我有一個具有以下屬性的 POJO 類患者：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;....現在，從我的休息控制器中，我只需要在 json 中發送患者的姓名、電話號碼、電子郵件和地址。我希望 json 輸出為{ "check":"Success", "details":{ "name":"Test User", "phoneNo":"9876544321", "email":"[email protected]", "address":"Address" }}此處檢查成功/失敗僅作為標志添加。
3 回答
蝴蝶刀刀
TA貢獻1801條經驗 獲得超8個贊
Spring boot 使用 Jackson 進行 JSON 序列化和反序列化，嘗試使用 @JSONIgnore (com.fasterxml.jackson.annotation.JsonIgnore)。
qq_笑_17
TA貢獻1818條經驗 獲得超7個贊
為您的要求
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 狀態
@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("[email protected]");
patient.setPassword(null); // set to null to ignore password
return ResponseEntity.ok(patient);
}
}
明月笑刀無情
TA貢獻1828條經驗 獲得超4個贊
只需創建另一個對象并將其用作您的 restful 控制器的響應；
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
}
}
添加回答
舉報
0/150
提交
取消