Subject: Re: problems with transforming xml with xsl! sample inside
From: Joerg Heinicke <joerg.heinicke@xxxxxx>
Date: Thu, 14 Mar 2002 14:33:14 +0100
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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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