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  • Max 144 posts 166 karma points
    Mar 28, 2011 @ 09:24
    Max
    0

    xslt sorting based on a Value Parameter

    Hi I have passed a paramter which is a field data value of a list of items in an xslt file i need to sort the data based on a field <xsl:value-of select="employeeDepartment"/>
    where i need to check if the employeeDepartment=$DeptName (as parmeter form request query string)

    any code example of the sorting wil be useful

     

     

  • Daniel Bardi 927 posts 2562 karma points
    Mar 28, 2011 @ 09:29
    Daniel Bardi
    0

    you can sort within a for-each block  (http://www.w3schools.com/xsl/el_sort.asp)

    <xsl:for-each select="$currentPage/*[department=$deptparam]">

    <xsl:sort select="department" />
    ... do something...

    </xsl:for-each>

    This will sort but you won't see much since the department would be the same for all

    Are you actually wanting to only filter and not sort?

  • Max 144 posts 166 karma points
    Mar 28, 2011 @ 10:05
    Max
    0

    i thinks its filter cos there are many employees form different departments i wann filter according to paramter the department name

     

  • Daniel Bardi 927 posts 2562 karma points
    Mar 28, 2011 @ 10:12
    Daniel Bardi
    0

    Then use what I suggested but remove the sort line:

    <xsl:for-each select="$currentPage/*[department=$deptparam]">

    ... do something...

    </xsl:for-each>

  • Max 144 posts 166 karma points
    Mar 28, 2011 @ 10:52
    Max
    0

    Thanks it works how to put a condition there if there is no paramter provided then the for-each statement wont be initiated that way

     

  • Daniel Bardi 927 posts 2562 karma points
    Mar 28, 2011 @ 11:04
    Daniel Bardi
    0

     

    <xsl:for-each select="$currentPage/*">
    <xsl:choose>
      <xsl:when test="$deptparam != ''">
        <xsl:if test="department=$deptparam">
          ... do something with filtered department ...
        </xsl:if>
      </xsl:when>
      <xsl:otherwise>
        ... do something with all departments
      </xsl:otherwise>
    </xsl:choose>
    </xsl:for-each>

     

  • Max 144 posts 166 karma points
    Mar 28, 2011 @ 12:05
    Max
    0

    Thanks for the anwwers it was very useful

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