xsl-list
[Top] [All Lists]

Re: Outputting from more than one source

2002-10-13 03:39:37
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


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