Spring 使用 @ResponseBody 返回一个修改后的 JSONObject

Question

我正在使用 Spring MVC 并尝试从我的控制器返回一个 JSONObject 作为响应。我用 @ResponseBody 注释了该方法，以便它将控制器返回的 JSONObject 放入 ResponseBody。 这是我的控制器：

@GetMapping(value="/student/{roll}",produces="application/json")
@ResponseBody
private JSONObject getStudentDetails(@PathVariable(value="roll") String roll) {
    JSONObject response = new JSONObject();
    Student student = studentDAO.getStudent(roll);
    response.put("firstName",student.getFirstName());
    response.put("lastName",student.getLastName());
    response.put("roll",student.getRoll());
    response.put("email",student.getEmail());
    response.put("course",student.getCourse());
    response.put("stream",student.getStream());
    response.put("year",student.getYear());
    response.put("gender",student.getGender());
    String date = null;
    DateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
    try {
        date = String.valueOf(df.parse(student.getSignUpDate()).getTime());
    } catch (ParseException e) {
        e.printStackTrace();
    }
    response.put("signUpDate", date);
    System.out.println("Response Body::::: "+response.toString());
    return response;
}

一个有效的响应应该是这样的：

{
  "firstName": "John",
  "lastName": "Doe",
  "gender": "M",
  "stream": "cse",
  "year": 3,
  "roll": "2013BT2011",
  "course": "btech",
  "signUpDate": "1476224877000",
  "email": "john@doe.com"
}

但我得到了这个：

{
  "map": {
    "firstName": "John",
    "lastName": "Doe",
    "gender": "M",
    "stream": "cse",
    "year": 3,
    "roll": "2013BT2011",
    "course": "btech",
    "signUpDate": "1476224877000",
    "email": "john@doe.com"
  }
}

这里，我的控制器返回的对象被包装成一个ma​​p对象，然后由Spring返回。

谁能告诉我这里出了什么问题。 任何帮助，将不胜感激。 :)

Answer 1

spring mvc 使用 jackson databind 将 Object 序列化为 JSON /将 JSON 反序列化为 Object。 因此不需要使用@ResponseBody 返回 JSONObject。 有一些办法：

用需要字段定义一个类（视图对象），然后 new ，填充一个实例并返回它。

使用 java.util.Map。所以你的代码将如下所示：

@GetMapping(value="/student/{roll}",produces="application/json")
@ResponseBody
private Map<String, Object> getStudentDetails(@PathVariable(value="roll") String roll) {
      Map<String, Object> response = new HashMap<>();
  Student student = studentDAO.getStudent(roll);
  response.put("firstName",student.getFirstName());
  response.put("lastName",student.getLastName());
  response.put("roll",student.getRoll());
  response.put("email",student.getEmail());
  response.put("course",student.getCourse());
  response.put("stream",student.getStream());
  response.put("year",student.getYear());
  response.put("gender",student.getGender());
  String date = null;
  DateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
  try {
    date = String.valueOf(df.parse(student.getSignUpDate()).getTime());
  } catch (ParseException e) {
    e.printStackTrace();
  }
  response.put("signUpDate", date);
  System.out.println("Response Body::::: "+response.toString());
  return response;
}

您可以在 Student 类 (https://github.com/FasterXML/jackson-annotations/wiki/Jackson-Annotations) 上返回带有正确 Jackson 注释的 Student(PO)。如果要限制返回字段，可以添加JsonView注解。

public class Student {
  public static class Response {}
  private String firstName;
  private String lastName;
  private String roll;
  private String email;
  private String course;
  private String stream;
  private String year;
  private String gender;

  private Date getSignUpDate;

  @JsonView(Response.class)
  public String getFirstName() {
    return firstName;
  }

  @JsonView(Response.class)
  public String getLastName() {
    return lastName;
  }

  @JsonView(Response.class)
  public String getRoll() {
    return roll;
  }

  @JsonView(Response.class)
  public String getEmail() {
    return email;
  }

  @JsonView(Response.class)
  public String getCourse() {
    return course;
  }

  @JsonView(Response.class)
  public String getStream() {
    return stream;
  }

  @JsonView(Response.class)
  public String getYear() {
    return year;
  }

  @JsonView(Response.class)
  public String getGender() {
    return gender;
  }

  @JsonView(Response.class)
  @JsonFormat(pattern="yyyy-MM-dd HH:mm:ss")
  public Date getGetSignUpDate() {
    return getSignUpDate;
  }
  /// setters are not written.
}


public class StudentCtl {

  @GetMapping(value="/student/{roll}",produces="application/json")
  @ResponseBody
  @JsonView(Student.Response.class)
  private Map<String, Object> getStudentDetails(@PathVariable(value="roll") String roll) {
    return studentDAO.getStudent(roll);
  }
}

Answer 2

假设您已经有具有这些属性的学生班级。因此，您可以做的就是简单地返回对象，如下所示，希望这应该可以工作。

private JSONObject getStudentDetails(@PathVariable(value="roll") String roll) {
    Student student = studentDAO.getStudent(roll);
    System.out.println("Response Body::::: "+response.toString());
    return student ;
}

即使你想更改这些字段的名称，你也可以像这样注释它们

class Student {
 @JsonProperty("firstName")
 private String fname;

 @JsonProperty("signUpDate") 
 @JsonFormat(shape=JsonFormat.Shape.STRING, pattern="yyyy-MM-dd HH:mm:ss") 
  private Date date; 
 //getters
 //setters
}

Answer 3

您可以简单地从 Controller 方法中返回 student，spring 将进行编组。

   @GetMapping(value="/student/{roll}",produces="application/json")
    @ResponseBody
    private Student getStudentDetails(@PathVariable(value="roll") String roll) {
       return studentDAO.getStudent(roll);            
    }

但由于您正在进行日期格式设置，我还建议您创建一个单独的类，例如 StudentView 并将学生映射到 StudentView

  class StudentView{
    private String signUpDate;
    ......
    ......
    }

并从控制器返回 studentView。所以它会是这样的

@GetMapping(value="/student/{roll}",produces="application/json")
@ResponseBody
private StudentView getStudentDetails(@PathVariable(value="roll") String roll) {
   Student student=studentDAO.getStudent(roll);            
   return toStudentView(student);
}

private static StudentView toStudentView(Student stu){
  ......
}