get all children nodes

maratg · November 5th, 2003, 07:33 PM

I am trying to get the names of all chidren nodes, however I only get some of them. Please, help.

here is my sample XML file :

<?xml version="1.0" encoding="ISO-8859-1"?>
<?xml-stylesheet type="text/xsl" href="testtransform.xsl"?>
<Object>
       <Test>
        <Node>
            <case1>it </case1>
            <case2>works </case2>
        </Node>
    </Test>
    <Test>
        <Node>
            <case3>here</case3>
            <case4> we go</case4>
        </Node>
    </Test>
</Object>

Here is my XSL file:

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="//Node">

<xsl:value-of select = "name(descendant::*)"/>

</xsl:template>
</xsl:stylesheet>

Here is the output:

case1case3

Question:
Why isn't XSL returning case2 and case 4 which are the other nodes in the XML file?

Thanks

armmarti · November 6th, 2003, 02:06 AM

First of all, change "//Node" to "Node" in the match attribute of the xsl:template.
Second, you didn't get "case2" and "case4" since the "name" function returns the element name of the first node in the argument node-set. You can use xsl:for-each to get all element names in the node set.

Regards,
Armen

maratg · November 6th, 2003, 01:07 PM

Armen,
what should be the select expression in the xsl:for-each? When I use
xsl:for-each select="Node/*", nothing is returned in the output.
Thanks,

Quote:

quote:Originally posted by armmarti
First of all, change "//Node" to "Node" in the match attribute of the xsl:template.
Second, you didn't get "case2" and "case4" since the "name" function returns the element name of the first node in the argument node-set. You can use xsl:for-each to get all element names in the node set.

Regards,
Armen

armmarti · November 7th, 2003, 11:27 AM

Code:

<xsl:template match="Node">
  <xsl:for-each select="descendant::*">

     <xsl:value-of select="name()"/>
     <xsl:if test="position() != last()">,</xsl:if>
  </xsl:for-each>
</xsl:template>

Regards,
Armen

maratg · November 7th, 2003, 02:07 PM

Armen,

Thank you. That worked.