Subject: [xsl] Pass the value for xmlns:schemaLocation as a variable? Possible? From: "Khorasani, Houman" <houman_khorasani@xxxxxxxxxxxxxx> Date: Wed, 19 Oct 2005 15:56:33 +0100 |
I have this following XSL. <ReportRequest xmlns="http://www.test.com/bta" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:schemaLocation="http://www.test.com/bta /home/PS_ReportRequest.xsd"> ... </ReportRequest> I would like to pass the path for the XSD file as a parameter to the XSLT. For example I have this XSL file (called abc.xsl), which contains the path to the XSD: XML: <?xml version="1.0" encoding="UTF-8"?> <param>/home/PS_ReportRequest.xsd</param> ------------------------------------------------- XSLT: <ReportRequest xmlns="http://www.test.com/bta" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:schemaLocation="$param"> ... </ReportRequest> I can't do a <xsl:value-of "param"/> instead of $param. Declaring a variable is also not possible because the spot I can use XSL is after the header. So how can I do that? Any idea? Mnay thanks Houman
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