xHow can I have @Id string for CrudRepository in Spring with Spring Data JPA?

Question

The problem is that I am getting an exception using @RepositoryRestResource for my UserRepository that extends JpaRepository.

The reason for that is that findById is only accepting Long or Int types by default, even I have

@Id String id; and not @Id Int id in my entity definition.

I have tried searching StackOverflow and Google, but haven't found any solutions.

The error message is as follows:

"Failed to convert from type [java.lang.String] to type [java.lang.Integer] for value '3175433272470683'; nested exception is java.lang.NumberFormatException: For input string: \"3175433272470683\""

I want to make it work with a

@Id String id;

Any suggestions?

Many thanks in advances. It's a big privilege to ask questions here.

The Entity class:

@Entity // This tells Hibernate to make a table out of this class
@Table(name = "users")
public class XmppUser {
    @Id
    private java.lang.String username;

    private String password;
    private String serverkey;
    private String salt;
    private int iterationcount;
    private Date created_at;

    //    @Formula("ST_ASTEXT(coordinates)")
//    @Column(columnDefinition = "geometry")
//    private Point coordinates;
    //    private Point coordinates;
    private String full_name;

    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "username", nullable = true)
    private XmppLast xmppLast;

Could you post the entity class code? – Sham Fiorin Aug 10 '19 at 01:11 — Sham Fiorin, Aug 10 '19 at 01:11
Just posted an entity class – Andrey Ruzhentsev Aug 10 '19 at 01:17 — Andrey Ruzhentsev, Aug 10 '19 at 01:17

admlz635 · Accepted Answer · 2022-12-05T17:24:25.317

9

You must change the type of the ID type parameter in the repository to match with the id attribute type on your entity.

From the Spring docs:

Interface Repository<T,ID>
Type Parameters:
  T - the domain type the repository manages    
  ID - the type of the id of the entity the repository manages

Based on

@Entity // This tells Hibernate to make a table out of this class
@Table(name = "users")
public class XmppUser {
    @Id
    private java.lang.String username;
    //...

    }

It should be

public interface UserRepository extends CrudRepository<XmppUser, String> {
    //..
    }

See:

edited Dec 05 '22 at 17:24

answered Aug 10 '19 at 01:26

admlz635

1,001
1
9
18

Just added this to the repository: ```@Override Optional findById(String integer);``` – Andrey Ruzhentsev Aug 10 '19 at 01:33
It says: `method does not override method from its superclass` – Andrey Ruzhentsev Aug 10 '19 at 01:33
@DavidJones , I’ve added an example – admlz635 Aug 10 '19 at 01:35
@admiz635 that's exactly what I have, and inside it: `@Override Optional findById(String integer);` – Andrey Ruzhentsev Aug 10 '19 at 01:38
Oh, my bad. Just noticed that the second type parameter to `CrudRepository` is a String! That solved the problem! Great job @admiz635! – Andrey Ruzhentsev Aug 10 '19 at 02:48

score 0 · Answer 2 · answered Aug 10 '19 at 01:21

0

You could try something like this:

@Id
@GeneratedValue(generator = "uuid")
@GenericGenerator(name = "uuid", strategy = "uuid2")
@Column(name = "PR_KEY")
private String prKey;

If you want to read more about this subject you could begin looking throw here or here

answered Aug 10 '19 at 01:21

Sham Fiorin

403
4
16

Tried. It doesn't work. Same error. Just tried again. – Andrey Ruzhentsev Aug 10 '19 at 01:23

Stone Feng · Answer 3 · 2019-08-10T01:34:48.637

0

according to the latest version of spring data jpa(2.1.10 GA), you can use it like

here is the link

edited Aug 10 '19 at 01:34

answered Aug 10 '19 at 01:27

Stone Feng

69
1
3

@DavidJones so your spring data jpa version is? – Stone Feng Aug 10 '19 at 01:38
Version: 2.1.9.RELEASE – Andrey Ruzhentsev Aug 10 '19 at 01:41
@DavidJones try it like this:@Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name = "id") private String id; – Stone Feng Aug 10 '19 at 01:46
Doesn't solve the problem. I expected Spring to have fewer issues. That's the second bug I found today. – Andrey Ruzhentsev Aug 10 '19 at 02:19

score 0 · Answer 4 · answered Mar 10 '21 at 00:09

JpaRepository is a special case of CrudRepository. Both JpaRepository and CrudRepository declare two type parameters, T and ID. You will need to supply these two class types. For example,

public interface UserRepository extends CrudRepository<XmppUser, java.lang.String> {
//..
}

or

public interface UserRepository extends JpaRepository<XmppUser, java.lang.String> {
//..
}

Notice that the second type java.lang.String must match the type of the primary key attribute. In this case, you cannot specify it as String or Integer, but it is java.lang.String instead.

Try not to name a customized class as String. It is a bad practice to use the same class name as already present in the JDK.

score 0 · Answer 5 · answered Nov 15 '21 at 05:33

0

I think there is an approach to solve this problem.

Let's say, Site is our @Entity.

@Id
private String id;

getters setters

then you can invoke findById as follow

 Optional<Site> site = getSite(id);

Note: this worked for me, I hope it will help someone.

answered Nov 15 '21 at 05:33

Sayed Hussainullah Sadat

178
2
17

xHow can I have @Id string for CrudRepository in Spring with Spring Data JPA?

5 Answers5