导读:1建站知识SQL 比较一个集合是否在另一个集合里存在的方法分享，需要的朋友可以参考下。网站建设哪家好网站优化seo培训。

复制代码 代码如下:

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这个函数如下： 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 ; D网seo优化趋势ECLARE @NextIndex INT ; DECLARE @ReturnText NVARCHAR(MAX) ; SELECT @CurrentInseo网站关键词优化dex = 1 ; WHILE ( @CurrentIndex <= LEN(@SplitString) ) BEGIN SELECT @NextIndex = CHARINDEX(@Separator, @SplitString, @CurrentIndex) ; IF ( @NextIndex = 0 OR @NextIndex IS NULL ) SELECT @NextIndex = LEN(@SplitStrinseo网站优化培训g) + 1 ; SELECT @ReturnText = SUBSTRING(@SplitString, @CurrentIndex, @NextIndex - @CurrentIndex) ; INSERT INTO @SplitStringsTable ( [value] ) VALUES ( @ReturnText ) ; SELECT @CurrentIndex = @NextIndex + 1 ; END RETURN ; END

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