Question

I have some problems excluding unwanted output from my xlst transformation.
I already know about the default rules behind match etc, but i’m not able to use match in template/apply-templates properly.
Can you help me fixing this please?

So I have an XML file structured this way:

<movies>
    <movie id="0">
        <title>Title</title>
        <year>2007</year>
        <duration>113</duration>
        <country>Country</country>
        <plot>Plot</plot>
        <poster>img/posters/0.jpg</poster>
        <genres>
            <genre>Genre1</genre>
            <genre>Genre2</genre>
        </genres>
        ...
    </movie>
    ...
</movies>

And I want to create a html UL list with a LI for each movie that belongs to a genre ‘#######'(replaced at runtime by my perl script) which is a link to a page(named by its id).

Right now i’m doing it this way:

<xsl:template match="/">
    <h2> List </h2>
    <ul>
        <xsl:apply-templates match="movie[genres/genre='#######']"/>
            <li>
                <a>
                    <xsl:attribute name="href">     
                        /movies/<xsl:value-of select= "@id" />.html
                    </xsl:attribute>
                    <xsl:value-of select= "title"/>
                </a>
            </li>
    </ul>
</xsl:template>

Obviously this way it shows me all the elements of the movies that matches the chosen genre.
Do I have to add tons of <xsl:template match="..."> to remove all the extra output?
Can you please teach me the correct way to create an html snippet like this?

List

• Title0
• Title2
• Title7

Thank you in advance!

Answer 1

You are almost there – your use of apply-templates is causing you the problem.

Instead, structure your XSLT this way:

  <xsl:template match="/">
    <h2> List </h2>
    <ul>
      <xsl:apply-templates select="movie[genres/genre='#######']"/>
    </ul>
  </xsl:template>

  <xsl:template match="movie">
    <li>
      <a>
        <xsl:attribute name="href">/movies/<xsl:value-of select= "@id" />.html</xsl:attribute>
        <xsl:value-of select= "title"/>
      </a>
    </li>
  </xsl:template>

It will apply the spcific template (match=”movie”) to your movie element. In your original attempt, you will be using the default template which will bring back everything contained within a movie element.