如何从playframework中的超类继承模型 [英] How to inherit a model from superclass in playframework

问题描述

我试图了解继承如何发挥作用！但是还没有成功。

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

所以，我有这样的超类：

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;
}

2个继承的类：

@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;
    }
}

此外，我还有2个用于继承类的控制器：

Also I have 2 controllers for inherited classes:

public class Subs and Sub1s extends CRUD {

}

应用程序启动后，我在MySQL数据库中为我的模型（Sub和Sub1）收到了2个表格，结构如下： id bigint（20），名称 varchar（255）。没有 testVal ，它位于超类中。

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.

当我尝试在CRUD界面中创建 Sub 类的新对象时我收到了这样的错误：
模板{module：crud} /app/views/tags/crud/form.html中发生了执行错误。引发的异常是 MissingPropertyException：没有这样的属性：testVal for class：models.Sub。

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（第64行）
＃{crud.numberField名称：field.name，value :( currentObject？currentObject [field.name]：null）/}

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

1. 如何正确生成继承模型的MySQL表并修复错误？


2. 是否可以为几个继承的类设置一个superController？

推荐答案

好吧，多亏了 sdespolit ，我做了一些实验。这就是我所拥有的：

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

超类：

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

继承类：

@Entity 
public class Sub extends SuperClass {
}

我以这种方式制作的超级控制器：

"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）用于告诉拦截器它应该工作的类

@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();
        }
    }
}

我不确定 doSomething（）方法非常好用Java风格/ Play！风格。但它对我有用。
请告诉我是否有可能以更本土的方式了解模特的课程。

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.

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