我有一个名为“Documents”的表,其中包含如下列:
DocumentID
我的数据格式为@DocID = 1,2,3,4
如何使用单个查询在单独的行中插入这些documentID?
答案 0 :(得分:6)
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。这就是如何使用拆分功能:
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,请参阅上一个链接,该链接解释了每个链接的PRO和CON。
要使Numbers Table方法起作用,您需要进行一次性表设置,这将创建一个包含1到10,000行的表Numbers
:
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
输出:
Value
----------------------
1
22
333
4444
(4 row(s) affected)
让你新表使用它:
--set up tables:
DECLARE @Documents table (DocumentID varchar(500), SomeValue varchar(5))
INSERT @Documents VALUES ('1,2,3,4','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
输出:
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,则可以使用此拆分功能(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,CHARINDEX(@SplitOn,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
答案 1 :(得分:3)
+1。这将使工作快速完成,但可能不一定最有效(再次参见KM对所有选项的响应)
我的快速回复:
安装SQL#(免费且非常有用)
然后
INSERT INTO Documents (documentId)
SELECT SplitVal FROM SQL#.String_Split(@DocId, ',', 1)