java如何根据映射URL从RestController获取实体

Question

我有MyEntity课程：

@Entity
@Table("entities)
public class MyEntity {

     @ID
     private String name;
     @Column(name="age")
     private int age;
     @Column(name="weight")
     private int weight;

     ...getters and setters..

}

在@RestController中有2个@GetMapping方法。 第一：

@GetMapping
public MyEntity get(){
   ...
   return myEntity;
} 

第二点：

@GetMapping("url")   
public List<MyEntity> getAll(){
   ...
   return entities;
}

需要提供：
1.@GetMapping返回实体，如MyEntity类中所述
2.@GetMapping（“url”）返回实体，如它的一个字段带有@JsonIgnore

更新：

当我返回myEntity时，客户端将获得，例如：

{
"name":"Alex",
"age":30,
"weight":70
}

我希望同时使用相同的实体有机会，具体取决于URL发送到客户端：

一,

{
    "name":"Alex",
    "age":30,
    "weight":70
}

二,

{
    "name":"Alex",
    "age":30
    }

Answer 1

您可以创建两个DTO类，将实体转换为适当的DTO类并返回它

public class MyEntity {
    private String name;
    private int age;
    private int weight;

    public PersonDetailedDTO toPersonDetailedDTO() {
        PersonDetailedDTO person = PersonDetailedDTO();
        //...
        return person;  
    }

    public PersonDTO toPersonDTO() {
        PersonDTO person = PersonDTO();
        //...
        return person;  
    }
}

public class PersonDetailedDTO {
    private String name;
    private int age;
    private int weight;
}

public class PersonDTO {
    private String name;
    private int age;
}

@GetMapping
public PersonDTO get() {
   //...
   return personService.getPerson().toPersonDTO();
}

@GetMapping("/my_url")
public PersonDetailedDTO get() {
   //...
   return personService.getPerson().toPersonDetailedDTO();
}

Answer 2

编辑：

您可以将实体对象序列化为映射，其中映射键表示属性名称，而不是返回实体对象。因此，可以根据include参数将值添加到地图中

@ResponseBody
public Map<String, Object> getUser(@PathVariable("name") String name, String include) {

    User user = service.loadUser(name);
    // check the `include` parameter and create a map containing only the required attributes
    Map<String, Object> userMap = service.convertUserToMap(user, include);

    return userMap;
}

As an example, if you have a Map like this and want All Details

userMap.put("name", user.getName());
userMap.put("age", user.getAge());
userMap.put("weight", user.getWeight());

Now if You do not want to display weight then you can put only two parameters

userMap.put("name", user.getName());
userMap.put("age", user.getAge());

---

有用参考123

Answer 3

您还可以使用JsonView注释，这使它更简洁。 定义视图

public class View {
    static class Public { }
    static class ExtendedPublic extends Public { }
    static class Private extends ExtendedPublic { }
}

实体

    @Entity
@Table("entities)
public class MyEntity {

     @ID
     private String name;
     @Column(name="age")
     private int age;
     @JsonView(View.Private.class)
     @Column(name="weight")
     private int weight;

     ...getters and setters..

}

在你的休息控制器里

    @JsonView(View.Private.class)
    @GetMapping
    public MyEntity get(){
       ...
       return myEntity;
    } 

    @JsonView(View.Public.class)
    @GetMapping("url")   
    public List<MyEntity> getAll(){
       ...
      return entities;
    }

这里已经解释过： https://stackoverflow.com/a/49207551/3005093