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List:       xmlbeans-user
Subject:    RE: RE: How to get Java Class of .xsd at runtime dynamically
From:       "Radu Preotiuc-Pietro" <radu.preotiuc-pietro () oracle ! com>
Date:       2008-10-24 3:25:48
Message-ID: 20081023202548031.00000005192 () RADUP02
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Yeah, keep in mind that there can be multiple root elements in a Schema, so use SchemaTypeSystem.documentTypes() to enumerate all of them on the SchemaTypeSystem object returned from compileXmlBeans.

Radu

> -----Original Message-----
> From: Julian Kölle [mailto:aliban@gmx.net] 
> Sent: Thursday, October 23, 2008 2:22 AM
> To: user@xmlbeans.apache.org
> Subject: Re: RE: How to get Java Class of .xsd at runtime 
> dynamically on-the-fly?
> 
> Hi again,
> 
> I have another question: when XML Beans creates the java 
> source files with a .xsd, is it possible to identify/know the 
> java class name of the root element defined int the XML 
> Schema? Because I must load this class at runtime i must know 
> its name. Can you point me to a solution for this?
> 
> regards
> 
> 
> -------- Original-Nachricht --------
> > Datum: Tue, 21 Oct 2008 17:52:00 -0700
> > Von: "Radu Preotiuc-Pietro" <radu.preotiuc-pietro@oracle.com>
> > An: "user@xmlbeans.apache.org" <user@xmlbeans.apache.org>
> > Betreff: RE: How to get Java Class of .xsd at runtime 
> dynamically on-the-fly?
> 
> > Well, this is somewhat a Java question also. Each Java class has to 
> > have a classloader associated to it. The only way to get a 
> Java class 
> > is to have a classloader that is able to find the .class 
> > representation of a class and load it.
> >
> > What this means for XMLBeans is that you need to compile 
> the Schemas 
> > to Java source (XmlBeans.compileXsd() or call "scomp" in a 
> different 
> > process), then compile and jar that source and then create a 
> > classloader that has access to the freshly created .jar 
> file. XmlBeans 
> > can't create classes without source .java files and .jar files.
> >
> > What I should also point out is that you will only be able 
> to use such 
> > dynamically generated classes via reflection (since they 
> don't exist 
> > when your code is compiled), so this is why most people 
> prefer to use 
> > DOM or XmlCursor when dealing with Schemas that they only 
> know about 
> > at runtime. The XML documents themselves will still be bound to a 
> > Schema, so you'll be able to do Schema validation on them 
> and get the actual Schema types.
> >
> > Radu
> >
> > > -----Original Message-----
> > > From: aliban@gmx.net [mailto:aliban@gmx.net]
> > > Sent: Wednesday, October 08, 2008 2:23 AM
> > > To: user@xmlbeans.apache.org
> > > Subject: How to get Java Class of .xsd at runtime dynamically 
> > > on-the-fly?
> > >
> > > Hello,
> > >
> > > I need the (freshly compiled?) Java Class from a .xsd at runtime 
> > > on-the-fly. Without creating java source code files or .jar files.
> > >
> > > I tried org.apache.xmlbeans.XmlBeans.compileXsd()
> > > and
> > > org.apache.xmlbeans.XmlBeans.loadXsd()
> > >
> > > but I could not figure out how to spawn a Class from it. 
> I tried to 
> > > use the SchemaTypeSystem.getClassLoader() but this returns null 
> > > always. I also tried SchemaType.getJavaClass() but 
> somehow it does 
> > > not work/I don't understand it/I don't know how to find the "root 
> > > item Class" with it.
> > >
> > > Can you tell me: how do I get the Java Class of the "root item" 
> > > within the .xsd at runtime (on-the-fly, without generating code 
> > > files or .jar) ?
> > >
> > > regards!
> > >
> > > ali
> > > --
> > > Ist Ihr Browser Vista-kompatibel? Jetzt die neuesten 
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> > >
> > > 
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> > >
> > >
> >
> >
> > 
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