Subject: Re: [xsl] Combining stylesheets for baseclass-subclass type documents From: "Daniel Brockman" <daniel.brockman@xxxxxxxxxxxxxxxx> Date: Mon, 15 Apr 2002 01:21:12 +0200 |
Alright, let's make it practical. (I realized all occurences of <c:foo> and <c:bar> were erroneous in my original message, but I've corrected them here.) Here are the stylesheets that transform the following: --- source 1 --- <?xml-stylesheet type="text/xsl" href="a.xsl"?> <a:a xmlns:a="urn:a" /> into the following: --- result 1 --- <c:c xmlns:c="urn:c"> <c:a-specific-information /> </c:c> and the following: --- source 2 --- <?xml-stylesheet type="text/xsl" href="b.xsl"?> <a:a xmlns:a="urn:a" /> into the following: --- result 2c --- <c:c xmlns:c="urn:c"> <c:b-specific-information> <c:a-specific-information /> </c:b-specific-information> </c:c> --- in a.xsl --- <xsl:template match="a:a"> <c:c> <xsl:call-template name="a:output-a-specific-information" /> </c:c> </xsl:template> <xsl:template name="a:output-a-specific-information"> <c:a-specific-information /> </xsl:template> --- in b.xsl --- <xsl:import href="a.xsl" /> <xsl:template match="a:a"> <c:c> <c:b-specific-information> <xsl:call-template name="a:output-a-specific-information" /> </c:b-specific-information> </c:c> </xsl:template> Now, changing only the two stylesheets -- a.xsl and b.xsl -- and maintaining all the restrictions, your mission is to transform source 2 into one of the following (source 1 must still transform into result 1): --- result 2a --- <c:c xmlns:c="urn:c"> <c:a-specific-information /> <c:b-specific-information /> </c:c> or the following: --- result 2b --- <c:c xmlns:c="urn:c"> <c:a-specific-information> <c:b-specific-information /> </c:a-specific-information> </c:c> Thanks in advance, - Daniel Brockman mailto:daniel.brockman@xxxxxxxxxxxxxxxx "To spot the expert, pick the one who predicts the job will take the longest and cost the most." - Murphy XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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