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RE: xsl to echo input xml
- To: "'xsl-list at lists dot mulberrytech dot com'" <xsl-list at lists dot mulberrytech dot com>
- Subject: RE: [xsl] xsl to echo input xml
- From: "Kearney, Bryan" <bkearney at solant dot com>
- Date: Thu, 25 Jan 2001 11:49:19 -0700
- Reply-To: xsl-list at lists dot mulberrytech dot com
You want the identity transformation:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<!-- Match every node whether it is an element or an attribute -->
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/">
<xsl:apply-templates select="node()|@*"/>
</xsl:template>
</xsl:stylesheet>
>> -----Original Message-----
>> From: Michael Kay [mailto:mhkay@iclway.co.uk]
>> Sent: Thursday, January 25, 2001 2:58 AM
>> To: xsl-list@lists.mulberrytech.com
>> Subject: RE: [xsl] xsl to echo input xml
>>
>>
>> > If I want to pass an xml to an xsl, and then have the xsl
>> > simply repeat
>> > what's in the xml, how would I go about that?
>>
>> <xsl:copy-of select="*"/>
>>
>> > I thought the following would do it:
>> >
>> > <xsl:template match="*">
>> > <xsl:apply-templates/>
>> > </xsl:template>
>> >
>> No, this will copy the text nodes to the output but not the
>> element nodes.
>> I.E, the output will contain no tags.
>>
>> Mike Kay
>>
>>
>> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
>>
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