CODEWITHSUNDEEP
springhibernatespring-bootjpa
Maifee Ul Asad
Maifee Ul Asad

Reputation: 4607

TransientPropertyValueException in Spring Boot

I have LoginCredential class:

@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;
}

And here is my User class:

@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;
}

And my LoginCredentialController's POST-ing method is simple :

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

Now when I tried this command : curl -X POST -H "Content-Type: application/json" -d "{ \"email\": \"1\"}" http://localhost:8080/login

I get a LoginCredential without any error, to be mentioned user field is null.

That's why I tried this command curl -X POST -H "Content-Type: application/json" -d "{ \"email\": \"1\",\"user\":{} }" http://localhost:8080/login

Which gives me error saying :

{
  "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

Upvotes: 1

Views: 2085

Answers (3)

buræquete
buræquete

Reputation: 14688

You have three problems

  1. Lacking a cascade option to trigger User entity creation from LoginCredential save.
  2. Lacking @MapsId annotation on User so that they share the same id, otherwise LoginCredential and its created User will have different id values for both of them has @GeneratedValue(strategy = GenerationType.AUTO) on their @Id columns
  3. Not setting both sides of the relationship...

To fix all you need to change your entities to below (I also removed some useless annotations and values);

@Data
@Entity
public class User {

    @Id
    Long userID;

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

and 

@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;
}

Also need to set both sides of the relationship before finalizing your endpoint;

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

Upvotes: 2

Shardendu
Shardendu

Reputation: 3588

Make User class Serializable as well.

Upvotes: 0

MyTwoCents
MyTwoCents

Reputation: 7624

You don't need @GeneratedValue for User object

Add CascadeType for OneToOne

By doing this you tell hibernate to save User to the database when saving LoginCredential.

@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;
}

Refer this for more detail

Upvotes: 0

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