I can't hard code the static path in xslt. I am asking dynamic path. I can
use any path for input file for the transformation.

I need to use dynamic input file path on this document method.

URL for the document method is input file's path.

-----Original Message-----
From: Andrew Welch [mailto:andrew.j.welch@xxxxxxxxx]
Sent: Thursday, December 03, 2009 8:16 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re:  Dynamic Input file path on document method

document('comments.xml', /)


2009/12/3 Ramkumar <ramkumar@xxxxxxxxxxxxxxxxxx>:
> How to use as a second argument in my code? Please share your idea.
>
> -----Original Message-----
> From: Andrew Welch [mailto:andrew.j.welch@xxxxxxxxx]
> Sent: Thursday, December 03, 2009 7:57 PM
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re:  Dynamic Input file path on document method
>
> 2009/12/3 Ramkumar <ramkumar@xxxxxxxxxxxxxxxxxx>:
>> Hi List,
>>
>> When using "document('comments.xml')", xslt looking default path for
>> comments.xml' on xslt path. Actually I have it in input file path.
>>
>> So how to implement the dynamic input path on this code. Please suggest
> your
>> idea.put
>
> You can supply a node from the input as second arg to the document()
> function, and then it will be resolved against that.
>
> http://www.w3.org/TR/xslt20/#function-document
>
>
>
>
> --
> Andrew Welch
> http://andrewjwelch.com
> Kernow: http://kernowforsaxon.sf.net/
>
>



--
Andrew Welch
http://andrewjwelch.com
Kernow: http://kernowforsaxon.sf.net/