返回一些對象的json

Question

我userController.java：

@RestController 
public class UserController { 

    private static final Logger LOGGER = LoggerFactory.getLogger(UserController.class); 
    private final UserService userService; 

    @Inject 
    public UserController(final UserService userService) { 
     this.userService = userService; 
    } 

    @RequestMapping(value = "/user", method = RequestMethod.GET) 
    public List<User> listUsers() { 
     LOGGER.debug("Received request to list all users"); 
     return userService.getList(); 
    } 



    @ExceptionHandler 
    @ResponseStatus(HttpStatus.CONFLICT) 
    public String handleUserAlreadyExistsException(UserAlreadyExistsException e) { 
     return e.getMessage(); 
    } 

} 

User.java：

@Entity 
public class User { 

    @Id 
    @NotNull 
    @Size(max = 64) 
    @Column(name = "id", nullable = false, updatable = false) 
    private String id; 

    @NotNull 
    @Size(max = 64) 
    @Column(name = "name", nullable = false) 
    private String name; 

    @NotNull 
    @Size(max = 64) 
    @Column(name = "firstname", nullable = false) 
    private String firstname; 

    @NotNull 
    @Size(max = 64) 
    @Column(name = "email", nullable = false) 
    private String email; 

    @NotNull 
    @Size(max = 64) 
    @Column(name = "password", nullable = false) 
    private String password; 

    @Transient 
    private List<Events> events; 

    public User() { 
    } 

    public User(String id, String name, String firstname, String email, String password) { 
     super(); 
     this.id = id; 
     this.name = name; 
     this.firstname = firstname; 
     this.email = email; 
     this.password = password; 
    } 

    public void setId(String id) { 
     this.id = id; 
    } 

    public String getId() { 
     return id; 
    } 

    public String getPassword() { 
     return password; 
    } 

    public void setPassword(String password) { 
     this.password = password; 
    } 

    public String getName() { 
     return name; 
    } 

    public void setName(String name) { 
     this.name = name; 
    } 

    public String getFirstname() { 
     return firstname; 
    } 

    public void setFirstname(String firstname) { 
     this.firstname = firstname; 
    } 

    public String getEmail() { 
     return email; 
    } 

    public void setEmail(String email) { 
     this.email = email; 
    } 

    public List<Events> getEvents() { 
     return events; 
    } 

    public void setEvents(List<Events> events) { 
     this.events = events; 
    } 

    @Override 
    public String toString() { 
     return Objects.toStringHelper(this).add("id", id).add("name", name).add("firstname", firstname) 
       .add("email", email).add("password", password).add("events", events).toString(); 
    } 
} 

有可能在我的JSON只是name到retun和firstname

我的pom.xml：

<!-- Spring Boot --> 
<parent> 
    <groupId>org.springframework.boot</groupId> 
    <artifactId>spring-boot-starter-parent</artifactId> 
    <version>1.0.1.RELEASE</version> 
</parent> 
... 
    <dependency> 
     <groupId>org.springframework.boot</groupId> 
     <artifactId>spring-boot-starter</artifactId> 
    </dependency> 

    <dependency> 
     <groupId>org.springframework.boot</groupId> 
     <artifactId>spring-boot-starter-test</artifactId> 
     <scope>test</scope> 
    </dependency> 

    <dependency> 
     <groupId>org.springframework.boot</groupId> 
     <artifactId>spring-boot-starter-web</artifactId> 
    </dependency> 

    <dependency> 
     <groupId>org.springframework.boot</groupId> 
     <artifactId>spring-boot-starter-data-jpa</artifactId> 
    </dependency> 
... 

Answer 1

你可以使用@JsonView控制哪些呈現。你可以在這個春天的博客文章https://spring.io/blog/2014/12/02/latest-jackson-integration-improvements-in-spring

更多細節假設你有這樣的View類接口Summary

public class View { 
    interface Summary {} 
} 

然後，您可以註釋你的屬性，像這樣：

@JsonView(View.Summary.class) 
private String name; 

@JsonView(View.Summary.class) 
private String firstname; 

然後你請求映射：

@JsonView(View.Summary.class) 
@RequestMapping(value = "/user", method = RequestMethod.GET) 

這隻會在生成的JSON中返回name和firstname。