Java - Spring Boot - Hibernate - JPA (Error with post @ManyToOne)

Question

I have two models, one for the customer and one for the seller, in the customer table I need to have only one seller, already in the seller table, I can have several customers for a single seller.

But when I try to add a new customer, setting the salesperson's id, it just doesn't work.

Here is the code:

Seller Model:


    package com.crud.spring.jpa.postgresql.model;
    
    import lombok.AllArgsConstructor;
    import lombok.Getter;
    import lombok.Setter;
    
    import javax.persistence.*;
    
    @Entity
    @AllArgsConstructor
    @Getter
    @Setter
    @Table(name = "sellers", uniqueConstraints = @UniqueConstraint(columnNames = "code"))
    public class Seller {
    
      @Id
      @GeneratedValue(strategy = GenerationType.AUTO)
      private Long id;
    
      @Column(name = "code")
      private long code;
    
      @Column(name = "name")
      private String name;
    
      public Seller(){
      }
    }

Client Model:


    package com.crud.spring.jpa.postgresql.model;
    
    import com.fasterxml.jackson.annotation.JsonIgnore;
    import lombok.AllArgsConstructor;
    import lombok.Getter;
    import lombok.Setter;
    import org.hibernate.annotations.OnDelete;
    import org.hibernate.annotations.OnDeleteAction;
    
    import javax.persistence.*;
    
    @Entity
    @AllArgsConstructor
    @Getter
    @Setter
    @Table(name = "clients", uniqueConstraints = @UniqueConstraint(columnNames = "code"))
    public class Client {
    
      @Id
      @GeneratedValue(strategy = GenerationType.AUTO)
      private Long id;
    
      @Column(name = "code")
      private long code;
    
      @Column(name = "name")
      private String name;
    
      @ManyToOne(fetch = FetchType.LAZY, optional = false)
      @JoinColumn(name = "seller_id", referencedColumnName = "id", nullable = false)
      @OnDelete(action = OnDeleteAction.CASCADE)
      @JsonIgnore
      private Seller seller;
    
      public Client(){
      }
    }

Client Controller:


    @PostMapping("/clients")
    public ResponseEntity createClients(@RequestBody Client client) {
      return this.service.createClients(client);
    }

Client Service:


    public ResponseEntity createClients(Client client) {
      try {
        client = this.repository.save(client);
    
        return new ResponseEntity(client, HttpStatus.OK);
      } catch (Exception e){
        return new ResponseEntity(e, HttpStatus.INTERNAL_SERVER_ERROR);
      }
    }

Post Seller:

Post Client:

Error is:

"entityName": "com.crud.spring.jpa.postgresql.model.Client", "propertyName": "seller", "message": "not-null property references a null or transient value : com.crud.spring.jpa.postgresql.model.Client.seller", "localizedMessage": "not-null property references a null or transient value : com.crud.spring.jpa.postgresql.model.Client.seller", "suppressed": [] "message": "not-null property references a null or transient value : com.crud.spring.jpa.postgresql.model.Client.seller; nested exception is org.hibernate.PropertyValueException: not-null property references a null or transient value : com.crud.spring.jpa.postgresql.model.Client.seller", enter code here"localizedMessage": "not-null property references a null or transient value : com.crud.spring.jpa.postgresql.model.Client.seller; nested exception is org.hibernate.PropertyValueException: not-null property references a null or transient value : com.crud.spring.jpa.postgresql.model.Client.seller",

Belaid BOUAOUALTEN · Accepted Answer · 2021-05-04 15:16:53Z

2

try in json "seller":{"id":15} insted of "seller_id":15

answered May 4, 2021 at 15:16

Belaid BOUAOUALTEN

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3 Comments

maistack123 Over a year ago

Don't work this. How do I work like this: { "code": 1, "name": "Eduardo", "seller_id": 15 }

Belaid BOUAOUALTEN Over a year ago

{ "code": 1, "name": "Eduardo","seller":{"id":15}} you should try it like this ,because when you reference to seller you reference to a class that you will find by it id

maistack123 Over a year ago

I managed to improve the visualization of the code in the post, can you check again?

Dharman · Accepted Answer · 2021-05-04 17:58:21Z

I've simulated your code & have the below observations to make it work.

Remove @JsonIgnore Annotation from the Client Model
Jpa works based on the object references & not the values alone. So you might have to modify the Client service code.

 try {
           Optional<Seller> seller = sellerRepository.findById(client.getSeller().getId());
           if(seller.isPresent()){
               client.setSeller(seller.get());
               client = this.repository.save(client);
           }
            return new ResponseEntity(client, HttpStatus.OK);
        } catch (Exception e) {
            return new ResponseEntity(e, HttpStatus.INTERNAL_SERVER_ERROR);
        }

In the above code, we use the seller id from our request client model & get the seller object from db. Then we set the reference to the client model.

Pass the Json input for createClient as shown

{
    "id": 1,
    "code": 2,
    "name": "Client_1",
    "seller": {
        "id": 1
    }
}

Note: You'll have to create a valid seller first before creating the client by passing the seller id.

Surjeet Singh Rathore · Accepted Answer · 2021-05-04 18:13:27Z

1

So , there are few corrections : When your payload contains existing seller's id :

Send existing seller id with payload

{
    "code" : 216,
    "name" : "addclientforseller5",
    "seller" :
        {
            "id" : 5
        }   
}

Add this code in your service Implementation


    public Client createClients(Client client) {
        Long sellerId = client.getSeller().getId();
        Optional<Seller> existingSeller = sellerRepository.findById(sellerId);
        if (existingSeller.isPresent()) {
            Seller savedSellerObject = existingSeller.get();
            client.setSeller(savedSellerObject);
        } else {
            // throw exception
        }

        return clientSellerRepository.save(client); // you can return ResponseEntity also
    }

When your payload contains Seller object to create new Seller do this :

payload should look like this

{
    "code" : 112,
    "name" : "Pure",
    "seller" :{
        "code" : 1111,
        "name" : "Guava"
    }
}

add this code in your service Implementation

    @Autowired
    ClientRepository clientRepository;

    @Autowired
    SellerRepository sellerRepository;

public Client createClients(Client client) {
        Seller seller = new Seller();
        seller.setCode(client.getSeller().getCode());
        seller.setName(client.getSeller().getName());
        Seller savedSeller = sellerRepository.save(seller);
        client.setSeller(savedSeller);
        return clientRepository.save(client);
    }

remove @JsonIngnore from seller field in Client class.

    @ManyToOne(fetch = FetchType.LAZY, optional = false)
    @JoinColumn(name = "seller_id", referencedColumnName = "id")
    @OnDelete(action = OnDeleteAction.CASCADE)
    //@JsonIgnore
    private Seller seller;

edited May 4, 2021 at 18:13

answered May 4, 2021 at 15:18

Surjeet Singh Rathore

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5 Comments

maistack123 Over a year ago

Don't work this. How do I work like this: ``` { "code": 1, "name": "Eduardo", "seller_id": 15 } ```

Surjeet Singh Rathore Over a year ago

it is not possible to send request as you want in your payload.

maistack123 Over a year ago

I managed to improve the visualization of the code in the post, can you check again?

Surjeet Singh Rathore Over a year ago

ok. if my answer resolve your issue . please vote it.

maistack123 Over a year ago

Am I unable to reference only the ID instead of the entire seller object within the client?

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