Re: How to enrich a message?

If you wish to insert Xml into the destination that did not originate in the
source you need to use <xsl:value-of> and a string representation of the Xml

A scripting functoid like
public string GetXml(XmlDocument doc)
return doc.OuterXml;

then use an inline XSLT
<xsl:value-of disable-output-escaping="yes"


"Obiwan Jacobi" <obiwanjacobi@xxxxxxxxxxx> wrote in message

I'm a BizTalk newbee and I am trying to enrich a message. Most of the
fields can be copied direcly and a database lookup should provide the
rest of the information. The information that is to be added to the
destination message is complex; a hierarchy with multiple repeating

I've tried to use a Map and a Database Lookup functoid. But this only
works well for single field (one row at best) enrichment. Then I tried
to use the scripting functoid to call an external assembly that looks
up the extra information. But there was no way I got the extra
information to be copied into the destination schema. I've tried
returning a custom Xml serializable class, an XmlDocument, an XmlNode
and a string containing the xml. Either the new element was not added
at all or I got an error the element could not contain text (as defined
in the destiniation schema).

I also tried to pass the output of the scripting functoid that calls
the external assembly into another scripting functoid that contained an
xslt call template. But I didn't know how to copy the XmlDocument in
the <xsl:param> out to the destination. I've tried <xsl:copy-of
select="document($param1)"/> but that didn't work.

So if anyone has any idea how to get an xml fragment into the
destination schema. I would be very gratefull.

Marc Jacobi


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