Re: [xsl] XML subset selection-a simpler way? (message 27746)

["Alan Varga" <aevarga@xxxxxx>]

If I have declared a processor and I want to pass parameters collected from 
a form in an HTML page to my XSL stylesheet, isn't it as simple as the 
example below?

~~~~~ start of Javascript snippet ~~~~~
var sCalType = checkRadio(theForm.grpOptions);
var sSelection = theForm.lstCalendar.value;
var xslStylesheet = "test.xsl";
...
var xslDoc = Sarissa.getDomDocument();
xslDoc.async = false;
xslDoc.load(xslStylesheet);

var processor = new XSLTProcessor();
processor.importStylesheet(xslDoc);
processor.setParameter(null, "type", sCalType);
processor.setParameter(null, "selection", sSelection);
...
function checkRadio(group){
 for (x=0; x<group.length; x++) {
   if (group[x].checked == true) {
     val = group[x].value;
//      alert(val);
   }
 }
 return val;
}
~~~~~ end of Javascript snippet ~~~~~


~~~~~ start of XSL snippet ~~~~~
<xsl:param name="type"/>
<xsl:param name="selection"/>
...
<xsl:template match="Events">
<tbody>
 <xsl:choose>
   <xsl:when test="$type='Month'">
     <xsl:apply-templates select="Event[MonthName=$selection]" />
   </xsl:when>
   <xsl:when test="$type='Band'">
     <xsl:apply-templates select="Event[Band=$selection]" />
   </xsl:when>
 </xsl:choose>
</tbody>
</xsl:template>
...
~~~~~ end of XSL snippet ~~~~~

If I use alert(processor.getParameter(null, "type")) the parameter value is 
null, and the transformation fails because nothing was passed into the XSL 
stylesheet.