I basically stole the code from one of your examples and made a few changes:
public class Main {
public static void main( String[] args ) {
try {
JAXBContext jc = JAXBContext.newInstance( "mypackage" );
Unmarshaller u = jc.createUnmarshaller();
MyType mt = (MyType)u.unmarshal( new FileInputStream( "myXml.xml" ) );
Marshaller m = jc.createMarshaller();
m.setProperty( Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE );
m.setProperty( "jaxb.schemaLocation", "myNamespace
http://my.com");
m.marshal( mt, new java.io.FileOutputStream("myXml_new.xml") );
} catch( JAXBException je ) {
je.printStackTrace();
} catch( IOException ioe ) {
ioe.printStackTrace();
}
}
}
Eric
Kohsuke Kawaguchi
<Kohsuke.Kawaguch To: JAXB-INTEREST_at_JAVA.SUN.COM
i_at_Sun.COM> cc:
Sent by: Subject: Re: how to obtain namespace of the unmarshalled object?
Discussion list
for the Java
Architecture for
XML Binding
<JAXB-INTEREST_at_JA
VA.SUN.COM>
04/15/03 12:51 PM
Please respond to
JAXB-INTEREST
Could you post the relevant part of the source code?
regards,
--
Kohsuke KAWAGUCHI 408-276-7063 (x17063)
Sun Microsystems kohsuke.kawaguchi_at_sun.com
--
This e-mail may contain confidential and/or privileged information. If you are not the intended recipient (or have received this e-mail in error) please notify the sender immediately and destroy this e-mail. Any unauthorized copying, disclosure or distribution of the material in this e-mail is strictly forbidden.