Re: [xsl] problems with transforming xml with xsl! sample inside

Subject: Re: [xsl] problems with transforming xml with xsl! sample inside
From: Joerg Heinicke <joerg.heinicke@xxxxxx>
Date: Thu, 14 Mar 2002 14:33:14 +0100
on a validated XML (with DTD):

<xsl:template match="paragraph">
    <xsl:apply-templates/>
</xsl:template>

<xsl:template match="linkRef">
    <a href="id(@id)/@value"><xsl:value-of select="id(@id)/@name"/></a>
</xsl:template>

with keys:

<xsl:key name="links" match="link" use="@id"/>

<a href="key('links',@id)/@value"><xsl:value-of select="key('links',@id)/@name"/></a>

"pure XPATH":

<a href="/doc/links/link[@id=current()/@id]/@value"><xsl:value-of select="/doc/links/link[@id=current()/@id]/@name"/></a>

Regards,

Joerg

Jan Krattiger wrote:
i have the following xml:

<doc>
    <links>
      <link id="lr1" name="xyz" value="xyz.com" />
      <link id="lr2" name="abc" value="abc.com" />
    </links>
    <paragraphs>
      <paragraph title="foo">
      some text here, some text here <linkRef id="lr1"/> another text,
another text <linkRef id="lr2"/> and even more text.
      </paragraph>
    </paragraphs>
</doc>

the result should look like this:

foo
some text here, some text here xyz.com another text, another text abc.com
and even more text.


the goal is to transform this xml with a xsl. now the problem is, that i'm
not able to replace the <linkRef> with the specfied link. i tried to do it
within xsl and tried to do it with dom in asp. i don't know how i can
realize that. anyone knows a solution? thanks for any help.
i heard something about transorming it twice.. but i dunno how to do it.


J.M.K

Software Developer

Unit.Net AG
Thurgauerstrasse 54
CH - 8050 Zurich

Email: jan.krattiger@xxxxxxxx Web: http://unit.net


--

System Development
VIRBUS AG
Fon  +49(0)341-979-7411
Fax  +49(0)341-979-7409
joerg.heinicke@xxxxxxxxx
www.virbus.de


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