How to enable hibernate to set manually primary key?

Question

Having these entity:

@Entity
@Data
@Builder
public class User {
    @Id
    private int id;
    private String name;
}

If i try to set the id:

@Bean
    CommandLineRunner dataLoader(UserRepository userRepo){
        return new CommandLineRunner() {
            @Override
            public void run(String... args) throws Exception {
                User u = User.builder()
                        .id(1)
                        .name("First User")
                        .build();
                userRepo.save(u);
            }
        };
    }

I got

java.lang.IllegalStateException: Failed to execute CommandLineRunner
    at org.springframework.boot.SpringApplication.callRunner(SpringApplication.java:794) ~[spring-boot-2.5.3.jar:2.5.3]
    at org.springframework.boot.SpringApplication.callRunners(SpringApplication.java:775) ~[spring-boot-2.5.3.jar:2.5.3]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:345) ~[spring-boot-2.5.3.jar:2.5.3]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:1343) ~[spring-boot-2.5.3.jar:2.5.3]
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:1332) ~[spring-boot-2.5.3.jar:2.5.3]
    at com.example.demo.DemoApplication.main(DemoApplication.java:16) ~[classes/:na]
Caused by: org.springframework.orm.jpa.JpaSystemException: No default constructor for entity:  : com.example.demo.domain.User; nested exception is org.hibernate.InstantiationException: No default constructor for entity:  : com.example.demo.domain.User
...

If i don't set the id, then no problem. So how do I can set the primary manually?

Looks like you have the same issue as shown here No default constructor for entity stackoverflow.com/questions/34241718/… — Pilpo
– Pilpo, Commented Sep 2, 2021 at 13:46

Simon Martinelli · Accepted Answer · 2021-09-02 14:14:41Z

2

In General: You sholdn't use @Data with Entities because the generated equals/hashCode and toString can lead to StackOverflowError if you have bi-directional entities.

Coming back to your question JPA needs a default constructor (no args constructor)

So I would recommend this:

@Entity
@Getter
@Setter
@NoArgsConstructor
@AllArgsConstructor
@Builder
public class User {
    @Id
    private int id;
    private String name;
}

edited Sep 2, 2021 at 14:14

answered Sep 2, 2021 at 13:56

Simon Martinelli

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10 Comments

milanHrabos Over a year ago

1) I don't understand your point of StackoverfloError, could you give an example please? 2) Why does not @Builder provide default (no-arg) constructor? Why does this behvaiour the @Builder overwrites? 3) Why does this issue (regarding the lack of default constructor) solves by adding @AllArgsConstrutor, which is not default constructor?

Simon Martinelli Over a year ago

1) equals/hashCode/toString will follow the references and will produce an endless loop 2 and 3) Please read the API Doc projectlombok.org/api/lombok/Builder.html

milanHrabos Over a year ago

So when it is safe to use @Data, when in case of @Entity, it would create infinite loop? When it doesn't ?

milanHrabos Over a year ago

Btw, your annotations are not enough. It will fail anyway, unless @AllArgsConstrutor is provided. The error otherwise:

constructor User in class com.example.demo.model.User cannot be applied to given types;   required: no arguments   found: int,java.lang.String   reason: actual and formal argument lists differ in length

Simon Martinelli Over a year ago

Ah you're right the Builder requires the AllArgsConstructor.

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