XSL copy without values is it possible? - xslt

Question

I want to compare two xmls.
1. First compare XML strucutre/schema.
2. Compare values.
I am using beyond compare tool to compare. Since these two xmls are different values, there are lot many differences in comparison report, for which I am not interested. Since, my focus now is to only compare structure/schema.
I tried to copy the xmls by following template, and other as well. But every time it is with values.
I surfed on google, xsl-copy command itself copies everything for selected node/element..
Is there any ways with which I can filter out values and only schema is copied ?
My Data :
<root>
<Child1>xxxx</Child1>
<Child2>yyy</Child2>
<Child3>
<GrandChild1>dddd<GrandChild1>
<GrandChild2>erer<GrandChild2>
</Child3>
</root>
Template used :
<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>
<!-- for all elements (tags) -->
<xsl:template match="*">
<!-- create a copy of the tag (without attributes and children) in the output -->
<xsl:copy>
<!-- For all attributes of the current tag -->
<xsl:for-each select="#*">
<xsl:sort select="name( . )" order="ascending" case-order="lower-first" />
<xsl:copy/>
</xsl:for-each>
<!-- recurse through all child tags -->
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="text()|comment()|processing-instruction()">
<xsl:copy/>
</xsl:template>
OutPut Required :
Something like..
<root>
<Child1></Child1>
<Child2></Child2>
<Child3>
<GrandChild1><GrandChild1>
<GrandChild2><GrandChild2>
</Child3>
</root>

Answer 1

At the moment, you have a template matching text() to copy it. What you need to do is remove this match from that template, and have a separate template match, that matches only non-whitespace text, and remove it.
<xsl:template match="comment()|processing-instruction()">
<xsl:copy/>
</xsl:template>
<xsl:template match="text()[normalize-space()]" />
For white-space only text (as used in indentation), these will be matched by XSLT'S built-in templates.
For attributes, use xsl:attribute to create a new attribute, without a value, rather than using xsl:copy which will copy the whole attribute.
<xsl:attribute name="{name()}" />
Note the use of Attribute Value Templates (the curly braces) to indicate the expression is to be evaluated to get the string to use.
Try this XSLT
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<!-- for all elements (tags) -->
<xsl:template match="*">
<!-- create a copy of the tag (without attributes and children) in the output -->
<xsl:copy>
<!-- For all attributes of the current tag -->
<xsl:for-each select="#*">
<xsl:sort select="name( . )" order="ascending" case-order="lower-first" />
<xsl:attribute name="{name()}" />
</xsl:for-each>
<!-- recurse through all child tags -->
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="comment()|processing-instruction()">
<xsl:copy/>
</xsl:template>
<xsl:template match="text()[normalize-space()]" />
</xsl:stylesheet>
Also note that attributes are considered to be unordered in XML, so although you have code to sort the attributes, and they probably will appear in the right order, you can't guarantee it.