Question

How can I take an xml that looks like:

<request>
  <User>
    <name>name value</name>
    <age>13</age>
  </User>
</request>

And then I can deserialize this into a User object that has matching properties.

public class User {
  private String name;
  private Integer age;

  // getter and setters
}

Is there anything simple I can do or do I have to parse the xml manually for this?

I’m using spring mvc, and this is in a method where the xml will be posted.

Answer 1

Your controller method should take an argument of type User, which is annotated with @RequestBody. Then you need to configure a MarshallingHttpMessageConverter with an appropriate marshaller/unmarshaller. An example, straight from the reference guide:

<bean class="org.springframework.web.servlet.mvc.method.annotation.RequestMappingHandlerAdapter">
    <property name="messageConverters">
      <util:list id="beanList">
        <ref bean="stringHttpMessageConverter"/>
        <ref bean="marshallingHttpMessageConverter"/>
      </util:list>
    </property
</bean>

<bean id="stringHttpMessageConverter" 
       class="org.springframework.http.converter.StringHttpMessageConverter"/>

<bean id="marshallingHttpMessageConverter" 
      class="org.springframework.http.converter.xml.MarshallingHttpMessageConverter">
  <property name="marshaller" ref="castorMarshaller" />
  <property name="unmarshaller" ref="castorMarshaller" />
</bean>

<bean id="castorMarshaller" class="org.springframework.oxm.castor.CastorMarshaller"/>

Take a look at “Supported method argument types” and “Mapping the request body with the @RequestBody annotation” in the reference guide for more details.