Hello everyone,
i have got a big problem, which I haven't resolved yet.

I have to transform from one xml file to another, but it seems that xsl(t) is not able to resolve my transformations.
i have got following xml-source (explanatory):

<table name="table1" schema="schema1"/>
<table name="table2" schema="schema2"/>
<table name="table3" schema="schema1"/>
<table name="table4" schema="schema1"/>


the output should be:

<Schema name="schema1" tables="table1 table3 table4"/>
<Schema name="schema2" tables="table2"/>

How to transform this kind of structure
1. without depending of any xslt-processor
2. and do the step in one transformation!
(not two xsl transformation files)

cheers for any response.
I am desparate with that problem..

You can solve this using the xslt2 "for-each-group" element (hopefully your processor can handle that):
<xsl:template match="/tables">
 <Database>
  <xsl:for-each-group select="table" group-by="@schema">
   <schema>
    <xsl:attribute name="name">
      <xsl:value-of select="current-grouping-key()"/>
    </xsl:attribute>
    <xsl:attribute name="tables">
     <xsl:value-of select="current-group()/@name"/>
    </xsl:attribute>
   </schema>
  </xsl:for-each-group>
 </Database>
</xsl:template>

Thanks,
Francine.

This is a simple grouping problem.

Simple grouping problems can be solved in XSLT 2.0 using the xsl:for-each-group construct, or in 1.0 using the Muenchian grouping technique described at http://www.jenitennison.com/xslt/grouping



Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference