Not sure what you are looking for. If you have an XmlObject o and you don't know what type it is, do
SchemaType type = o.schemaType(); Does your Schema have 200 types that are all extending a base type such as Shape in the example below? Can all of those types result from the xquery? Do you need specific processing for each of them? Radu -----Original Message----- From: Christian Kaiser [mailto:[EMAIL PROTECTED] Sent: Wednesday, August 02, 2006 11:42 AM To: user@xmlbeans.apache.org Subject: Re: RE: AbstractTypes problem Hello, i am new to xmlbeans and got stuck with the same problem. my schema consists of 200 types I dont want to compare all types with the xquery result. *********************************************** final Shape shape = Shape.Factory.parse(someShapteInput); if(shape instanceof Circle) { final Circle = (Circle) shape; // do whatever you want with a circle } else if(shape instanceof Square) { final Square square = (Square)shape; . . . ********************************************} i dont see how andrejs way could work for me. (SchemaType instanceType = xo.type) Does it work? Is there another solution? Thanks, Christian --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] _______________________________________________________________________ Notice: This email message, together with any attachments, may contain information of BEA Systems, Inc., its subsidiaries and affiliated entities, that may be confidential, proprietary, copyrighted and/or legally privileged, and is intended solely for the use of the individual or entity named in this message. If you are not the intended recipient, and have received this message in error, please immediately return this by email and then delete it. --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
