sam wrote:
> I need to know a way for knowing tag position in a xml like this:
>
> <AAA>
> <BBB>
> <CCC>
> <DDD_1/>
> <DDD_2/>
> <DDD_3/>
> </CCC>
> <BBB/>
> </AAA>
> I need to get that AAA = 1, BBB=2,CCC=3,
> DDD_1=4, DDD_2=4,DDD_3=4

> Martin Honnen
> count($n/ancestor::node()) where $n is your node. So with a stylesheet

> sam wrote:
>> I need to know a way for knowing tag position in a xml like this:
>>
>> <AAA>
>> <BBB>
>> <CCC>
>> <DDD_1/>
>> <DDD_2/>
>> <DDD_3/>
>> </CCC>
>> <BBB/>
>> </AAA>
>> I need to get that AAA = 1, BBB=2,CCC=3,
>> DDD_1=4, DDD_2=4,DDD_3=4
>
> count($n/ancestor::node()) where $n is your node. So with a stylesheet you
> could do:
>
> <xsl:stylesheet
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
> version="1.0">
>
> <xsl:output method="text"/>
>
> <xsl:template match="/">
> <xsl:apply-templates select="//*"/>
> </xsl:template>
>
> <xsl:template match="*">
> <xsl:value-of select="concat(name(), ' = ',
> count(./ancestor::node()))"/>
> <xsl:if test="position() != last()"><xsl:text>, </xsl:text></xsl:if>
> </xsl:template>
>
> </xsl:stylesheet>
>
>
> --
>
> Martin Honnen --- MVP XML
> http://JavaScript.FAQTs.com/

sam wrote:
>> Martin Honnen
>> count($n/ancestor::node()) where $n is your node. So with a stylesheet
> Okey, but I am using XML .Net control for handling xml files.
> So, how to get this value in csharp code ?