You can do one thing:
Create a servlet which will return xml reponse.
Take the xml response in xsl and use the response xml as an arg to the
template.
Meaning you could pass name value pair to servlet and then you
could use the resulting xml reponse inside xsl as an argument.
Hope this solves your problem.
sundar
-----Original Message-----
From: Karthickraj N - CTD, Chennai.
[mailto:karthickrajn(_at_)ctd(_dot_)hcltech(_dot_)com]
Sent: Wednesday, May 14, 2003 11:56 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] Passing inputs to an XSLT
Is there any concept like passing command line arguments in XSLT
-----Original Message-----
From: Sundar Shanmugasundaram
[mailto:SSHANMUGASUNDARAM(_at_)selectica(_dot_)com]
Sent: Wednesday, May 14, 2003 11:07 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] Passing inputs to an XSLT
Karthick,
You can give <xsl:param> inside <xsl:template>. Include your xsl file in
the
file which has to call this xsl file. And use the call-template with
with-param.
sundar
-----Original Message-----
From: Karthickraj N - CTD, Chennai.
[mailto:karthickrajn(_at_)ctd(_dot_)hcltech(_dot_)com]
Sent: Wednesday, May 14, 2003 11:08 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Passing inputs to an XSLT
Hello friends,
I wanted to pass a string as input to an XSLT file, which will then be used
as a variable in the XSLT file.
If anyone could find a solution please help me out.
Thanx in advance,
Regards,
Karthickraj
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list