Sorry I had messed up the header. But I got what you meant.
Thanks
Houman
-----Original Message-----
From: Michael Kay [mailto:mike(_at_)saxonica(_dot_)com]
Sent: 19 October 2005 16:17
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] Pass the value for xmlns:schemaLocation as a
variable? Possible?
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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