Re: Linking to a 6 level tag content

   <xsl:template match="/">
       <xsl:value-of select="office:document-content/office:body
           /office:spreadsheet/table:table/table:table-row/table:table-cell
           /text:p"
           />
   </xsl:template>
</xsl:stylesheet>

<?xml version="1.0" encoding="UTF-8"?>
23

From: Kahlil Johnson <jzarecta@xxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re:  Linking to a 6 level tag content
Date: Wed, 22 Jun 2005 23:54:35 +0300

Sorry I meant 23.. I didn't want to put the whole tree just the
relevant tags what I want is to know how exactly to get to the 23
value.

On 6/22/05, Aron Bock <aronbock@xxxxxxxxxxx> wrote:
> Khalil,
>
> I don't see a "test:p > 21" in your input. The closest is
> <text:p>23</text:p>
>
> If you could post a more complete snippet, with namespace decrarations and
> end-tags, and specify if you want top copy the element or get its text
> value, I assume people could respond better.
>
> Regards,
>
> --A
>
> >From: Kahlil Johnson <jzarecta@xxxxxxxxx>
> >Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> >To: XSL <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
> >Subject: Linking to a 6 level tag content
> >Date: Wed, 22 Jun 2005 13:03:32 +0300
> >
> >I have this xml and I want to capture the content test:p > 21 from
> >the XSLT however I have problems targeting the branch:
> >
> >XML: output
> ><office:document-content office:version="1.0">
> > <office:body>
> > <office:spreadsheet>
> > <table:table >
> > <table:table-column />
> > <table:table-row table:style-name="ro1">
> > <table:table-cell office:value-type="float"
> >office:value="23">
> > <text:p>23</text:p>
> >
> >I am using xsltproc to generate an output.