如果有一个字符串 eg: “sun,star,moon,clouds”,想要在MS SQL中根据给定的分隔符’,’把这个字符串分解成各个元素[sun] [star] [moon] [clouds]，如何实现呢？为此，创建一个Function,代码如下:
复制代码 代码如下:
CREATE FUNCTION [dbo].[Split_StrByDelimiter](@String VARCHAR(8000), @Delimiter CHAR(1))
RETURNS @temptable TABLE (items VARCHAR(8000))
AS
BEGIN
DECLARE @idx INT
DECLARE @slice VARCHAR(8000)
SELECT @idx = 1
IF len(@String)<1 OR @String IS NULL RETURN
while @idx!= 0
BEGIN
SET @idx = charindex(@Delimiter,@String)
IF @idx!=0
SET @slice = LEFT(@String,@idx – 1)
ELSE
SET @slice = @String
IF(len(@slice)>0)
INSERT INTO @temptable(Items) VALUES(@slice)
SET @String = RIGHT(@String,len(@String) – @idx)
IF len(@String) = 0 break
END
RETURN
END
示例：如果输入
SELECT * FROM dbo.Split_StrByDelimiter(‘sun,star,moon,clouds’,’,’)
结果返回
sun
star
moon
clouds
在上面的代码做变形，返回有多少个元素
复制代码 代码如下:
CREATE FUNCTION [dbo].[GetCount_Split_StrByDelimiter](@String VARCHAR(8000), @Delimiter CHAR(1))
RETURNS INT
AS
BEGIN
DECLARE @temptable TABLE (items VARCHAR(8000))
DECLARE @SplitCount INT
DECLARE @idx INT
DECLARE @slice VARCHAR(8000)
SELECT @idx = 1
IF len(@String)<1 OR @String IS NULL RETURN 0
while @idx!= 0
BEGIN
SET @idx = charindex(@Delimiter,@String)
IF @idx!=0
SET @slice = LEFT(@String,@idx – 1)
ELSE
SET @slice = @String
IF(len(@slice)>0)
INSERT INTO @temptable(Items) VALUES(@slice)
SET @String = RIGHT(@String,len(@String) – @idx)
IF len(@String) = 0 break
END
SET @SplitCount=(SELECT COUNT(*) FROM @temptable)
RETURN @SplitCount
END
示例
SELECT dbo.GetCount_Split_StrByDelimiter(‘sun,star,moon,clouds’,’,’)
结果返回
4
原创文章，作者：bd101bd101，如若转载，请注明出处：https://blog.ytso.com/234294.html