Re: 2 XML source

You could use a xsl transform to create xsl transforms for the files (Sure
I've seen this in one of the books somewhere. I'll try to dig it out).

If you (don't know how, if on unix I'd write a shell script to rip the names
from the source directory) load the xml file names into an xml doc like

<files>
    <file>030312_01.xml</file>
  ....
  ....
    <file>030312_100.xml</file>
</files>

Then write a simple transform that takes your main transform and inserts the
names from the file.  Then a simple script again that runs each of these
files one at a time in the xslt engine of your choice.

There may be a more elegant way, but I ain't got Jennis skill at this xsl
stuff (yet :-)

Cheers

Simon - "If you can't do it on a command line, it's not worth doing" - Kelly


----- Original Message -----
From: "David Alcantara" <wwwdavid@xxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Wednesday, March 12, 2003 2:34 PM
Subject: AW:  2 XML source


Thank you,

but it´s  not so simple. :-(

every day i get approx. 100 XML files like 030312_01.xml (XML name =
date+position) and i want to have only one XSL :-)
Because every day other application creates 100's of xml files from which i
need to filter some data out of these files. As you said in your mail that
to incorporate the xml file names into the XSL File, i think in my case it
is difficult to type all the file names into the XSL File. So in this case I
request some other solution which suite my need given above.
maybe is easily make a C program who integreate the 100 xml files in other
one.
Is there a way to make C or C++ program.

Regards
david



-----Ursprüngliche Nachricht-----
Von: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]Im Auftrag von Jarkko
Moilanen
Gesendet: Mittwoch, 12. März 2003 12:40
An: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Betreff: Re:  2 XML source


On Wed, 12 Mar 2003, David Alcantara wrote:

> Hi,
>
> i need to make a new XML file from 2 diferent XML source.
> It´s possible?
>
> Regards
> david
>
> I´m working with xalan 1.4.0


Yes it is and simple. =)
Use document() function.

For example:

<xsl:value-of select="document('source.xml')/XPath"/>

would look inside a file source.xml and traverse the tree
by the given XPath.

For more information see:
http://www.w3.org/TR/xslt#document

Cheers,
Jarkko.

>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>


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Researcher                mutta todella suuret mokat
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