如何自定义 spring rest 控制器的 json 输出

Question

我有一个具有以下属性的 POJO 类患者：public class Patient implements Serializable{    private static final long serialVersionUID = 2L;    private long id;    private String name;    private Date dob;    private String phoneNo;    private String email;    private Address address;    private String username;    private String password;....现在，从我的休息控制器中，我只需要在 json 中发送患者的姓名、电话号码、电子邮件和地址。我希望 json 输出为{   "check":"Success",   "details":{      "name":"Test User",      "phoneNo":"9876544321",      "email":"[e-mail]",      "address":"Address"   }}此处检查成功/失败仅作为标志添加。

Answer 1

Spring boot 使用 Jackson 进行 JSON 序列化和反序列化，尝试使用 @JSONIgnore (com.fasterxml.jackson.annotation.JsonIgnore)。

Answer 2

为您的要求

PatientDTO.java

public class PatientDTO {

    private Check check;

    @JsonIgnoreProperties(value = {"id", "dob", "username", "password"})

    private Object details;

    /* Getter & Setter */

    public enum Check {

        SUCCESS("Success"),

        FAILURE("Failure");

        private String name;

        Check(String name) {

            this.name = name;

        }

        @JsonValue

        public String getName() {

            return name;

        }

    }

}

控制器演示：

@RestController

@RequestMapping("/test")

public class TestController {

    @GetMapping("/patient")

    public PatientDTO getPatient() {

        PatientDTO patientDTO = new PatientDTO();

        patientDTO.setCheck(PatientDTO.Check.SUCCESS);

        patientDTO.setDetails(new Patient());

        return patientDTO;

    }

}

更好的方法

使用 http 状态

@JsonInclude(JsonInclude.Include.NON_NULL)

public class Patient {

    private long id;

    private String name;

    private Date dob;

    private String phoneNo;

    private String email;

    private Address address;

    private String username;

    private String password;

    /* Getter & Setter */

}

控制器演示：

@RestController

@RequestMapping("/test")

public class TestController {

    @GetMapping("patient")

    // or @ResponseStatus(HttpStatus.OK)

    public ResponseEntity<Patient> patient() {

        Patient patient = new Patient();

        patient.setId(123);

        patient.setName("123");

        patient.setEmail("[e-mail]");

        patient.setPassword(null); // set to null to ignore password

        return ResponseEntity.ok(patient);

    }

}

Answer 3

只需创建另一个对象并将其用作您的 restful 控制器的响应；

public class PatientResponse implements Serializable {

    private static final long serialVersionUID = 2L;

    private Check check;

    private Detail details;

    // getter, setter, etc

    public static class Detail {

        private String name;

        private String phoneNo;

        private String email;

        private String address;

        // getters, setters, etc

    }

    public enum Check {

        Success, Failure

    }

}

& 在控制器中

@RestController

public class PatientController {

    @GetMapping(...)

    public PatientResponse get(...) {

        Patient patient = ... // get patient somehow

        return mapPatientToResponse(patient);  // map Patient to PatientResponse here

    }

}