匿名通过本文主要向大家介绍了集合等相关知识,希望本文的分享对您有所帮助
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。
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 ;
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