Subject: RE: [xsl] Delete XML Node From: "KIENLE, STEVEN C [IT/0200]" <steven.c.kienle@xxxxxxxxxxxxx> Date: Thu, 31 Oct 2002 14:27:42 -0500 |
Deepak, In case what you really want to do is output Y if and only if it has a Z element node, the following transform will work: <?xml version="1.0"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/X"> <X> <xsl:for-each select="A"> <A> <xsl:value-of select="."/> </A> </xsl:for-each> <xsl:for-each select="Y[Z]"> <Y> <xsl:for-each select="Z"> <Z> <xsl:value-of select="."/> </Z> </xsl:for-each> </Y> </xsl:for-each> </X> </xsl:template> </xsl:stylesheet> This key is to use the XPath selection of "Y[Z]" which means all Y which have a Z element. With the input XML of: <X> <A>A</A> <Y></Y> </X> The output will be: <X> <A>A</A> </X> With the input XML of: <X> <A>A</A> <Y> <Z>B</Z> </Y> </X> The output will be: <X> <A>A</A> <Y> <Z>B</Z> </Y> </X> I hope this may help. Steve XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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