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how to get the file location within an xsl transformation

 
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h-gent-o
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Joined: 01 Oct 2006
Posts: 7

PostPosted: Mon Apr 07, 2008 7:44 am    Post subject: how to get the file location within an xsl transformation Reply with quote

Hi,

is there a way to get the filename as well as the location of the input source?

I setup an import xsl filter but I need to get the file location of the xml source. How can I get it done. As xalan has no support for xslt2.0 function fn:base-uri() I think I have to use the "java" way. I already got it working with xalan when I test the xsl within XML editor. But when I try to import within OO the function gives me no result.

So this is working outside of OO with xalan-j

<xsl:stylesheet extension-element-prefixes="NodeInfo" xmlns:NodeInfo="org.apache.xalan.lib.NodeInfo" ... >
.
.
<xsl:value-of select="NodeInfo:systemId()"/>
.
.
</xsl:stylesheet>

the NodeInfo:systemId() function is available as function-available('NodeInfo:systemId') return true within OO

So where is the trick to get the filename and its location? Rolling Eyes
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