XSL Transformation Error

[matthias_prller]

Hello,

I try to do an inline transformation in an xacute query, but always get the error "Unable to create inline transformation".

I am using MII 12.0 SP3.

Here is the result of the query without transformation:

<?xml version="1.0" encoding="UTF-8" ?> 
<Rowsets DateCreated="2008-07-03T11:06:17" EndDate="2008-07-03T11:06:17" StartDate="2008-07-03T10:06:17" Version="12.0.3 Build(106)">
<Rowset>
<Columns>
  <Column Description="Ausgabe" MaxRange="0" MinRange="0" Name="Ausgabe" SQLDataType="1" SourceColumn="Ausgabe" /> 
  </Columns>
<Row>
  <Ausgabe>111.111.113 insert</Ausgabe> 
  </Row>
  </Rowset>
  </Rowsets>

and here is the (smallest) xsl file but it doesn't work:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:template match="/">

<xsl:value-of select="."/>

</xsl:template>
</xsl:stylesheet>

What is wrong with this transformation - i tried other (more complicated) transformations but they don't work, too.

Thanks for your help.

Best regards,

Matthias

[Former Member]

Deleted

Edited by: Christian Libich on Jul 3, 2008 2:47 PM

Edited by: Christian Libich on Jul 3, 2008 2:48 PM

[matthias_prller]

Hi again,

Finally after looking around the whole internet and asking google I found the solution rigth here in this forum:

[;

If anyone is intrested in the solution:

Inline transformation in an xacute query is not the right place for the xsl stylesheet.

You have to make the transformation in the url like:

<server:port>

/XMII/Illuminator?QueryTemplate=<link_to_query>=web://<link_to_sytlesheet>&Content-Type=text/xml

Bye

Matthias