Question

我使用Spring HATEOAS在我的应用程序中创建REST HATEOAS API。它到目前为止运行良好，但在嵌套资源方面我很困难。将像这样的类层次结构映射到REST HATEOAS资源的正确方法是什么：

public class MyEntity {

    private int id;

    private List<ChildEntity> children;

}


public class ChildEntity {

    private int id;

    private AnotherEntity entity;

}


public class AnotherEntity {
}

我为所有这些实体创建了Resource类，但是在序列化MyEntity时，所有包含的实体都被序列化为POJO，尽管我也需要将它们序列化为资源（使用链接等）。有没有办法将资源添加到父资源（而不是使用Resources类）？或者我是否必须向子项添加@JsonIgnore，然后在我的ResourceAssembler中手动添加子项作为资源？那么使用ResourceSupport代替资源会更有意义吗？

Answer 1

扩展ResourceSupport：

public class MyEntityResource extends ResourceSupport {

    private int identificator;

    private List<ChildEntityResource> children;

    public int getIdentificator() {
        return identificator;
    }

    public void setIdentificator(int id) {
        this.identificator = identificator;
    }

    public List<ChildEntityResource> getChildren() {
        return children;
    }

    public void setChildren(List<ChildEntityResource> children) {
        this.children = children;
    }

}

public class ChildEntityResource extends ResourceSupport {

    private int identificator;

    private AnotherEntityResource entity;

    public int getIdentificator() {
        return identificator;
    }

    public void setIdentificator(int identificator) {
        this.identificator = identificator;
    }

    public AnotherEntityResource getEntity() {
        return entity;
    }

    public void setEntity(AnotherEntityResource entity) {
        this.entity = entity;
    }
}

public class AnotherEntityResource extends ResourceSupport {

    private String value;


    public String getValue() {
        return value;
    }

    public void setValue(String value) {
        this.value = value;
    }
}

@RestController
public class EntityController {
    @RequestMapping(value = "/entity", method = RequestMethod.GET)
    public ResponseEntity<MyEntityResource> index() {

        AnotherEntityResource anotherEntityResource  = new AnotherEntityResource();
        anotherEntityResource.add(new Link("link-for-another-entity-resource", "rel-1"));

        anotherEntityResource.setValue("value for Another Entity","rel-2");

        ChildEntityResource childEntityResource = new ChildEntityResource();
        childEntityResource.setIdentificator(20);
        childEntityResource.setEntity(anotherEntityResource);
        childEntityResource.add(new Link("link-for-child-entity-resource", "rel-3"));

        MyEntityResource entityResource = new MyEntityResource();

        entityResource.setIdentificator(100);
        entityResource.setChildren(Arrays.asList(childEntityResource));
        entityResource.add(new Link("link-for-entity-resource"));

        return new ResponseEntity<MyEntityResource>(entityResource,  HttpStatus.OK);
    }

}

@SpringBootApplication
public class Application {

    public static void main(String[] args) {
         SpringApplication.run(Application.class, args);
    }

}

结果是：

{
     "identificator": 100,
     "children": [
         {
             "identificator": 20,
             "entity": {
                 "value": "value for Another Entity",
                 "_links": {
                     "rel-1": {
                         "href": "link-for-another-entity-resource"
                    }
                }
            },
            "_links": {
                "rel-2": {
                    "href": "link-for-child-entity-resource"
                } 
            }
        }
    ],
    "_links": {
        "rel-3": {
            "href": "link-for-entity-resource"
        }
    }
}

但是你必须考虑这是否是连接不同资源的正确选择。除非您在控制器中提供方法获得此嵌入资源，否则您将无法单独访问它们。一个解决方案是使用HAL。使用HAL，您可以使用_links属性指向资源，或将此资源嵌入_embedded属性。

REST HATEOAS：如何序列化嵌套资源（使用Spring HATEOAS）？

1 个答案: