CODEWITHSUNDEEP
xmljspxslt
Harry
Harry

Reputation: 313

Pass parameter from jsp into xsl

My code below seems logical however i don't know why the sorting doesn't work with the error "Variable or parameter 'sort' is undefined.'"? Im suspecting there are something wrong with declaring param in xsl. Could anyone point my mistake? thanks

java code to pass parameter

String sort = "rating";
transformer.setParameter("sort", sort); /It will control the sort in xsl

xml file

    <?xml version="1.0" ?>

<cd>
  <title>A Funk Odyssey</title>
  <artist>Jamiroquai</artist>

  <tracklist>
    <track id="1">
      <title>Feels So Good</title>
      <time>4:38</time>
      <rating>2</rating>
    </track>

    <track id="2">
      <title>Little L</title>
      <time>4:10</time>
      <rating>5</rating>
    </track>

    <track id="3">
      <title>You Give Me Something</title>
      <time>5:02</time>
      <rating>3</rating>
    </track>

    <track id="4">
      <title>Corner of the Earth</title>
      <time>3:57</time>
      <rating>1</rating>
    </track>
  </tracklist>


</cd>

This is my xsl

            <?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="UTF-8" omit-xml-declaration="yes"/>
    <xsl:param name="sort" select="title"/>
    <xsl:template match="/">
        <table border="1">
            <thead>
                <tr>
                    <th><a href="#">Title</a></th>
                    <th><a href="#">Time</a></th>
                    <th><a href="#">Rating</a></th>
                </tr>
            </thead>
            <tbody>
                <xsl:for-each select="cd/tracklist/track">
                    <xsl:sort select="$sort"/>
                    <tr>
                        <td><xsl:value-of select="title" /></td>
                        <td><xsl:value-of select="time" /></td>
                        <td><xsl:value-of select="rating" /></td>
                    </tr>
                </xsl:for-each>
            </tbody>
        </table>
    </xsl:template>
</xsl:stylesheet>

Upvotes: 1

Views: 1800

Answers (1)

Pavel Veller
Pavel Veller

Reputation: 6105

In your xsl:param declaration you are trying to default to $sort which is not defined at the time xsl:param is evaluated. It does look like a reference to itself.

If you don't need a default then just change your parameter declaration to:

<xsl:param name="sort"/>

or default to a string value:

<xsl:param name="sort" select="'title'"/>

or

<xsl:param name="sort">title</xsl:param>

That said, we have only addressed the parameter declaration issue. Now on to sorting. The xsl:sort needs an expression, it won't convert a string value into XPath like you expect it to.

Here's a solution: Using Variables in <xsl:sort select=""/>.

You would basically do something like:

<xsl:sort select="*[name() = $sort]"/>

Upvotes: 4

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