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