[xsl] Simple (external XML) internationalization with XSLT?

[knocte <knocte@xxxxxxxxx>]

Hello.

I am beginning to deal with XSLT at more complicated problems. There
is one now that I cannot figure out how to solve:

This is my XML file:

<root>
  <MyTitle>CODE-XXX</Title>
</root>

This is my XSLT file:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

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

  <xsl:template match="/*">

    <page>

      <title><xsl:value-of select="MyTitle" /></title>

      <content>
        <div class="Image">
          <i18n:text key="1" />
          <img>
            <xsl:attribute
name="href">http://www.example.com/img.gif</xsl:attribute>
            <xsl:attribute name="alt"><i18n:text key="2" /></xsl:attribute>
          </img>
        </div>
      </content>

    </page>

  </xsl:template>

</xsl:stylesheet>


This is an external XML file (called "en-US.xml" for example):

<i18n:language code="en-US">
  <i18n:hash-list>
    <i18n:string key="1">Rendered image</string>
    <i18n:string key="2">Alt text</string>
  </i18n:hash-list>
</i18n:language>


My question is: what do I have to add to my XSLT stylesheet to obtain
the following expected XML result?:

<page>

  <title>CODE-XXX</title>

  <content>
    <div class="Image">
      Rendered image
      <img href="http://www.example.com/img.gif"; alt="Alt text" />
    </div>
  </content>

</page>


And another question: can I obtain this result without the need of
applying an XSLT pre-transformation to the XSLT itself? (I mean,
without the need of an extra XSLT stylesheet.)

Thanks in advance.

  Andrew  [ knocte ]

--