Subject: Re: [xsl] Using copy to change a node, whilst retaining the attributes From: David Carlisle <davidc@xxxxxxxxx> Date: Tue, 5 Aug 2003 09:47:47 +0100 |
> Tried the below but this does not copy across the attributes. T that's because you removed the <xsl:copy-of select="@*"/> line that were there before. xsl:copy just generates an element node with the name of the current element you need to copy all attributes and children if you need them copied. So you could just add them but... <xsl:template match="player/dates/yes"> <xsl:copy> <xsl:choose> <xsl:when test="ancestor::player/@pword=$pmkey"> there's not really a lot of point in going down and then coming back with ancester to test, you may as well test on the way down.. You have two templates like this: <xsl:template match="player/dates/yes/@ID"> but nowhere are you applying templates to attribute nodes so these will never activate. You don't want to apply templates to attributes anyway, you want to copy them as above. David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star Internet. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] Using copy to change a no, John Reid | Thread | RE: [xsl] Using copy to change a no, Américo Albuquerque |
Re: AW: [xsl] Using key() from outs, David Carlisle | Date | RE: [xsl] Using copy to change a no, Américo Albuquerque |
Month |