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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: can you select name() of attributes?
 I got somewhat the same solution, but then I run into a different problem, which is why I want to use it as a nodeset. The ones in the input document (let us say $ Doc1) must be in $Doc2, but if they are not in $Doc1, then the (default) value in $Doc2 should be used. <xsl:variable name="Attr1" select="$Doc1/@*"/> <xsl:variable name="Attr2" select="$Doc2/@*"/> <xsl:for-each select="$Attr2"> <xsl:variable name="Attr" select="."/> <xsl:if test="not($Attr1)"> <xsl:attribute name="{name($Attr)}"><xsl:value-of select="$Attr"/></xsl:attribute> </xsl:if> <xsl:for-each select="$Attr1[name()=name($Attr)]"> <xsl:choose> <xsl:when test="name()=name($Attr)"> <xsl:attribute name="{name(.)}"><xsl:value-of select="."/></xsl:attribute> </xsl:when> <xsl:otherwise> <xsl:attribute name="{name($Attr)}"><xsl:value-of select="$Attr"/></xsl:attribute> </xsl:otherwise> </xsl:choose> </xsl:for-each> </xsl:for-each> This works if there are no attributes in $Doc1 (if-sentence takes care of that) but if some of the attributes are in $Doc1, then the attributes - by nature of being nested in two for-each loops - will be overwritten (same name), ending in the last iteration creating the output. Is there something else I can do? Unfortunately I do not have access to a version which uses XSLT2, and I do not know if MK's solution would work in my case anyway (same problem as above). Thank you for your time Ragulf Pickaxe :-) _________________________________________________________________ STOP MORE SPAM with the new MSN 8 and get 2 months FREE* http://join.msn.com/?page=features/junkmail 
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