SQL 比擬一個聚集能否在另外一個聚集裡存在的辦法分享。本站提示廣大學習愛好者：（SQL 比擬一個聚集能否在另外一個聚集裡存在的辦法分享）文章只能為提供參考，不一定能成為您想要的結果。以下是SQL 比擬一個聚集能否在另外一個聚集裡存在的辦法分享正文
DECLARE @c INT
DECLARE @c2 INT
SELECT @c = COUNT(1)
FROM dbo.SplitToTable('1|2|3|4', '|')
SELECT @c2=COUNT(1)
FROM dbo.SplitToTable('1|2|3|4', '|') a
INNER JOIN dbo.SplitToTable('1|2|3|', '|') b ON a.value = b.value
IF @c = @c2
SELECT 'ok'
ELSE
SELECT 'no'
SplitToTable這個函數以下：
set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
go
ALTER FUNCTION [dbo].[SplitToTable]
(
@SplitString NVARCHAR(MAX) ,
@Separator NVARCHAR(10) = ' '
)
RETURNS @SplitStringsTable TABLE
(
[id] INT IDENTITY(1, 1) ,
[value] NVARCHAR(MAX)
)
AS
BEGIN
DECLARE @CurrentIndex INT ;
DECLARE @NextIndex INT ;
DECLARE @ReturnText NVARCHAR(MAX) ;
SELECT @CurrentIndex = 1 ;
WHILE ( @CurrentIndex <= LEN(@SplitString) )
BEGIN
SELECT @NextIndex = CHARINDEX(@Separator, @SplitString,
@CurrentIndex) ;
IF ( @NextIndex = 0
OR @NextIndex IS NULL
)
SELECT @NextIndex = LEN(@SplitString) + 1 ;
SELECT @ReturnText = SUBSTRING(@SplitString,
@CurrentIndex,
@NextIndex - @CurrentIndex) ;
INSERT INTO @SplitStringsTable
( [value] )
VALUES ( @ReturnText ) ;
SELECT @CurrentIndex = @NextIndex + 1 ;
END
RETURN ;
END
