[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: Can anyone point me in the right direction?
> -----Original Message----- > From: Patricia LaRue > > "How do I display only that worksheet section of the > XML file?" > So (guessing a bit more) the source XML has subelements on the worksheet nodes (--correct?): Very simplified typical example <root> <worksheet name="ws_name1"> <row id="..."> <column></column> <!-- more columns --> </row> <!-- more rows --> </worksheet> <!-- more worksheets --> </root> But then I'm losing it a bit... You need to display the subelements of the worksheet nodes below the hyperlink layed-out in an HTML table? If so, to transform the above into a very simple HTML table placed below the hyperlink, using three templates: <xsl:template match="worksheet"> <a> <xsl:attribute name="href"> <xsl:value-of select="@name" /> </xsl:attribute> </a> <table> <xsl:apply-templates select="row" /> </table> </xsl:template> <xsl:template match="row"> <tr><xsl:apply-templates select="column" /></tr> </xsl:template> <xsl:template match="column"> <td><xsl:value-of select="." /></td> </xsl:template> > So far, everything I've read says that I need XPath, > XLink, XPointer, etc. I've been reading for days on > this and I'm still confused about how these work > together or even if they work in the current version. > Can anyone tell me whether I should be looking at > XPath and XSLT? or XLink and XPointer? or should I > break up the XML file into seperate worksheet and XSLT > files? I'm just looking for someone to point me in > the right direction and I'm greatfull for any > suggestions. Depends on what exactly you want to do --which you still haven't told us, so it's difficult to tell... ;) Anyway, if the above is sufficient, then XSLT/XPath is definitely *the* way to go here. Cheerz, Andreas XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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