Re: Outputting from more than one source

Joerg: It brings the same, even with this variable.
I'm using MSXML 4.0 (IE6) and Mozilla.
In both the result is the same.
Do you have another suggestion?

Liao: I could change the Java code to produce only one document.
But I don't want to change the specification of my project.
Just for the record, this project is a parcial requisite for my graduation.
Let's say I'll change my Jave code as a last option.

----- Original Message -----
From: "Joerg Heinicke" <joerg.heinicke@xxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Sunday, October 13, 2002 7:39 AM
Subject: Re:  Outputting from more than one source

Hmm, it should work - at least it looks ok. What processor are you
using? What happens if you do it in the following way:

<xsl:variable name="reg" select="/clients_ora/reg |
             document('other.xml')/clients_sqlsrv/reg |
             document('another.xml')/clients_db2/reg"/>

<xsl:apply-templates select="$reg">
    <xsl:sort select="name"/>
</xsl:apply-templates>

Regards,

Joerg

Gustavo Moreira wrote:

Yes, but what if there are three or more documents?

This, for instance, would this work?:

xsl:apply-templates select="/clients_ora/reg |
 document('other.xml')/clients_sqlsrv/reg |
 document('another.xml')/clients_db2/reg">
     <xsl:sort select="name"/>
</xsl:apply-templates>

I am using this:

     <xsl:apply-templates
        select="document('clients_sqlsrv.xml')//clients_sqlsrv/reg |
                document('clients_ora.xml')//clients_ora/reg">
       <xsl:sort select="name"/>
     </xsl:apply-templates>

...and it doesn't work because it brings first all the clients from the

1st

file (ordered), then it follows with all the clients from the 2nd file
(ordered).
What I want is all clients from all files ordered, no matter what file

they

come from.


Gustavo Moreira