I searched for this all over and couldn't find a solution. I
want to take the following XML:
<root>
<file>
<element/>
</file>
<file id="a">
<element id="a"/>
</file>
<file class="b">
<element class="b"/>
</file>
<file id="c" class="d" not="1" this="2">
<element id="c" class="d" not="1" this="2"/>
</file>
</root>
#### and output something this ####
root
root file
root file element
root file #a
root file #a element #a
root file .b
root file .b element .b
root file #c .d
root file #c .d element #c .d
XSLT 1.0 or 2.0? Here's a 2.0 solution, untested:
<xsl:template match="*">
<xsl:apply-templates select="ancestor-or-self::*/(.|@class|@id)"
mode="token"/>
<xsl:text>
</xsl:text>
</xsl:template>
<xsl:template match="*" mode="token">
<xsl:value-of select="concat(name(), ' ')"/>
</xsl:template>
<xsl:template match="@class" mode="token">
<xsl:value-of select="concat('.', ., ' ')"/>
</xsl:template>
<xsl:template match="@id" mode="token">
<xsl:value-of select="concat('#', ., ' ')"/>
</xsl:template>
Michael Kay
http://www.saxonica.com/
#### ignoring any attributes that are not "id" or "class" and
containing all ancestor elements and id's and classes no
matter what the element name is or how many levels deep. I'd
like to use this for auto generating CSS declarations from
XHTML snipits, I can deal with removing redundant rows
manually. Below is the closest I've gotten which only strips
the XML to the elements and attributes I'm interested in, but
doesn't format it or list ancestor elements, id's and classes
(I'm a beginner at XSLT) thanks! ####
<xsl:template match="node()|@id|@class"> <xsl:copy>
<xsl:apply-templates select="node()|@id|@class"/> </xsl:copy>
</xsl:template>
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