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Re: [xsl] An (almost) identity stylesheet

2008-04-10 01:46:27
think it should be like this you're looking for.


<xsl:template match="*">
<xsl:copy><xsl:apply-templates select="@*"/><xsl:apply-templates/></xsl:copy>
</xsl:template>
<xsl:template match="@*">
<xsl:copy/>
</xsl:template>
<xsl:template match="@attrib2">
<xsl:attribute name="@attrib2">0</xsl:attribute>
</xsl:template>

cheers,
Bryan Rasmussen
On Thu, Apr 10, 2008 at 10:40 AM, Karol Krenski 
<mimooh(_at_)inf(_dot_)sgsp(_dot_)edu(_dot_)pl> wrote:
Hi list,

 I am a novice to XSLT, learned mainly form online tutorials. This XSLT
 thing is killing me. Can't figure out how to transform my doc.xml:

 <?xml version="1.0" encoding="UTF-8"?>
 <foo>
    <bar>
        <date attrib1='aaaaaaa' attrib2='bbbbbb'>25-02-08</date>
                                         ^
        <name attrib3='null'>AAAAAAAAAAAAAAA</name>
    </bar>
    <bar>
        <date attrib1='cccccc' attrib2='dddddd'>26-02-08</date>
                                        ^
        <name attrib4='null'>BBBBBBBBBBBBBBB</name>
    </bar>
 </foo>


 into


 <?xml version="1.0" encoding="UTF-8"?>
 <foo>
    <bar>
        <date attrib1='aaaaaaa' attrib2='0'>25-02-08</date>
                                         ^
        <name attrib3='null'>AAAAAAAAAAAAAAA</name>
    </bar>
    <bar>
        <date attrib1='cccccc' attrib2='0'>26-02-08</date>
                                        ^
        <name attrib4='null'>BBBBBBBBBBBBBBB</name>
    </bar>
 </foo>

 Can somebody please suggest the correct stylesheet?

 I am blindly typing mainly 'copy' or 'copy-of' in my templates and the
 output generation mechanics make little sense to me. The docs, comparing
 to other programing documentation seems hard to understand. Could
 somebody suggest good docs, please?

 Regards,
 Karol

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