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I have two xml files with exactly similar hierarchy, but for a few different element nodes. Something like the following:
I have a variable $path that holds an element's (in File1) xpath as if it was assinged by one the following:
<xsl:variable name="path" select="'element/child/a'"/>
How do I obtain and copy the node <adesc> of File2 using the above $path variable? Something like the below one:
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
[xsl] variable in xpath
Subject: [xsl] variable in xpath From: "Kalyan Kumar Mudumbai" <kalyan_tech@xxxxxxxxxxxxxx> Date: 15 Nov 2002 05:33:36 -0000 |
I have two xml files with exactly similar hierarchy, but for a few different element nodes. Something like the following:
File1.xml <element> <child> <a> <b> </child> </element>
File2.xml <element> <child> <a> <adesc> </a> </child> </element>
I have a variable $path that holds an element's (in File1) xpath as if it was assinged by one the following:
<xsl:variable name="path" select="'element/child/a'"/>
How do I obtain and copy the node <adesc> of File2 using the above $path variable? Something like the below one:
<xsl:template match="node()"> <xsl:copy-of select="document(File2.xml)/$path.."/> </xsl:template>
Output has to be smth like this: <element> <child> <a> <adesc> </a> <b> </child> </element>
Thanks, Kalyan
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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