用于rest Get调用的POJO，以返回java中嵌套的复杂json。

Question

嗨，我需要工作流(如何创建pojo结构和如何通信)在json结构下面创建REST调用。

请帮帮我

[
  {
   "id":"1", 
   "Nmae":"gourav",
    "Gender":"Male",
     "usertype":{
                 "Typeone":"Admin",
                 "Status":"Active"
                 }
                {
                  "Typetwo":"Agent",
                  "Status":"Disabled"
                  }
  },
  {
   "id":"2", 
   "Nmae":"satya",
    "Gender":"Male",
     "usertype":{
                 "Typeone":"Admin",
                 "Status":"disabled"
                 }
                {
                  "Typetwo":"Agent",
                  "Status":"active"
                  }
  }
]

Answer 1

你可以这样做，

控制器类，您将在其中定义您的路线-

@GET
@Path("/users")
@Produces({ javax.ws.rs.core.MediaType.APPLICATION_JSON})
public Response getUser(){

    // Create Object of your DAO or Service class and use getUserDetails() to return a list of users, maybe from the Database.
    List<User> usersList = serviceExampleObject.getUserDetails();

    //Set the list field to the response object
    return Response.status(Status.OK).entity(usersList).build();
}

如果您想遵循SpringBoot的方法，您的控制器将是-

@RestController
public class UserController {

    @Autowired
    private UserService userService;

    @GetMapping("/users")
    public ResponseEntity<?> getUsers(){

        List<Users> usersList = userService.getUsersDetail();

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

创建一个用户PoJo (没有添加所有的getter/setter来缩短发布)-

class User{

    private Integer id;
    private String name;
    private String gender;
    private List<UserType> userType;

    public Integer getId() {
        return id;
    }
    ...
    ...

    public void setUserType(List<UserType> userType) {
        this.userType = userType;
    }
}

创建一个UserType PoJo -

class UserType{
    private String typeNumber;
    private String typeRole;
    private String status;

    public String getTypeNumber() {
        return typeNumber;
    }
    ...
    ...

    public void setStatus(String status) {
        this.status = status;
    }
}

另外，还需要修改json响应。在userType字段中，可以有不同名称的字段，因此，我建议遵循以下方法：

"usertype": [
    {
        "typeNumber": "one",
        "typeRole": "Admin",
        "status": "Active"
    },
    {
        "typeNumber": "two",
        "typeRole": "Agent",
        "status": "Disabled"
    }
]

Answer 2

您可以使用您想要的字段创建一个名为User的类，并使用javax.ws.rs.core.Response，如下所示：

@Produces({ javax.ws.rs.core.MediaType.APPLICATION_JSON})
Response getUser(){
    User user = new User();
    //set fields
    return Response.status(Status.OK).entity(user).build();
}

Answer 3

您只需创建两个类，一个用于user，另一个用于UserType

Public Class UserType {

  private String type;
  private String status;

//Constructor, Getters and Setters, etc.
}




Public Class User {

  private long id;
  private String name;
  private String gender;
  private UserType userType;

//Constructor, Getters and Setters, etc.
}

然后，当您返回User时，它还将返回与您前面描述的json对应的UserType。它应该是这样的：

@RequestMapping(value = "/url", method = RequestMethod.GET)
public User getUser(PARAMS_IF_ANY) {

  // Application logic to find the user
  return user;

}

这招就完成了