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Answer 1

+1 A:

Combining <xsl:import> and <xsl:param> seems to work fine for me:

main.xsl:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:param name="title" select="'Page title'" />
  <xsl:include href="head.xsl"/>
</xsl:stylesheet>

head.xsl:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:template match="head">
    <title>
      <xsl:value-of select="$title"/>
    </title>
  </xsl:template>
</xsl:stylesheet>

Applying main.xsl on a file containing <head /> produces <title>Page title</title> as expected. If this doesn't help you, you should post some code where you actually use <xsl:import> and <xsl:param> (and that doesn't work).

svick 2010-08-01 14:31:17

hi @svick, I've updated the question with a more specific code about what I'm doing. I tried to do what you said but I get the error mentioned on the question.Thank you in advance!

Isern Palaus 2010-08-01 20:06:50

Answer 2

A:

I see. The problem is that each <xsl:include>d file has to be valid XSLT, which means it has to be valid XML. And you can't have unclosed tags in valid XML, but you need for example unclosed <html> in head.xsl. So, I don't think you can do this in XSLT.

svick 2010-08-01 20:38:54

I also tried `<xsl:include href="head.xsl" />` that contains the code from _NOTE1_ from question, and I can't get it working to. I get this error:XSLTProcessor::importStylesheet() [xsltprocessor.importstylesheet]: compilation error: file /home/xsl/index.xsl line 9 element include in /home/index.php on line 9XSLTProcessor::importStylesheet() [xsltprocessor.importstylesheet]: element include only allowed as child of stylesheet in /home/index.php on line 9

Isern Palaus 2010-08-01 22:54:48

Sorry for 'spamming'... but I see that it needs to be direct child of `<xsl:stylesheet />`... so dificult to fit it :(

Isern Palaus 2010-08-01 22:56:58

Answer 3

+1 A:

So, something is wrong with splitting in the way I'm doing and then including it. How can I solve it?

Yes, something is wrong. You want to split a brick template...

First, you need to have something to split, so this stylesheet:

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:dummy="dummy"
exclude-result-prefixes="dummy">
    <xsl:param name="menu" select="document('menu.xml')"/>
    <xsl:param name="sidebar" select="document('sidebar.xml')"/>
    <xsl:param name="footer" select="document('footer.xml')"/>
    <dummy:attSet>
        <footer class="footer"/>
        <menu class="menu" id="tabs"/>
        <sidebar class="body_dcha"/>
    </dummy:attSet>
    <xsl:template match="/page">
        <html lang="es">
            <head>
                <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/>
                <title>
                    <xsl:value-of select="title" />
                </title>
                <link rel="stylesheet" type="text/css" href="css/main.css" />
                <script type="text/javascript" src="js/custom.js"></script>
            </head>
            <body>
                <div class="content">
                    <div class="header">
                        <xsl:apply-templates select="$menu"/>
                    </div>
                    <div class="body">
                        <xsl:apply-templates select="news/contents"/>
                        <xsl:apply-templates select="$sidebar"/>
                        <div class="clear"></div>
                    </div>
                    <xsl:apply-templates select="$footer"/>
                </div>
            </body>
        </html>
    </xsl:template>
    <xsl:template match="page/news/contents">
        <div class="body_izqda">
            <xsl:apply-templates/>
        </div>
    </xsl:template>
    <xsl:template match="contents/entry/title">
        <h2>
            <xsl:value-of select="."/>
        </h2>
    </xsl:template>
    <xsl:template match="contents/entry/text">
        <p>
            <xsl:value-of select="."/>
        </p>
    </xsl:template>
    <xsl:template match="menu/category|sidebar/results/category|footer/entry">
        <li>
            <a href="{link}">
                <xsl:value-of select="name"/>
            </a>
        </li>
    </xsl:template>
    <xsl:template match="/footer|/menu|/sidebar">
        <div>
            <xsl:copy-of select="document('')/*/dummy:*/*[name()=name(current())]/@*"/>
            <ul>
                <xsl:apply-templates/>
            </ul>
        </div>
    </xsl:template>
</xsl:stylesheet>

With this input:

<page>
    <title>Some Page</title>
    <news>
        <contents>
            <entry>
                <title>Title1</title>
                <text>Text1</text>
            </entry>
            <entry>
                <title>Title2</title>
                <text>Text2</text>
            </entry>
            <entry>
                <title>Title3</title>
                <text>Text3</text>
            </entry>
            <entry>
                <title>Title4</title>
                <text>Text4</text>
            </entry>
        </contents>
    </news>
</page>

And this documents:

menu.xml

<menu>
    <category>
        <link>http://www.example.com/link1&lt;/link&gt;
        <name>Link1</name>
    </category>
    <category>
        <link>http://www.example.com/link2&lt;/link&gt;
        <name>Link2</name>
    </category>
    <category>
        <link>http://www.example.com/link3&lt;/link&gt;
        <name>Link3</name>
    </category>
</menu>

sidebar.xml

<sidebar>
    <results>
        <category>
            <link>http://www.example.com/link4&lt;/link&gt;
            <name>Link4</name>
        </category>
        <category>
            <link>http://www.example.com/link5&lt;/link&gt;
            <name>Link5</name>
        </category>
        <category>
            <link>http://www.example.com/link6&lt;/link&gt;
            <name>Link6</name>
        </category>
    </results>
</sidebar>

and footer.xml

<footer>
    <entry>
        <link>http://www.example.com/link7&lt;/link&gt;
        <name>Link7</name>
    </entry>
    <entry>
        <link>http://www.example.com/link8&lt;/link&gt;
        <name>Link8</name>
    </entry>
    <entry>
        <link>http://www.example.com/link9&lt;/link&gt;
        <name>Link9</name>
    </entry>
</footer>

Output the same result that provided stylesheet.

So, now you can split the stylesheet in several modules.

Alejandro 2010-08-02 13:03:26

And then, with the templates, I can put it them into a separated .xsl file and load it with `<xsl:include href="sidebar.xsl" />`. Thank you!

Isern Palaus 2010-08-02 16:15:10

@Isern Palaus: You're wellcome! Also, I would not recommend to split this stylesheet because the same template is apply for elements from diferents source. If you'll split the patterns in order to group them by the source, then you'd end up repeating the templates or calling a template in main stylesheet modelu for example.

Alejandro 2010-08-02 16:33:41

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Combining <xsl:param> and <xsl:include>

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