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java - 更新@OneToMany 关系（Spring Boot、Spring Data）

问题描述

我有一个OneToMany关系（两个表双向）。当我保存医生的专业时，它确实有效，但是当我删除任何专业并更新医生时，它不起作用。

医生

@Entity
@Table(name = "doctors")
public class Doctor implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer doctorId;

    @Column(length = 20)
    private String doctorName;

    @Column(length = 9)
    private String doctorPhoneNo;

    @Column(length = 30)
    private String doctorEmailAddress;

    private String doctorProfileImage;

    @Enumerated(EnumType.STRING)
    private Status status;

    @Column(length = 6)
    private String doctorCmp;

    @OneToMany(mappedBy = "doctor", cascade = CascadeType.ALL)
//  @JsonIgnore
    private Set<DoctorSpecialties> doctorSpecialties;

    public Doctor() {
        this.doctorSpecialties = new HashSet<>();
    }

    public Doctor(Integer id){
        this();
        this.doctorId = id;
    }

    // getters y setters
}

专业

@Entity
@Table(name = "specialties")
public class Specialty implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer specialtyId;

    private String specialtyName;

    @OneToMany(mappedBy = "specialty")
    @JsonIgnore
    private Set<DoctorSpecialties> doctorSpecialties;

    public Specialty() {
    }

    public Specialty(Integer id) {
        this.specialtyId = id;
    }

    // getters and setters
  }

博士专科

@Entity
@Table(name = "doctor_specialties")
public class DoctorSpecialties implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer id;

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

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "specialty_id")
    private Specialty specialty;

    @OneToMany
    @JoinColumn(name = "doctor_specialties_id")
    private Set<Appointment> appointments;

    @OneToMany
    @JoinColumn(name = "doctor_specialties_id")
    private Set<DoctorSchedule> schedules;

    public DoctorSpecialties(){

    }

    public DoctorSpecialties(Specialty specialty, Doctor doctor){
        this.specialty = specialty;
        this.doctor = doctor;
    }
      getters / setters
 }

控制器

@PostMapping(value = "/saveSpecialties/{id}")
public String saveDoctorSpecialties(@RequestParam(required = false) String[] specialtiesId,
                                    @PathVariable Integer id, RedirectAttributes message) {

    if (id != null && id > 0) {
        Doctor doctor = doctorService.findOne(id);

        if (doctor != null) {

            // It does not work
            doctor.getDoctorSpecialties().forEach(ds -> doctorSpecialtiesService.delete(ds.getId()));

            doctor.getDoctorSpecialties().clear();

            if (specialtiesId != null) {
                for (String specialtyId : specialtiesId) {
                    DoctorSpecialties ds = new DoctorSpecialties();
                    ds.setSpecialty(new Specialty(Integer.parseInt(specialtyId)));
                       
                    ds.setDoctor(doctor);

                    doctor.getDoctorSpecialties()
                            .add(ds);
                }
                
            }
            doctorService.update(doctor);
            message.addFlashAttribute("success", "Specialties successfully saved.");
            return "redirect:/doctors/profile/{id}/specialties";
        }
    }
    // specialtiesId = new String[]{};
    message.addFlashAttribute("error", "Doctor doesn't exists");
    return "redirect:/doctors/list";
}

安慰：

2021-10-30 21:19:13.330 DEBUG 44504 --- [nio-8080-exec-7] org.hibernate.SQL：选择医生0_.doctor_id 作为医生_i1_3_0_，医生0_.doctor_cmp 作为医生_c2_3_0_，医生0_.doctor_email_address 作为医生_e3_3_0_，医生0_ .doctor_name 作为医生_n4_3_0_，医生0_.doctor_phone_no 作为医生_p5_3_0_，医生0_.doctor_profile_image 作为医生_p6_3_0_，医生0_.status 作为状态7_3_0_ 来自医生医生0_ where doctor0_.doctor_id=? 2021-10-30 21:19:13.339 DEBUG 44504 --- [nio-8080-exec-7] org.hibernate.SQL：选择医生规范0_.doctor_id 作为医生_i2_2_0_，医生规范0_.id 作为id1_2_0_，医生规范0_.id 作为id1_2_1_，医生规范0_ .doctor_id 作为医生_i2_2_1_，医生spec0_.specialty_id 作为specialt3_2_1_ from doctor_specialties doctorspec0_ where doctorspec0_.doctor_id=? 2021-10-30 21:19:13。401 调试 44504 --- [nio-8080-exec-8] org.hibernate.SQL：选择医生0_.doctor_id 作为医生_i1_3_0_，医生0_.doctor_cmp 作为医生_c2_3_0_，医生0_.doctor_email_address 作为医生_e3_3_0_，医生0_.医生名称作为医生_n4_3_0_，医生0_.医生电话号码作为医生_p5_3_0_，医生0_.医生个人资料_图像作为医生_p6_3_0_，医生0_.状态作为状态7_3_0_来自医生医生0_哪里医生0_.医生ID=？2021-10-30 21:19:13.404 DEBUG 44504 --- [nio-8080-exec-8] org.hibernate.SQL：选择 special0_.specialty_id 作为specialt1_7_0_，doctorspec1_.id 作为 id1_2_1_，doctor2_.doctor_id 作为医生_i1_3_2_，specialty0_ .specialty_name 为 specialt2_7_0_，doctorspec1_.doctor_id 为 doctor_i2_2_1_，doctorspec1_.specialty_id 为 specialt3_2_1_，doctorspec1_.specialty_id 为 specialt3_2_0__，doctorspec1_.id 为 id1_2_0__，doctor2_.doctor_cmp as doctor_c2_3_2_, doctor2_.doctor_email_address as doctor_e3_3_2_, doctor2_.doctor_name as doctor_n4_3_2_, doctor2_.doctor_phone_no as doctor_p5_3_2_, doctor2_.doctor_profile_image as doctor_p6_3_2_, doctor2_.status as status7_3_2_ from specialties specialty0_ inner join doctor_specialties doctorspec1_ on specialty0_.specialty_id=doctorspec1_.specialty_id内部加入医生 doctor2_ on doctorspec1_.doctor_id=doctor2_.doctor_id where doctor2_.doctor_id=? 2021-10-30 21:19:13.565 调试 44504 --- [nio-8080-exec-4] org.hibernate.SQL：选择 special0_.specialty_id 作为specialt1_7_0_，doctorspec1_.id 作为 id1_2_1_，doctor2_.doctor_id 作为医生_i1_3_2_，specialty0_ .specialty_name 为 specialt2_7_0_，doctorspec1_.doctor_id 为 doctor_i2_2_1_，doctorspec1_.specialty_id 为 specialt3_2_1_，doctorspec1_。

没有删除语句...

------------------------ 编辑 1 ------------

Doctor find = doctorRepository.findById(1).get();
    
DoctorSpecialties ds1 = new DoctorSpecialties();
ds1.setSpecialty(specialtyRepository.findById(1).get());
ds1.setDoctor(find);
    
DoctorSpecialties ds2 = new DoctorSpecialties();
ds2.setSpecialty(specialtyRepository.findById(2).get());
ds2.setDoctor(find);
    
find.getDoctorSpecialties().add(ds1);
find.getDoctorSpecialties().add(ds2);

doctorRepository.save(find);

我做了一些测试，我不能完全理解。我做到了，实际上我添加两个对象时它只保存一次。

insert into doctor_specialties (id, doctor_id, specialty_id) values (null, ?, ?)

------------------------ 编辑 2 ------------

DoctorSpecialties （修改构造函数）

@Entity
@Table(name = "doctor_specialties")
public class DoctorSpecialties implements Serializable {

    public DoctorSpecialties(Integer specialtyId, Doctor doctor) {
        this.specialty = new Specialty(specialtyId);
        this.doctor = doctor;
    }

 }

控制器

@PostMapping(value = "/saveSpecialties/{id}")
public String saveDoctorSpecialties(@RequestParam(required = false) String[] specialtiesId,
                                    @PathVariable Integer id, RedirectAttributes message) {
    if (id != null && id > 0) {

        doctorService.saveDelete(id);

        Doctor doctor = doctorService.findOne(id);
        if (specialtiesId != null && specialtiesId.length > 0) {
            for(String idSpecialty : specialtiesId){
                doctorSpecialtiesService.save(new DoctorSpecialties(Integer.parseInt(idSpecialty), doctor));
            }
        }
        message.addFlashAttribute("success", "Specialties successfully saved.");
        return "redirect:/doctors/profile/{id}/specialties";
    }
    message.addFlashAttribute("error", "Doctor doesn't exists");
    return "redirect:/doctors/list";
}

服务

@Override
@Transactional
public void saveDelete(Integer doctorId) {
    Doctor doctor = this.doctorRepository
            .findById(doctorId).get();
    doctor.getDoctorSpecialties().clear();
}

安慰：

select doctor0_.doctor_id as doctor_i1_3_0_, doctor0_.doctor_cmp as doctor_c2_3_0_, doctor0_.doctor_email_address as doctor_e3_3_0_, doctor0_.doctor_name as doctor_n4_3_0_, doctor0_.doctor_phone_no as doctor_p5_3_0_, doctor0_.doctor_profile_image as doctor_p6_3_0_, doctor0_.status as status7_3_0_ from doctors doctor0_ where doctor0_.doctor_id= ? 选择医生规范0_.doctor_id 作为医生_i2_2_0_，医生规范0_.id 作为id1_2_0_，医生规范0_.id 作为id1_2_1_，医生规范0_.doctor_id 作为医生_i2_2_1_，医生规范0_.specialty_id 作为specialt3_2_1_ from doctor_specialties doctorspec0_ where doctorspec0_.doctor_id=?

更新约会设置医生_专业_id = null where doctor_specialties_id =？

更新医生_时间表设置医生_专科_id = null where doctor_specialties_id =？

从 id=? 的医生专业中删除

标签： javaspring-bootspring-data-jpa

解决方案

对于事务中的一对多关系，一旦您获得父级（Doctor）并遍历其子级（DoctorSpecialties）[换句话说，一旦您将整个父级及其子级加载到持久状态]，您将无法DoctorSpecialties直接删除你的repository.

您可以尝试类似下面的示例来查看它：

@Transactional
public void removeLine(Long doctorId, Long specId) {
    Doctor doctor = this.doctorRepository  // (1)
        .findById(doctorId)
        .orElseThrow(IllegalArgumentException::new);
    this.doctorSpecialtiesRepository.deleteById(specId); // (2)
}

在 (1) 处，我们加载doctor到持久状态。所以在这里，如果fetch = FetchType.EAGER，这意味着它将加载doctor并全部doctorSpecialties进入持久状态，并且这个原因（2）不会给你任何影响。

否则，如果fetch = FetchType.LAZY，它只加载doctor到持久状态，并且在（2）它将被成功删除。

尽管您的情况类似，但是您通过使用并加载到持久状态fetch = FetchType.LAZY来遍历子级。这就是为什么你不能删除它们。forEachdoctorSpecialties

建议：orphanRemoval = true在您的父实体中使用

@OneToMany(mappedBy = "doctor", cascade = CascadeType.ALL, orphanRemoval = true)
    private Set<DoctorSpecialties> dss;

只需在您的方法中清除其子项（在 @Transactional 方法中）

doctor.getDoctorSpecialties().clear();

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