I guess, a simple XSL like written below(its complete
in all respects) -- might serve the purpose.
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html" version="1.0"
encoding="UTF-8" indent="yes"/>
<xsl:template match="/ads/ad">
<ads>
<ad>
<hello>world</hello>
<htmlcontent>
<xsl:apply-templates select="paragraph"/>
</htmlcontent>
</ad>
</ads>
</xsl:template>
<xsl:template match="paragraph">
<p><xsl:apply-templates/></p>
</xsl:template>
<xsl:template match="bold">
<b><xsl:apply-templates/></b>
</xsl:template>
</xsl:stylesheet>
*I have'nt used CDATA* for the sake of simplicity. The
output requirements are being met, without using CDATA
section.
Please see Mike's answer also. He has given excellent
explanation, to the problem.
Regards,
Mukul
--- Thomas Rasmussen <tbr77(_at_)post(_dot_)tele(_dot_)dk> wrote:
Julian Reschke wrote:
Thomas Rasmussen wrote:
Hello
I am trying to make a xsl document that converts
one xml document
into another, but have run into some trouble.
I have this document:
<ads>
<ad>
<hello>world</hello>
<paragraph>Some <bold>html</bold> formatted
text</paragraph>
<paragraph>Another line of
<bold>html</bold></paragraph>
</ad>
</ads>
And I want this out:
<ads>
<ad>
<hello>world</hello>
<htmlcontent>
<![CDATA[
<p>Som Some <b>html</b> formatted text<p>
<p>Another line of <b>html</b></p>
]]>
</htmlcontent>
</ad>
</ads>
*Why* do you want that output? In general, it's
better to keep
wellformed XHTML content as XML instead of plain
text (escaped markup).
...
I agree, but it's for an external company and theire
requirements says
it has to be CDATA.
But it is wellformed XHTML so I can't really see a
reason either?
Maybe it's because they import from many different
companies and some og
those can't suplly wellformed XHTML.
XSL-List info and archive:
http://www.mulberrytech.com/xsl/xsl-list
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