Re: [xsl] How to process a list of files

Subject: Re: [xsl] How to process a list of files
From: Ragulf Pickaxe <ragulf.pickaxe@xxxxxxxxx>
Date: Tue, 8 Nov 2005 08:53:55 +0100

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

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:

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

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

Ragulf Pickaxe :-/

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