Spring Boot 中的 TransientPropertyValueException

Question

我有LoginCredential 类：

@Data
@Entity
public class LoginCredential implements Serializable {
   @Id
   @GeneratedValue(strategy = GenerationType.AUTO)
   Long userID;
   String eMail;
   String passwordHash;
   @GeneratedValue(strategy = GenerationType.AUTO)
   @OneToOne(mappedBy = "loginCredential", fetch = FetchType.LAZY)
   User user;
}

这是我的User 课程：

@Data
@Entity
public class User {
   @Id
   @GeneratedValue(strategy = GenerationType.AUTO)
   Long userID;
   @OneToOne(fetch = FetchType.LAZY,targetEntity = LoginCredential.class)
   @JoinColumn(name = "userID",referencedColumnName = "userID")
   private LoginCredential loginCredential;
}

而我的LoginCredentialController 的POST 方法很简单：

@PostMapping("/login")
LoginCredential newLoginCredential(@RequestBody LoginCredential newLoginCredential) {
    logger.debug(newLoginCredential);
    LoginCredential a=repository.save(newLoginCredential);
    logger.debug(a);
    return a;
}

现在当我尝试这个命令时：curl -X POST -H "Content-Type: application/json" -d "{ \"email\": \"1\"}" http://localhost:8080/login

我得到一个LoginCredential 没有任何错误，要提到的user 字段是null。

这就是我尝试这个命令的原因curl -X POST -H "Content-Type: application/json" -d "{ \"email\": \"1\",\"user\":{} }" http://localhost:8080/login

这给了我错误的说法：

{
  "status" : 500,
  "error" : "Internal Server Error",
  "message" : "org.hibernate.TransientPropertyValueException: object references an unsaved transient instance - save the transient instance before flushing : com.mua.cse616.Model.LoginCredential.user -> com.mua.cse616.Model.User; org.hibernate.TransientPropertyValueException: object references an unsaved transient instance - save the transient instance before flushing : com.mua.cse616.Model.LoginCredential.user -> com.mua.cse616.Model.User",
  "trace":....
}

Application.properties

pom.xml

Answer 1

你有三个问题

1. 缺少cascade 选项来触发User 实体创建从LoginCredential 保存。
2. 在User 上缺少@MapsId 注释以便它们共享相同的ID，否则LoginCredential 及其创建的User 将具有不同的ID 值，因为它们的@GeneratedValue(strategy = GenerationType.AUTO) 列上有@GeneratedValue(strategy = GenerationType.AUTO)
3. 不设置关系的双方...

要解决所有问题，您需要将实体更改为以下（我还删除了一些无用的注释和值）；

@Data
@Entity
public class User {

    @Id
    Long userID;

    @JsonBackReference
    @MapsId
    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "userID", referencedColumnName = "userID")
    @ToString.Exclude
    private LoginCredential loginCredential;
}

和

@Data
@Entity
public class LoginCredential {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    Long userID;
    String eMail;
    String passwordHash;

    @JsonManagedReference
    @OneToOne(mappedBy = "loginCredential", fetch = FetchType.LAZY, cascade = CascadeType.PERSIST)
    private User user;
}

还需要在最终确定端点之前设置关系的双方；

Optional.ofNullable(loginCredential.getUser())
        .ifPresent(user -> user.setLoginCredential(loginCredential));
loginCredentialRepo.save(loginCredential);

Answer 2

用户对象不需要@GeneratedValue

为 OneToOne 添加CascadeType

通过这样做，您告诉 hibernate 在保存 LoginCredential 时将用户保存到数据库中。

@Data
@Entity
public class LoginCredential implements Serializable {

   @Id
   @GeneratedValue(strategy = GenerationType.AUTO)
   Long userID;

   String eMail;

   String passwordHash;

   @OneToOne(cascade = {CascadeType.ALL}, mappedBy = "loginCredential", fetch = FetchType.LAZY)
   User user;
}

请参考more detail

Answer 3

使用户类也可序列化。