Ayoub k
Ayoub k

Reputation: 8868

No serializer found for class org.hibernate.proxy.pojo.bytebuddy.ByteBuddyInterceptor

When i try to navigate to an endpoint i get the following error

Type definition error: [simple type, class org.hibernate.proxy.pojo.bytebuddy.ByteBuddyInterceptor]; nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: No serializer found for class org.hibernate.proxy.pojo.bytebuddy.ByteBuddyInterceptor and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS)  

I checked all my models and all the attributes have getters and setters. So what's the problem ?

I can fix that by adding spring.jackson.serialization.fail-on-empty-beans=false but i think this is just a work around to hide the exception.

Edit

Product model:

@Entity
public class Product {
    private int id;
    private String name;
    private String photo;
    private double price;
    private int quantity;
    private Double rating;
    private Provider provider;
    private String description;
    private List<Category> categories = new ArrayList<>();
    private List<Photo> photos = new ArrayList<>();
    
    // Getters & Setters
}

PagedResponse class :

public class PagedResponse<T> {

    private List<T> content;
    private int page;
    private int size;
    private long totalElements;
    private int totalPages;
    private boolean last;
    
    // Getters & Setters
}

RestResponse Class :

public class RestResponse<T> {
    private String status;
    private int code;
    private String message;
    private T result;

    // Getters & Setters
}

In my controller i'm returning ResponseEntity<RestResponse<PagedResponse<Product>>>

Upvotes: 151

Views: 199547

Answers (18)

Gustavo Ribeiro
Gustavo Ribeiro

Reputation: 523

My approach was just use findById() with orElseThrow() method instead of getReferenceById().

I ended up like this:

  @PutMapping("/{id}")
    public Food updateFood(
            @PathVariable("id") Long id,
            @RequestBody
            @Validated RequestUpdateFoodDTO data
    ){
        Food food = this.repository.findById(id).orElseThrow();
        
        food.setTitle(data.title());
        food.setImage(data.image());
        food.setPrice(data.price());

        repository.save(food);

        return food;
    }

Upvotes: 0

cs7eric
cs7eric

Reputation: 1

i found the approach from other article, you can add the public setter and getter in your class, it can solve this problem

Upvotes: -1

Tarun
Tarun

Reputation: 838

Add the following dependency in your pom.xml for javax.*

<groupId>com.fasterxml.jackson.datatype</groupId>
<artifactId>jackson-datatype-hibernate5</artifactId>

Or following for jakarta.*

<groupId>com.fasterxml.jackson.datatype</groupId>
<artifactId>jackson-datatype-hibernate5-jakarta</artifactId>

And add the configuration as below for javax.*:

package microservices.book.multiplication.configuration;

import com.fasterxml.jackson.databind.Module;
import com.fasterxml.jackson.datatype.hibernate5.jakarta.Hibernate5JakartaModule;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;

@Configuration
public class JsonConfiguration {
    @Bean
    public Module hibernateModule() {
        return new Hibernate5JakartaModule();
    }
   
}

Or for jakarta.* package, add the configuration as below:

package microservices.book.multiplication.configuration;

import com.fasterxml.jackson.databind.Module;
import com.fasterxml.jackson.datatype.hibernate5.jakarta.Hibernate5JakartaModule;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;

@Configuration
public class JsonConfiguration {
    @Bean
    public Module hibernateModule() {
        return new Hibernate5JakartaModule();
    }
   
}

The above two steps should resolve this issue.

Upvotes: 4

Apoorv Vardhman
Apoorv Vardhman

Reputation: 119

If you are using spring 6 then it will work

spring:
  jackson:
    serialization:
      fail-on-empty-beans: false

Upvotes: 5

Malek Tubaisaht
Malek Tubaisaht

Reputation: 1387

I faced this problem once I moved to using jakarta.persistence library , once I changed back to javax.persistence the problem sovled.

Upvotes: 0

Chris Neve
Chris Neve

Reputation: 2424

Changing the FetchType from Lazy to Eager did the trick for me.

Upvotes: 29

Abhishek Agashe
Abhishek Agashe

Reputation: 52

As I worked on localhost I just needed a Server. The following worked for me

<dependency>
        <groupId>org.apache.tomcat</groupId>
        <artifactId>tomcat-jasper</artifactId>
        <version>9.0.65</version>
</dependency>

Upvotes: -2

Az.MaYo
Az.MaYo

Reputation: 1106

In my case I was facing same exception with @EntityGraph. Actually I was trying to fetch OneToMany (inventoryList) via EntityGraph and it's working but when I tried get another relation with ManyToOne (category) it was giving error. I don't know why it became lazy. As we know @ManyToOne is by default eager.

class Product {
 
    @ManyToOne
    @JoinColumn(name = "categoryId")
    private Category category;

    @OneToMany(mappedBy = "product", cascade = CascadeType.ALL)
    private List<Inventory> inventoryList;
}

Repository method:

@EntityGraph(attributePaths = { "inventoryList" })
List<Product> findAll();

I was getting same expection because I didn't add category in EntityGraph fetch. After adding category it's get fixed.

@EntityGraph(attributePaths = { "inventoryList", "category" })
List<Product> findAll();

Upvotes: 0

kvs
kvs

Reputation: 338

This answer comes from the book: "learn microservices with spring boot" Instead of supressing the error on empty bean which is suggested by the spring there are more preferable ways to handle this.

We configured our nested User entities to be fetched in LAZY mode, so they’re not being queried from the database. We also said that Hibernate creates proxies for our classes in runtime. That’s the reason behind the ByteBuddyInterceptor class. You can try switching the fetch mode to EAGER, and you will no longer get this error. But that’s not the proper solution to this problem since then we’ll be triggering many queries for data we don’t need. Let’s keep the lazy fetch mode and fix this accordingly. The first option we have is to customize our JSON serialization so it can handle Hibernate objects. Luckily, FasterXML,the provider of Jackson libraries, has a specific module for Hibernate that we can use in our ObjectMapper objects: jackson-datatype-hibernate:

<dependency>
        <groupId>com.fasterxml.jackson.datatype</groupId>
        <artifactId>jackson-datatype-hibernate5</artifactId>
</dependency>

We create a bean for our new Hibernate module for Jackson. Spring Boot’s Jackson2ObjectMapperBuilder will use it via autoconfiguration, and all our ObjectMapper instances will use the Spring Boot defaults plus our own customization.

import com.fasterxml.jackson.databind.Module;
import com.fasterxml.jackson.datatype.hibernate5.Hibernate5Module;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
@Configuration
public class JsonConfiguration {
    @Bean
    public Module hibernateModule() {
        return new Hibernate5Module();
    }
}

Upvotes: 20

Georges MOMO
Georges MOMO

Reputation: 41

Changing from

MyEntityClassRepositorie.getOne(id)

to

MyEntityClassRepositorie.findById(id).get()

work fine for me.

Upvotes: 4

Arber Berisha
Arber Berisha

Reputation: 91

This solved my issue.

 @JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})

Upvotes: 9

Meena Chaudhary
Meena Chaudhary

Reputation: 10675

For me, I got this error for a DTO object. The problem was I didn't provide getters for DTO properties. Therefore, Jackson was not able to fetch those values and assumed the bean is empty. Solution:

Add Getters to your DTO

Upvotes: 0

SNabi
SNabi

Reputation: 69

I also faced with this problem. @Szelek's answer helped me. But I did it with another way. Changed getOne() method to:

repository.findById(id).orElse(null)

Ensure you are taking care of the NullPointerException this will generate when it's not found.

Upvotes: 2

Szelek
Szelek

Reputation: 2837

I came across this error while doing a tutorial with spring repository. It turned out that the error was made at the stage of building the service class for my entity.

In your serviceImpl class, you probably have something like:

    @Override
    public YourEntityClass findYourEntityClassById(Long id) {
      return YourEntityClassRepositorie.getOne(id);
    }

Change this to:

    @Override
    public YourEntityClass findYourEntityClassById(Long id) {
      return YourEntityClassRepositorie.findById(id).get();
    }

Basically getOne is a lazy load operation. Thus you get only a reference (a proxy) to the entity. That means no DB access is actually made. Only when you call it's properties then it will query the DB. findByID does the call 'eagerly'/immediately when you call it, thus you have the actual entity fully populated.

Take a look at this: Link to the difference between getOne & findByID

Upvotes: 220

akhil Raj Kunwar
akhil Raj Kunwar

Reputation: 92

i also faced same problem. I was using repo.getOne(id); i changed it to repo.findById(id). It returned optional, but now error is gone

Upvotes: 3

Seldo97
Seldo97

Reputation: 719

@JsonIgnoreProperties({"hibernateLazyInitializer", "handler"}) work for me very well. It doesn't miss any reference objects and resolve the problem.

In my case:

@Entity
@Table(name = "applications")
public class Application implements Serializable {

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

    @NotBlank
    @Size(max = 36, min = 36)
    private String guid;

    @NotBlank
    @Size(max = 60)
    private String name;

    @Column(name = "refresh_delay")
    private int refreshDelay;

    @ManyToOne(fetch = LAZY)
    @JoinColumn(name = "id_production", referencedColumnName = "id")
    @JsonIgnoreProperties(value = {"applications", "hibernateLazyInitializer"})
    private Production production;

Upvotes: 15

Subarata Talukder
Subarata Talukder

Reputation: 6311

You can Ignore to produce JSON output of a property by

@JsonIgnore 

Or If you have any lazy loaded properties having a relationship. You can use this annotation at top of the property.

@JsonIgnoreProperties({"hibernateLazyInitializer", "handler"}) 

Example:

@Entity
public class Product implements Serializable{
   private int id;
   private String name;
   private String photo;
   private double price;
   private int quantity;
   private Double rating;
   private Provider provider;
   private String description;

   @JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})
   private List<Category> categories = new ArrayList<>();

   @JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})
   private List<Photo> photos = new ArrayList<>();

   // Getters & Setters
}

If you still have this error, please add this line of code in your application.properties file

spring.jackson.serialization.fail-on-empty-beans=false

I hope your problem will be solved. Thanks.

Upvotes: 142

Ivan Perales M.
Ivan Perales M.

Reputation: 355

Hmm are you traying to send entities from one instance of the jvm to another one which need to serialize them? if this is the case i think the error is because you fetched the entities somehow and hibernate is using its not serializable classes, you need to convert entities to pojo's (i mean use native types or objects that are serializables).

Upvotes: -1

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