Kirk
Markus Heussen wrote:
Ah now I understand your concern. Use <xsl:copy-of select="content"/> or <xsl:copy-of select="content/*"/>.
Markus
-----Original Message----- From: Kirk Storer [mailto:[EMAIL PROTECTED] Sent: Thursday, March 04, 2004 9:00 PM To: [EMAIL PROTECTED] Subject: Re: Easy way to include xhtml when transforming xml to xhtml
Ok, I will try to explain this better. I am using two files as sources, combined via aggregate in the sitemap. The content of one of the files will be contained in <menu></menu> The content of the other file will be contained in <content></content> Both of these are then inside of <page> </page>
I want to process the xml within <menu> into xhtml. I merely want to use what is in the <content> tags as it is, it is already xhtml.
this is an extremely simplified version of the xsl file ---------------------------------------------- <xsl:template match="page"> <xsl:for-each select="menu"> do whatever </xsl:for-each>
<xsl:value-of select="content"/> </xsl:template> ----------------------------------------------- If i use <xsl:value-of select="content"/> it displays the text, but not all of the markup tags
How can it get it to include all of the markup tags that are in the file, such as <p> </p> <ul><li></li></ul> etc.
Thanks, Kirk
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