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HomeDatabaseMysql Tutorial在SQL查询中使用LIKE来代替IN查询的方法

在SQL查询中根据已知ID的集合来查询结果我们通常会用到IN,直接在IN后面给出ID的集合或是在IN后面跟一个子查询。

如下:
代码如下:
SELECT * FROM Orders
WHERE OrderGUID IN('BC71D821-9E25-47DA-BF5E-009822A3FC1D','F2212304-51D4-42C9-AD35-5586A822258E')

可以看出直接在IN后面跟ID的集合需要将每一个ID都用单引号引起来。在实际应用中会遇到这么一种情况,在界面中收集的是一串GUID的拼接字符串,中间以逗号隔开,如果作为参数传到一个存储过程中执行,最终生成的语句会是下面这样:
代码如下:
SELECT * FROM Orders
WHERE OrderGUID IN('BC71D821-9E25-47DA-BF5E-009822A3FC1D,F2212304-51D4-42C9-AD35-5586A822258E')

这样就不能查询到正确的结果。

一般情况下我们解决此问题的思路是将传入的字符串用一个split函数来处理,最终处理的结果是一张表,然后将这个表做自查询即可,如下:
代码如下:
DECLARE @IDs VARCHAR(4000)
SET @IDs='BC71D821-9E25-47DA-BF5E-009822A3FC1D,F2212304-51D4-42C9-AD35-5586A822258E'
DECLARE @temp TABLE(str VARCHAR(50))
INSERT INTO @temp
SELECT * FROM dbo.Split(@IDs,',')
SELECT * FROM Orders WHERE OrderGUID IN (SELECT str FROM @temp)

当然split函数系统比不提供,需要我们自己写:
代码如下:
CREATE FUNCTION Split
(
@SourceSql varchar(8000),
@StrSeprate varchar(10)
)
RETURNS @temp TABLE(F1 VARCHAR(100))
AS
BEGIN
DECLARE @i INT
SET @SourceSql=rtrim(ltrim(@SourceSql))
SET @i=charindex(@StrSeprate,@SourceSql)
WHILE @i>=1
BEGIN
INSERT @temp VALUES(left(@SourceSql,@i-1))
SET @SourceSql=substring(@SourceSql,@i+1,len(@SourceSql)-@i)
SET @i=charindex(@StrSeprate,@SourceSql)
END
IF @SourceSql''
INSERT @temp VALUES(@SourceSql)
RETURN
END

像这样做非常麻烦,而且还需要借助函数来实现,下面介绍一种简单的方法,因为GUID是唯一的,所以在上面的例子中可以使用LIKE来代替IN也可以达到同样的查询效果:
代码如下:
SELECT * FROM Orders
WHERE 'BC71D821-9E25-47DA-BF5E-009822A3FC1D,F2212304-51D4-42C9-AD35-5586A822258E'
LIKE '%'+convert(VARCHAR(40),OrderGUID)+'%'
Statement
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