|
[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Simple (external XML) internationalization with XSLT?
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 ]
--
|
PURCHASE STYLUS STUDIO ONLINE TODAY!Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced! Download The World's Best XML IDE!Accelerate XML development with our award-winning XML IDE - Download a free trial today! Subscribe in XML format
|

Cart








