Subject: [xsl] Simple (external XML) internationalization with XSLT? From: knocte <knocte@xxxxxxxxx> Date: Mon, 3 Oct 2005 17:20:43 +0200 |
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 ] --
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