Spring CrudRepository - 如何通过外键 ID 插入记录？

Question

当使用 post 请求外键相关引用记录插入记录时未链接。

@RestController
@RequestMapping("auth")
public class PatientController {

    @Autowired
    private PatientService patientService;  

    @PostMapping(value = "patient/register", consumes = MediaType.APPLICATION_JSON_VALUE, produces = MediaType.APPLICATION_JSON_VALUE)
    public String registerPatient(@RequestBody Patient patient) {   
        String response = patientService.registerPatient(patient);
        return "{'result':" + response + "}";
    }
}

@Service
public class PatientService {

    @Autowired
    private PatientRepository patientRepo;  

    public String registerPatient(Patient patient) {                
        patient = patientRepo.save(patient);            
    }
}

@Repository
public interface PatientRepository extends CrudRepository<Patient, Integer> {

}

实体类:

@Entity
@Table(name = "patient")
public class Patient implements java.io.Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "patient_id")
    private int patientId;  

    @Column(name = "patient_name", length = 200) 
    private String patientName; 

    @Column(name = "problem", length = 200) 
    private String problem;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "doctor_id", nullable = false, insertable = false, updatable = false)
    private Doctor doctor;  

}

@Entity
@Table(name = "doctor")
public class Doctor implements java.io.Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "doctor_id")
    private int doctorId;   

    @Column(name = "doctor_name", length = 200) 
    private String doctorName;  

    @Column(name = "department", length = 200) 
    private String department;

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "doctor")
    private Set<Patient> patients = new HashSet<Patient>(0);

}

数据库-医生Table: doctor_id doctor_name 部门 12345678 Dfirstname Dlastname ENT

POST 请求 - JSON 正文 { "patientName":"Pfirstname Plastname", "problem:"能见度问题 - 弱光下的困难", "doctor":{"doctorId":"12345678"} }

当我发送此请求时，患者 table doctor_id 列未填充 docortId。

Answer 1

乍一看（因为没有提供服务层）你必须从 @JoinColumn 中删除 insertable=false 和 updatable=false

@JoinColumn(name = "doctor_id", nullable = false, insertable = false, updatable = false)

将其更改为：

@JoinColumn(name = "doctor_id", nullable = false)

因为这个指令不允许 jpa insert/update DOCTOR_ID 列

此外，我更喜欢使用 werappers 而不是原始类型，因为 @Id 将 int 更改为 Integer，如此处建议 Using wrapper Integer class or int primitive in hibernate mapping

而且你似乎已经坚持doctor（因为它已经分配了id）你应该首先select医生到数据库并在两端添加病人：

public void assignToDoctor(Doctor doctor) {
        doctor.patients.add(this);
        this.doctor = doctor;
}

这里是完整的例子：

    public static void main(String[] args) {
        SpringApplication.run(DemostackApplication.class, args);
    }


    @Component
    public static class AppRunner implements ApplicationRunner {

        @Autowired
        MainService mainService;

        @Override
        public void run(ApplicationArguments args) throws Exception {
            Doctor doctor = new Doctor();
            doctor.department = "a";
            doctor.doctorName = "Covid19 Ninja";
            doctor = mainService.saveDoctor(doctor);

            Patient patient = new Patient();
            patient.patientName = "test";
            patient.problem = "test";
            patient.assignToDoctor(doctor);
            Patient newPatient = mainService.savePatient(patient);
        }
    }

    @Service
    public static class MainService {
        @Autowired
        DoctorRepo doctorRepo;
        @Autowired
        PatientRepo patientRepo;

        @Transactional
        public Doctor saveDoctor(Doctor doctor) {
            return doctorRepo.save(doctor);
        }

        @Transactional
        public Patient savePatient(Patient patient) {
            return patientRepo.save(patient);
        }
    }

    @Entity
    @Table(name = "patient")
    public static class Patient implements java.io.Serializable {
        @Id
        @GeneratedValue(strategy = GenerationType.AUTO)
        @Column(name = "patient_id")
        private Integer patientId;

        @Column(name = "patient_name", length = 200)
        private String patientName;

        @Column(name = "problem", length = 200)
        private String problem;

        @ManyToOne(fetch = FetchType.LAZY)
        @JoinColumn(name = "doctor_id", nullable = false)
        private Doctor doctor;

        public void assignToDoctor(Doctor doctor) {
            doctor.patients.add(this);
            this.doctor = doctor;
        }
    }

    @Entity
    @Table(name = "doctor")
    public static class Doctor implements java.io.Serializable {
        @Id
        @GeneratedValue(strategy = GenerationType.AUTO)
        @Column(name = "doctor_id")
        private Integer doctorId;
        @Column(name = "doctor_name", length = 200)
        private String doctorName;

        @Column(name = "department", length = 200)
        private String department;

        @OneToMany(fetch = FetchType.LAZY, mappedBy = "doctor")
        private Set<Patient> patients = new HashSet<Patient>(0);
    }

我没有用过 getter/setters 但你应该:)

编辑

你的 registerPatient() 逻辑应该是这样的：

    @Transactional
    public String registerPatient(Patient patient) {
         Integer doctorId= patinet.getDoctor().getId();
         //fetch the doctor from database
         Doctor doctor = doctorRepository.findById(doctorId).orElseThrow(() -> new RuntimeException("doctor not found"));
         //create bidirectional reference between patient and doctor
         patient.setDoctor(doctor);
         doctor.getPatients().add(patient);
         //save patient
         patient = patientRepo.save(patient);
         return "OK";
    }

Spring CrudRepository - 如何通过外键 ID 插入记录？

Spring CrudRepository- How do I insert a record by foreign key ID?

java

hibernate-mapping

spring-data-jpa

spring-boot