Cart
|
[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] document tree fragments
 XSLT 1.0 (i.e. Firefox 3.0, etc.) XSLT 2.0 solutions not suitable. I have a working xsl stylesheet that transforms an xml file with the structure: < wines > < wine >...< /wine >+ < /wines > I want, now, to rewrite my xsl stylesheet to work with a collection of xml files, each of which contain an individual < wine > element--i.e. in place of the single xml < wines > file. I have succeeded, but with two problems. Firstly, I am unable to separately specify the directory path of the files. Here is my "base" xml file's content: <docs path="jb"> <doc filename="jb/wine1.xml" /> <doc filename="jb/wine2.xml" /> <doc filename="jb/wine3.xml" /> ... </docs> Note that I have provided the directory path within the @filename attribute, but I would prefer to retrieve it from the @path attribute. Here is the part of my xsl stylesheet that retrieves the contents of my collection of xml files: <xsl:template match="/docs"> <xsl:variable name="thePath" select="@path" /> <xsl:variable name="theWines" select="document(doc/@filename)/wine" /> ... Is there a way to use my $thePath variable, instead of including the directory path inside the @filename attribute? I tried to concatenate, but without success. I want to deal with all my xml files together, rather than using a for-each loop, because of several aggregate processes that I want to apply. Secondly, I'm not sure I fully understand what my $theWines variable now contains. I believe it is a collection of tree fragments--i.e. there's no single top-most node, is that right? Or is there an implicit / (document) node? The reason I ask is because I have some templates that rely on processing the preceding axis, and whereas my templates work fine for my < wines > xml file, they do not produce the anticipated results when using my $theWines variable. This template works with the < wines > xml file, and is called by another template that has already matched the "wines" node: <xsl:template name="winesByCountry"> <table border="1"> <xsl:for-each select="wine/countries/country[not(name=../../preceding-sibling::wine/countries/country/name)]"> <xsl:sort select="name" order="ascending" /> <xsl:variable name="countryName" select="name" /> <tr><th><xsl:value-of select="$countryName" /></th><td> <xsl:value-of select="count(/wines/wine/countries/country[name = $countryName])" /></td></tr> </xsl:for-each> </table> </xsl:template> The following template does not work. It outputs each country multiple times--e.g. 2 wines from Italy are output twice as: Italy 2 Italy 2 rather than the second time correctly seeing that that country is already present in a preceding node: <xsl:template name="winesByCountry"> <xsl:param name="theWines" /> <table border="1"> <xsl:for-each select="$theWines/countries/country[not(name=preceding::country/name)]"> <xsl:sort select="name" order="ascending" /> <xsl:variable name="countryName" select="name" /> <tr><th><xsl:value-of select="$countryName" /></th><td> <xsl:value-of select="count($theWines/countries/country[name=$countryName])" /></td></tr> </xsl:for-each> </table> </xsl:template> Do I need to wrap the sequence of tree fragments in my $theWines variable inside a node? I tried using the following: <xsl:variable name="theWines"> <wines> <xsl:value-of select="document(doc/@filename)/wine" /> </wines> </xsl:variable> but Firefox says: Error during XSLT transformation: An XPath expression was expected to return a NodeSet. Sorry for the length. Any suggestions? Cheers! Joe 
|
PURCHASE STYLUS STUDIO ONLINE TODAY!Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced! 
Download The World's Best XML IDE!Accelerate XML development with our award-winning XML IDE - Download a free trial today! Subscribe in XML format
|
