sql-server – 如何在单个语句中将CSV数据拆分并插入到新表中?
|
我有一个名为“Documents”的表,其中包含如下列: DocumentID 我有格式的数据 – @DIDID = 1,2,3,4 如何使用单个查询在单独的行中插入这些documentID? 解决方法您需要一种在TSQL中拆分和处理字符串的方法,有很多方法可以做到这一点.本文涵盖几乎所有方法的PRO和CON:Arrays and Lists in SQL Server 2005 and Beyond 您需要创建拆分功能.这是分割函数的使用方式: SELECT
*
FROM YourTable y
INNER JOIN dbo.yourSplitFunction(@Parameter) s ON y.ID=s.Value
I prefer the number table approach to split a string in TSQL – Using a Table of Numbers但是有很多方法可以在SQL Server中拆分字符串,请参阅前面的链接,它解释了每个链接的PRO和CON. 要使Numbers Table方法起作用,您需要执行一次性表设置,这将创建一个包含1到10,000行的表号: SELECT TOP 10000 IDENTITY(int,1,1) AS Number
INTO Numbers
FROM sys.objects s1
CROSS JOIN sys.objects s2
ALTER TABLE Numbers ADD CONSTRAINT PK_Numbers PRIMARY KEY CLUSTERED (Number)
设置Numbers表后,创建此拆分功能: CREATE FUNCTION inline_split_me (@SplitOn char(1),@param varchar(7998)) RETURNS TABLE AS
RETURN(SELECT substring(@SplitOn + @param + ',',Number + 1,charindex(@SplitOn,@SplitOn + @param + @SplitOn,Number + 1) - Number - 1)
AS Value
FROM Numbers
WHERE Number <= len(@SplitOn + @param + @SplitOn) - 1
AND substring(@SplitOn + @param + @SplitOn,Number,1) = @SplitOn)
GO
您现在可以轻松地将CSV字符串拆分为表格并加入其中: select * from dbo.inline_split_me(';','1;22;333;4444;;') where LEN(Value)>0
OUTPUT: Value ---------------------- 1 22 333 4444 (4 row(s) affected) 让你新表使用这个: --set up tables:
DECLARE @Documents table (DocumentID varchar(500),SomeValue varchar(5))
INSERT @Documents VALUES ('1,4','AAA')
INSERT @Documents VALUES ('5,6','BBBB')
DECLARE @NewDocuments table (DocumentID int,SomeValue varchar(5))
--populate NewDocuments
INSERT @NewDocuments
(DocumentID,SomeValue)
SELECT
c.value,a.SomeValue
FROM @Documents a
CROSS APPLY dbo.inline_split_me(',a.DocumentID) c
--show NewDocuments contents:
select * from @NewDocuments
OUTPUT: DocumentID SomeValue ----------- --------- 1 AAA 2 AAA 3 AAA 4 AAA 5 BBBB 6 BBBB (6 row(s) affected) 如果您不想创建Numbers表并且正在运行SQL Server 2005及更高版本,则可以使用此split函数(不需要Numbers表): CREATE FUNCTION inline_split_me (@SplitOn char(1),@String varchar(7998))
RETURNS TABLE AS
RETURN (WITH SplitSting AS
(SELECT
LEFT(@String,CHARINDEX(@SplitOn,@String)-1) AS Part,RIGHT(@String,LEN(@String)-CHARINDEX(@SplitOn,@String)) AS Remainder
WHERE @String IS NOT NULL AND CHARINDEX(@SplitOn,@String)>0
UNION ALL
SELECT
LEFT(Remainder,Remainder)-1),RIGHT(Remainder,LEN(Remainder)-CHARINDEX(@SplitOn,Remainder))
FROM SplitSting
WHERE Remainder IS NOT NULL AND CHARINDEX(@SplitOn,Remainder)>0
UNION ALL
SELECT
Remainder,null
FROM SplitSting
WHERE Remainder IS NOT NULL AND CHARINDEX(@SplitOn,Remainder)=0
)
SELECT Part FROM SplitSting
)
GO (编辑:鄂州站长网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
