Subject: [xsl] Identity Transformations revisited From: "Karl J. Stubsjoen" <karl@xxxxxxxxxxxxx> Date: Sat, 6 Mar 2004 10:39:08 -0700 |
Hello, Previously I had asked the question on how to identity transform the a XML that has been passed in as a parameter. There was some mixed conclusions whether you could or could not. For sure, I can treat this parameter as XML and as a result tree. Why can't I set up a for-each... and transform on each node and attribute of the result tree? Maybe I can not push an apply-templates... but I should be able to implicitly step through the source. So I have this: <xsl:apply-templates select="$c"/> A correct template match is made here: <xsl:template match="campaign"> <xsl:copy> <xsl:apply-templates select="@* | node()"/> </xsl:copy> </xsl:template> (the above doesn't seem to work...) but couldn't I: <xsl:template match="campaign"> <xsl:for-each select="*"> <xsl:copy-of select="."/> </xsl:for-each </xsl:template> (but this is where I am not sure how to appropriately write the for-each..) Any help would be appreciated. Karl XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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