Hi,

Actually I need to know whether Multiple input to multiple output can be
done? How to do this.

Here multiple input means my input.xml which contains 

<files>
<file fileloc="a/gen.xml"/>
<file fileloc="a/gen1.xml"/>
</files>

Thanks,
Shailesh
-----Original Message-----
From: Ragulf Pickaxe [mailto:ragulf.pickaxe@xxxxxxxxx] 
Sent: Tuesday, November 08, 2005 1:24 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re:  How to process a list of files

Shailesh,

You have not received any answers to this question (nor to the last
one with a different title, which is identical). Rather than just
reposting, you should perhaps see if there is something with your
mail, that makes it not answered.

When I look at the input, the desired output, and the XSL that you
give, there is absolutely no correlation between the XSL and the
input/output.

Taking one thing, a line like this - which is a solution to a question
you asked some days previously - gives the impression that you do not
read the answers that you are given:
<xsl:value-of
select="concat(substring-before(substring-after(@fileloc,'/xml/'),'.'),'.htm
')"/>

A line like that would not in any way, given "A/gen1.xml" give you gen1.xml.

The XSL also contains code that is not relevant to the problem, which
1) makes it harder to read, 2) gives the impression that you have only
a small subset of the problem, and a solution to your question would
not answer your real problem.

As you ask your question, a solution would be:

<xsl:template match="/">
  <xsl:apply-templates select="Root"/>
</xsl:template>

<xsl:template match="Root">
  <xsl:for-each select="file"/>
    <xsl:value-of select="substring-after(@fileloc,'A/')"/><xsl:text>
</xsl:text>
  </xsl:for-each>
</xsl:template>

I suspect that this is not what you really want, but this is the
answer to the question that you posed.

Regards,
Ragulf Pickaxe :-/