How creating an XmlObject only knowing the name of the generated document class?

Hi, how can I get the XmlObject like I would get with the following call XmlObject xmlobject = OwnReportDocument.Factory.newInstance(); but without using the generated Document-Class 'OwnReportDocument'. I tried using the following code and got a NoSuchMethodException (no

Re: How creating an XmlObject only knowing the name of the generated document class?

Have you tried XmlObject xmlobject = XmlObject.Factory.newInstance()? -jacobd On Tue, Jun 16, 2009 at 12:01 AM, Hildegunde Weinzierlhildegunde.weinzi...@esg.de wrote: Hi, how can I get the XmlObject  like I would get with the following call         XmlObject xmlobject =

Antwort: Re: How creating an XmlObject only knowing the name of the generated document class?

Hi, yes, but I need the SchemaType of the corresponding MyReportDocument. Best regards Hildegunde Weinzierl

Adding/Removing xml documents as xs:any

hi I am running into problem while trying to extract an xs:any as a typed XMLObject. I am writing code that is not aware of the actual concrete types but responsible for inserting and extracting XmlObjects as xs:any (and dynamically converting to concrete type while extracting) I have following: