看不太懂一个xml函数,请看懂的朋友告诉下
-- Return the value of the element by given the name of the element

CREATE FUNCTION dbo.fnXmlParserGetValueByName (@XmlString nvarchar(4000) , @Tag nvarchar(100))
RETURNS nvarchar(4000) AS
BEGIN
DECLARE @TagStart nvarchar(102) , @TagEnd nvarchar(103)
SET @TagStart = ' < ' + @Tag + '> '
SET @TagEnd = ' </ ' + @Tag + '> '

DECLARE @StartNo int, @EndNo int
SET @StartNo = CHARINDEX(@TagStart, @XmlString)
SET @EndNo = CHARINDEX(@TagEnd, @XmlString)

IF @StartNo = 0 OR @EndNo = 0
RETURN NULL

DECLARE @Value nvarchar(4000)
SET @Value = SUBSTRING(@XmlString, @StartNo + LEN(@TagStart) , @EndNo - @StartNo - LEN(@TagStart) )

RETURN @Value
END


测试代码:
declare @xmlstring nvarchar(4000)

DECLARE @TagStart nvarchar(102) , @TagEnd nvarchar(103),@Tag nvarchar(105),@StartNo nvarchar(10),@EndNo nvarchar (20)

set @xmlstring = ' <grp> Empty </grp> <params> <param> <name> RowNum </name> <CorpPhone> 52 </CorpPhone> </param> </params> <hgt> 85 </hgt> <description> 3 </description> 公司 '
set @tag = 'CorpPhone '
select @TagStart = ' < ' + @Tag + '> '
select @TagEnd = ' </ ' + @Tag + '> '
select @StartNo = CHARINDEX(@TagStart, @XmlString)
select @EndNo = CHARINDEX(@TagEnd, @XmlString)

select @TagStart
select @TagEnd
select len( 'CorpPhone ')
select @EndNo




[解决办法]
功能非常简单，就是返回xml里边一个元素的值.如> <CorpPhone> 52 </CorpPhone> 的值是52则

declare @xmlstring nvarchar(4000)

DECLARE @TagStart nvarchar(102),@Tag nvarchar(105)

set @xmlstring = ' <grp> Empty </grp> <params> <param> <name> RowNum </name> <CorpPhone> 52 </CorpPhone> </param> </params> <hgt> 85 </hgt> <description> 3 </description> 公司 '
set @tag = 'CorpPhone '
select dbo.fnXmlParserGetValueByName(@xmlstring,@tag)

返回的结果就是52



[解决办法]
有时间帮你看一下
[解决办法]
hehe,学习再学习。