從Java Bean或使用JAXB

Question

XML我使用這個代碼

JAXBContext jc = JAXBContext.newInstance(Bookdata.class); 
Bookdata bookdata=new Bookdata(); 
Marshaller marshaller = jc.createMarshaller(); 
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true); 
marshaller.marshal(bookdata, (OutputStream) output); 

但它生成XML創建XSD但我需要XSD 我需要這樣的創建XSD：

<ArrayOfCommandInfoDTO> 
    <CommandInfoDTO> 
     <a:allowAddingGameCenterFriends>true</a:allowAddingGameCenterFriends> 
     <a:enter code here>allowAppInstallation>true</a:allowAppInstallation> 
     <a:allowAssistant>true</a:allowAssistant> 
     <a:allowAssistantWhileLocked>true</a:allowAssistantWhileLocked> 
     <a:allowCamera>true</a:allowCamera> 
    </CommandInfoDTO> 
</ArrayOfCommandInfoDTO> 

所以請告訴我如何從Java Beans或XML創建XSD

Answer 1

您可以使用JAXBContext上的generateSchema方法生成XML模式。

• http://wiki.eclipse.org/EclipseLink/Examples/MOXy/JAXB/GenerateSchema

Answer 2

public class ObjToSchema { 

    public static void main(String[] args) throws IOException, JAXBException { 
     // TODO Auto-generated method stub 
     (JAXBContext.newInstance(Bookdata.class)).generateSchema(new DataSchemaOutputResolver()); 
    } 
} 


public class DataSchemaOutputResolver extends SchemaOutputResolver { 
    @Override 
    public Result createOutput(String arg0, String arg1) throws IOException { 
     // TODO Auto-generated method stub 
     return new StreamResult(new File("d:/data.xsd")); 
    } 
} 

希望這將解決您的問題。