表完成 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)
SQLServer 簡單數據拆分