Question

我使用spring jpa数据。在多对多的关系中，我想保存一个关联。

@Entity
public class Lodger {
    ...
    @ManyToMany(cascade = {CascadeType.PERSIST, CascadeType.MERGE}, fetch = FetchType.LAZY, mappedBy = "lodger")
    private List<MedicalContact> medicalContactList;
}

@Entity
public class MedicalContact {
    ...
    @ManyToMany(cascade = {CascadeType.PERSIST, CascadeType.MERGE}, fetch = FetchType.LAZY)
    @JoinColumn(name = "lodger_id")
    private List<Lodger>lodger;
}

我的方法相互关联。

public void associateLodgerAndMedicalContact(Long lodgerId, Long medicalContactId){
    Lodger lodger = lodgerRepository.findOne(lodgerId);

    List<MedicalContact> medicalContactList = lodger.getMedicalContactList();
    if(medicalContactList==null){
        medicalContactList = new ArrayList<>();
    }

    MedicalContact medicalContact = medicalContactRepository.findOne(medicalContactId);
    medicalContactList.add(medicalContact);

    List<Lodger> lodgerList = medicalContact.getLodger();
    if(lodgerList==null){
        lodgerList = new ArrayList<>();
    }
    lodgerList.add(lodger);  

    lodgerRepository.save(lodger)
    medicalContactRepository.save(medicalContact);  
}

我是否需要添加双方的每个对象？

Answer 1

不，你不必双方都做，一方应该做。您需要JoinTable ManyToMany关系而不是JoinColumn

我建议您在Entity类上创建add方法，这样每次想要添加新元素时都不必重复它：

@Entity
public class Lodger {
    ...
    @ManyToMany( mappedBy = "lodger") 
    @JoinTable(name="yourTableName",joinColumns=@JoinColumn("LODGER_ID") , reverseJoinColumn=@JoinColumn(name="MEDICAL_CONTACT_ID"))
    private List<MedicalContact> medicalContactList;

    public void addToMediaContactList(MedicalContact medicalContact){
      if(medicalContact == null){
        //return;  or throw new IllegalArgumentException("MedicalContact can't be null)" or any Other Exception ....
        }
      if(medicalContactList == null){
        this.medicalContactList = new ArrayList<>();
        }
      medicalContactList.add(medicalContact);
    }

}

关于你的关联方法：

public void associateLodgerAndMedicalContact(Long lodgerId, Long medicalContactId){
    Lodger lodger = lodgerRepository.findOne(lodgerId);

    MedicalContact medicalContact =    medicalContactRepository.findOne(medicalContactId);

    lodger.addToMedicalContactList(medicalContact );  

    lodgerRepository.save(lodger)  
}

将关联保存到很多

1 个答案: