Simon,

No need for loop wrap the whole thing in xsl:choose statement do a test there namely if count(umbraco.library:Split(data [@alias = 'location'],',')/value)>0 in the otherwise clause show text no results found.

Regards

Isamil

Sorry buddy, im new to umbraco and im basicaly editing the code that our web developers created for us. Could you explain that in a little more detail? I only have a basic knowledge of XSLT from tweaking bits here and there.

Regards,

Simon

Simon,

Can you paste your xslt in full that will give better idea of what you are trying todo.

Regards

Ismail

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:Stylesheet [ <!ENTITY nbsp "&#x00A0;"> ]>
<xsl:stylesheet
 version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:msxml="urn:schemas-microsoft-com:xslt"
 xmlns:umbraco.library="urn:umbraco.library"
 exclude-result-prefixes="msxml umbraco.library">

<xsl:output method="xml" omit-xml-declaration="yes" indent="yes"/>

<xsl:param name="currentPage"/>
<xsl:variable name="source" select="/macro/source/node"/>

<!-- Input the documenttype you want here -->
<xsl:variable name="documentTypeAlias" select="string('Job')"/>
<xsl:variable name="numberOfItems" select="/macro/numberOfItems"/>
<xsl:template match="/">

<xsl:choose>
   <xsl:when test="umbraco.library:RequestQueryString('careerslist')!=''">
 <xsl:choose>
  <xsl:when test="$source/node[@nodeTypeAlias=$documentTypeAlias and contains(data[@alias='location'],umbraco.library:RequestQueryString('careerslist'))]">
   <!-- use year date filter -->
   <xsl:apply-templates select="$source/node[@nodeTypeAlias=$documentTypeAlias and contains(data[@alias='location'],umbraco.library:RequestQueryString('careerslist'))]"> 
    <xsl:sort select="@createDate" data-type = "number" order="descending"/>
    </xsl:apply-templates>
  </xsl:when>
  <xsl:otherwise>
   <xsl:value-of select="umbraco.library:GetDictionaryItem('lbl-no-jobs')"/><xsl:value-of select="umbraco.library:GetXmlNodeById(umbraco.library:RequestQueryString('careerslist'))/@nodeName"/>
  </xsl:otherwise>
 </xsl:choose>
   </xsl:when>
   <xsl:otherwise> 
 <xsl:apply-templates select="$source/node[@nodeTypeAlias=$documentTypeAlias]"> 
  <xsl:sort select="@createDate" data-type = "number" order="descending"/>
  </xsl:apply-templates>
   </xsl:otherwise>  
</xsl:choose> 
 
</xsl:template>

<xsl:template match="node">
<div class="job">
  
           <div class="joblink"><a href="{umbraco.library:NiceUrl(@id)}"><xsl:value-of select="@nodeName"/>&nbsp;
  

  (<xsl:value-of select="./data[@alias='jobRef']"/>)
  </a></div>
   <!--<div class="jobindustry"></div>-->
   <div class="joblocation">

   <xsl:apply-templates select="umbraco.library:Split(data [@alias = 'location'],',')/value"/>

  </div>
          <xsl:value-of select="data [@alias = 'teaser']"/><p></p>
  
</div>

</xsl:template>

<xsl:template match="value">
<xsl:variable name="regionNode" select="umbraco.library:GetXmlNodeById(.)"/>
<xsl:value-of select="$regionNode/@nodeName"/><xsl:if test="position()!=last()"><xsl:text>,</xsl:text></xsl:if>
</xsl:template>

</xsl:stylesheet>

Hi Nik, i'm part way there now, it seems to be ignoring the otherwise though. I left out <xsl:sort select="@createDate" data-type = "number" order="descending"/>
because i wasnt sure what it did. Can you explain this to me? As it stands at the moment if there are no jobs it just states the location with nothing underneith, can't figure out why it does this.

Hi Nik sorry for being an idiot then, i was looking at the wrong bit in my code, i have sorted this now with your help, i didnt realise the code for it was actually in there the whole time, just had to change the bit between otherwise. doh! thanks for your help.

Cool, glad I could help! 

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