How to inherit a model from superclass in playfram

问题:

I'm trying to understand how does the inheritance work in play! But unsuccessfully yet.

So, I have such superclass:

@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)  
abstract class SuperClass extends Model {  
    @Id  
    @GeneratedValue(strategy = GenerationType.TABLE, generator = "SEQ_TABLE")   
    @TableGenerator(name = "SEQ_TABLE")  
    Long id;  

    int testVal;
}

And 2 inherited classes:

@Entity
public class Sub extends SuperClass {        
    String name;

    @Override
    public String toString() {
            return name;
    }
}

@Entity
public class Sub1 extends SuperClass {        
    String name;

    @Override
    public String toString() {
            return name;
    }
}

Also I have 2 controllers for inherited classes:

public class Subs and Sub1s extends CRUD {

}

After application was started, I recieve 2 tables in MySQL db for my models (Sub and Sub1) with such structure: id bigint(20), name varchar(255). Without testVal which is in superclass.

And when I try to create new object of Sub class in CRUD interface I recieve such error: Execution error occured in template {module:crud}/app/views/tags/crud/form.html. Exception raised was MissingPropertyException : No such property: testVal for class: models.Sub.

In {module:crud}/app/views/tags/crud/form.html (around line 64) #{crud.numberField name:field.name, value:(currentObject ? currentObject[field.name] : null) /}

1. What should I do to generate MySQL tables for inherited models properly and fix the error?
2. Is it possible to have a single superController for several inherited classes?

回答1:

Well, thanks to sdespolit, I've made some experiments. And here is what I've got:

Superclass:

@MappedSuperclass
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class SuperClass extends Model {
}

Inherited class:

@Entity 
public class Sub extends SuperClass {
}

"Super Controller" I made in such way:

@With({Secure.class, SuperController.class})
@CRUD.For(Sub.class)
public class Subs extends CRUD {
}

@With({Secure.class, SuperController.class})
@CRUD.For(Sub1.class)
public class Sub1s extends CRUD {
}

@CRUD.For(Sub.class) is used to tell the interceptors with what class it should work

public class SuperController extends Controller {

    @After/Before/Whatever
    public static void doSomething() {
        String actionMethod = request.actionMethod;
        Class<? extends play.db.Model> model = getControllerAnnotation(CRUD.For.class).value();

        List<String> allowedActions = new ArrayList<String>();
        allowedActions.add("show");
        allowedActions.add("list");
        allowedActions.add("blank");

        if (allowedActions.contains(actionMethod)) {
            List<SuperClass> list = play.db.jpa.JPQL.instance.find(model.getSimpleName()).fetch();
        }
    }
}

I'm not sure about doSomething() approach is truly nice and Java-style/Play!-style. But it works for me. Please tell me if it's possible to catch out the model's class in more native way.

回答2:

"and table per class is an optional feature of the JPA spec, so not all providers may support it" from WikiBook.

Why don't you use @MappedSuperclass? Furthermore you should extend GenericModel. In your example you defined id twice, which could be the reason of you problem too.