Subject: RE: [xsl] Pass the value for xmlns:schemaLocation as a variable? Possible? From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Wed, 19 Oct 2005 16:16:40 +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> Are you sure it's "an XSL" (meaning a stylesheet, presumably?) - because it doesn't look like one. And it should be xsi:schemaLocation. xsi:schemaLocation is just another attribute as far as XSLT is concerned so you can generate it using an attribute value template, as xsi:schemaLocation="{$var}" Michael Kay http://www.saxonica.com/ > > > I would like to pass the path for the XSD file as a parameter to the > XSLT. > > For example I have this XSL file (called abc.xsl), which contains the > path to the XSD: > > > XML: > <?xml version="1.0" encoding="UTF-8"?> > <param>/home/PS_ReportRequest.xsd</param> > > ------------------------------------------------- > > XSLT: > <ReportRequest > xmlns="http://www.test.com/bta" > xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" > xmlns:schemaLocation="$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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