xsl-list
[Top] [All Lists]

RE: [xsl] Using a sequence of years to select documents to display in a table row

2007-11-23 15:03:12

   <xsl:variable name="docs" select="documents/document"/>
   <xsl:for-each select="1842 to 1862">
      <tr><td><xsl:value-of select="."/></td>
      <td>
         <xsl:apply-templates select="$docs[year = . and 
contains(title, 'Selectmen')"/>

I suspect this should be

  <xsl:apply-templates select="$docs[year = current() and contains(title,
'Selectmen')"/>

Michael Kay
http://www.saxonica.com/ 


--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--

<Prev in Thread] Current Thread [Next in Thread>