'How to fix org.hibernate.LazyInitializationException - could not initialize proxy - no Session

I get the following exception:

Exception in thread "main" org.hibernate.LazyInitializationException: could not initialize proxy - no Session
    at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:167)
    at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:215)
    at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:190)
    at sei.persistence.wf.entities.Element_$$_jvstc68_47.getNote(Element_$$_jvstc68_47.java)
    at JSON_to_XML.createBpmnRepresantation(JSON_to_XML.java:139)
    at JSON_to_XML.main(JSON_to_XML.java:84)

when I try to call from main the following lines:

Model subProcessModel = getModelByModelGroup(1112);
System.out.println(subProcessModel.getElement().getNote());

I implemented the getModelByModelGroup(int modelgroupid) method firstly like this :

public static Model getModelByModelGroup(int modelGroupId, boolean openTransaction) {

    Session session = SessionFactoryHelper.getSessionFactory().getCurrentSession();     
    Transaction tx = null;

    if (openTransaction) {
        tx = session.getTransaction();
    }

    String responseMessage = "";

    try {
        if (openTransaction) {
            tx.begin();
        }
        Query query = session.createQuery("from Model where modelGroup.id = :modelGroupId");
        query.setParameter("modelGroupId", modelGroupId);

        List<Model> modelList = (List<Model>)query.list(); 
        Model model = null;

        for (Model m : modelList) {
            if (m.getModelType().getId() == 3) {
                model = m;
                break;
            }
        }

        if (model == null) {
            Object[] arrModels = modelList.toArray();
            if (arrModels.length == 0) {
                throw new Exception("Non esiste ");
            }

            model = (Model)arrModels[0];
        }

        if (openTransaction) {
            tx.commit();
        }

        return model;

   } catch(Exception ex) {
       if (openTransaction) {
           tx.rollback();
       }
       ex.printStackTrace();
       if (responseMessage.compareTo("") == 0) {
           responseMessage = "Error" + ex.getMessage();
       }
       return null;
    }
}

and got the exception. Then a friend suggested me to always test the session and get the current session to avoid this error. So I did this:

public static Model getModelByModelGroup(int modelGroupId) {
    Session session = null;
    boolean openSession = session == null;
    Transaction tx = null;
    if (openSession) {
        session = SessionFactoryHelper.getSessionFactory().getCurrentSession(); 
        tx = session.getTransaction();
    }
    String responseMessage = "";

    try {
        if (openSession) {
            tx.begin();
        }
        Query query = session.createQuery("from Model where modelGroup.id = :modelGroupId");
        query.setParameter("modelGroupId", modelGroupId);

        List<Model> modelList = (List<Model>)query.list(); 
        Model model = null;

        for (Model m : modelList) {
            if (m.getModelType().getId() == 3) {
                model = m;
                break;
            }
        }

        if (model == null) {
            Object[] arrModels = modelList.toArray();
            if (arrModels.length == 0) {
                throw new RuntimeException("Non esiste");
            }

            model = (Model)arrModels[0];

            if (openSession) {
                tx.commit();
            }
            return model;
        } catch(RuntimeException ex) {
            if (openSession) {
                tx.rollback();
            }
            ex.printStackTrace();
            if (responseMessage.compareTo("") == 0) {
                responseMessage = "Error" + ex.getMessage();
            }
            return null;        
        }
    }
}

but still, get the same error. I have been reading a lot for this error and found some possible solutions. One of them was to set lazyLoad to false but I am not allowed to do this that's why I was suggested to control the session

Solution 1:^[1]

What is wrong here is that your session management configuration is set to close session when you commit transaction. Check if you have something like:

<property name="current_session_context_class">thread</property>

in your configuration.

In order to overcome this problem you could change the configuration of session factory or open another session and only than ask for those lazy loaded objects. But what I would suggest here is to initialize this lazy collection in getModelByModelGroup itself and call:

Hibernate.initialize(subProcessModel.getElement());

when you are still in active session.

And one last thing. A friendly advice. You have something like this in your method:

for (Model m : modelList) {
    if (m.getModelType().getId() == 3) {
        model = m;
        break;
    }
}

Please insted of this code just filter those models with type id equal to 3 in the query statement just couple of lines above.

Some more reading:

session factory configuration

problem with closed session

Solution 2:^[2]

If you using Spring mark the class as @Transactional, then Spring will handle session management.

@Transactional
public class MyClass {
    ...
}

By using @Transactional, many important aspects such as transaction propagation are handled automatically. In this case if another transactional method is called the method will have the option of joining the ongoing transaction avoiding the "no session" exception.

WARNING If you do use @Transactional, please be aware of the resulting behavior. See this article for common pitfalls. For example, updates to entities are persisted even if you don't explicitly call save

Solution 3:^[3]

You can try to set

<property name="hibernate.enable_lazy_load_no_trans">true</property>

in hibernate.cfg.xml or persistence.xml

The problem to keep in mind with this property are well explained here

Solution 4:^[4]

The best way to handle the LazyInitializationException is to use the JOIN FETCH directive:

Query query = session.createQuery("""
    select m
    from Model m
    join fetch m.modelType
    where modelGroup.id = :modelGroupId
    """
);

Anyway, DO NOT use the following Anti-Patterns as suggested by some of the answers:

Sometimes, a DTO projection is a better choice than fetching entities, and this way, you won't get any LazyInitializationException.

Solution 5:^[5]

I was getting the same error for a one to many relationships for below annotation.

@OneToMany(mappedBy="department", cascade = CascadeType.ALL)

Changed as below after adding fetch=FetchType.EAGER, it worked for me.

@OneToMany(mappedBy="department", cascade = CascadeType.ALL, fetch=FetchType.EAGER)

Solution 6:^[6]

if you use spring data jpa , spring boot you can add this line in application.properties

spring.jpa.properties.hibernate.enable_lazy_load_no_trans=true

Solution 7:^[7]

This exception because of when you call session.getEntityById(), the session will be closed. So you need to re-attach the entity to the session. Or Easy solution is just configure default-lazy="false" to your entity.hbm.xml or if you are using annotations just add @Proxy(lazy=false) to your entity class.

Solution 8:^[8]

I encountered the same issue. I think another way to fix this is that you can change the query to join fetch your Element from Model as follows:

Query query = session.createQuery("from Model m join fetch m.element where modelGroup.id = :modelGroupId")

Solution 9:^[9]

This means that the object which you are trying to access is not loaded, so write a query that makes a join fetch of the object which you are trying to access.

Eg:

If you are trying to get ObjectB from ObjectA where ObjectB is a foreign key in ObjectA.

Query :

SELECT objA FROM ObjectA obj JOIN FETCH obj.objectB objB

Solution 10:^[10]

Faced the same Exception in different use case.

Use Case : Try to read data from DB with DTO projection.

Solution: Use get method instead of load.

Generic Operation

public class HibernateTemplate {
public static Object loadObject(Class<?> cls, Serializable s) {
    Object o = null;
    Transaction tx = null;
    try {
        Session session = HibernateUtil.getSessionFactory().openSession();
        tx = session.beginTransaction();
        o = session.load(cls, s); /*change load to get*/
        tx.commit();
        session.close();
    } catch (Exception e) {
        e.printStackTrace();
    }
    return o;
}

}

Persistence Class

public class Customer {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "Id")
private int customerId;

@Column(name = "Name")
private String customerName;

@Column(name = "City")
private String city;

//constructors , setters and getters

}

CustomerDAO interface

public interface CustomerDAO 
     {
   public CustomerTO getCustomerById(int cid);
     }

Entity Transfer Object Class

public class CustomerTO {

private int customerId;

private String customerName;

private String city;

//constructors , setters and getters

}

Factory Class

public class DAOFactory {

static CustomerDAO customerDAO;
static {
    customerDAO = new HibernateCustomerDAO();
}

public static CustomerDAO getCustomerDAO() {
    return customerDAO;
}

}

Entity specific DAO

public class HibernateCustomerDAO implements CustomerDAO {

@Override
public CustomerTO getCustomerById(int cid) {
    Customer cust = (Customer) HibernateTemplate.loadObject(Customer.class, cid);
    CustomerTO cto = new CustomerTO(cust.getCustomerId(), cust.getCustomerName(), cust.getCity());
    return cto;
}

}

Retrieving data: Test Class

CustomerDAO cdao = DAOFactory.getCustomerDAO();
CustomerTO c1 = cdao.getCustomerById(2);
System.out.println("CustomerName -> " + c1.getCustomerName() + " ,CustomerCity -> " + c1.getCity());

Present Data

Query and output generated by Hibernate System

Hibernate: select customer0_.Id as Id1_0_0_, customer0_.City as City2_0_0_, customer0_.Name as Name3_0_0_ from CustomerLab31 customer0_ where customer0_.Id=?

CustomerName -> Cody ,CustomerCity -> LA

Solution 11:^[11]

There are several good answers here that handle this error in a broad scope. I ran into a specific situation with Spring Security which had a quick, although probably not optimal, fix.

During user authorization (immediately after logging in and passing authentication) I was testing a user entity for a specific authority in a custom class that extends SimpleUrlAuthenticationSuccessHandler.

My user entity implements UserDetails and has a Set of lazy loaded Roles which threw the "org.hibernate.LazyInitializationException - could not initialize proxy - no Session" exception. Changing that Set from "fetch=FetchType.LAZY" to "fetch=FetchType.EAGER" fixed this for me.

Solution 12:^[12]

This means you are using JPA or hibernate in your code and performing modifying operation on DB without making the business logic transaction. So simple solution for this is mark your piece of code @Transactional

Solution 13:^[13]

If you are using JPQL, use JOIN FETCH is the easiest way: http://www.objectdb.com/java/jpa/query/jpql/from#LEFT_OUTER_INNER_JOIN_FETCH_

Solution 14:^[14]

If you are using Grail's Framework, it's simple to resolve lazy initialization exception by using Lazy keyword on specific field in Domain Class.

For-example:

class Book {
    static belongsTo = [author: Author]
    static mapping = {
        author lazy: false
    }
}

Find further information here

Solution 15:^[15]

In my case a misplaced session.clear() was causing this problem.

Solution 16:^[16]

In Spring Application Just Add

@Transactional(readOnly = true)

on your Function.

Remind that import spring Transactional annotation

import org.springframework.transaction.annotation.Transactional;

Solution 17:^[17]

Use @NamedEntityGraph. Eagar fetch will deteriorate the performance. Refer https://thorben-janssen.com/lazyinitializationexception/ for in-depth explanation.

Solution 18:^[18]

This happened to me when I was already using @Transactional(value=...) and was using multiple transaction managers.

My forms were sending back data that already had @JsonIgnore on them, so the data being sent back from forms was incomplete.

Originally I used the anti pattern solution, but found it was incredibly slow. I disabled this by setting it to false.

spring.jpa.properties.hibernate.enable_lazy_load_no_trans=false

The fix was to ensure that any objects that had lazy-loaded data that weren't loading were retrieved from the database first.

Optional<Object> objectDBOpt = objectRepository.findById(object.getId());

if (objectDBOpt.isEmpty()) {
    // Throw error
} else {
    Object objectFromDB = objectDBOpt.get();
}

In short, if you've tried all of the other answers, just make sure you look back to check you're loading from the database first if you haven't provided all the @JsonIgnore properties and are using them in your database query.

Solution 19:^[19]

All answers about adding JOIN FETCH (or left join fetch) are correct, I want only to add this: if you have converter be sure the getAsObject sub uses a "find" than includes in the sql the Join Fetch too. I lost much time to fix a similar problem, and the problem was in the converter.

Solution 20:^[20]

springBootVersion = '2.6.7'
hibernate = 5.6.8.Final'

For me i get the error in:

MyEntity myEntity = myEntityRepository.getById(id);

I change to this:

MyEntity myEntity = myEntityRepository.findById(id).orElse(null);

and i add @ManyToOne(fetch = FetchType.EAGER) in entity

Solution 21:^[21]

uses session.get(*.class, id); but do not load function

Solution 22:^[22]

you could also solved it by adding lazy=false into into your *.hbm.xml file or you can init your object in Hibernate.init(Object) when you get object from db

Solution 23:^[23]

Do the following changes in servlet-context.xml

    <beans:property name="hibernateProperties">
        <beans:props>

            <beans:prop key="hibernate.enable_lazy_load_no_trans">true</beans:prop>

        </beans:props>
    </beans:property>

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	Max
Solution 2	mowwwalker
Solution 3	Community
Solution 4
Solution 5	Ahmed Ashour
Solution 6	Shaaban Ebrahim
Solution 7	Garf365
Solution 8	Tony Vu
Solution 9	rex roy
Solution 10	Kms
Solution 11	Night Owl
Solution 12	giannis christofakis
Solution 13	xihui
Solution 14	Zeb
Solution 15	absin
Solution 16	Mehdi Varse
Solution 17	Leena
Solution 18	Nightfa11
Solution 19
Solution 20	AissaDevLab
Solution 21	Artavazd Manukyan
Solution 22	Sandeep Roniyaar
Solution 23	ArunDhwaj IIITH

'How to fix org.hibernate.LazyInitializationException - could not initialize proxy - no Session

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