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

Question

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

score 213 · Answer 1 · edited Mar 19 '20 at 01:04

213

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

edited Mar 19 '20 at 01:04

mowwwalker

14,309
23
87
144

answered Aug 28 '15 at 17:45

user2601995

5,407
8
34
38

26

I cannot overstate the importance of this answer. I'd seriously recommend trying this option first. – sparkyspider Jul 29 '16 at 16:19
6

Also note that you have to add `@EnableTransactionManagement` to your configuration in order to enable transactions. "_if another transactional method is called the method will have the option of joining the ongoing transaction_" this behavior is different for the different ways transactions are implemented, i.e. interface proxy vs class proxy or AspectJ weaving. Refer to the [documentation](http://docs.spring.io/spring-framework/docs/current/spring-framework-reference/html/transaction.html#transaction-declarative-annotations). – Erik Hofer Aug 25 '16 at 11:30
1

Should we understand the Spring `Transactional` annotation is thus advised, not only for modifying transactions, but also for accessing only ones ? – Stephane May 28 '18 at 12:43
Note that Spring will also pick up the javax.persistence.Transactional annotation if you want to stick to JPA. – ThetaSinner Aug 29 '18 at 22:17
8

I'd seriously recommend to use this annotation on top of class only for testing. Real code should mark each method as transaction in class separately. Unless all the methods in class will require opened connection with transaction to database. – m1ld Dec 06 '18 at 13:10
1

Isn't it safe to put @Transactional(readOnly = true) instead of just @Transactional ? – Hamedz Jun 02 '19 at 23:33
@Transactional didn't work for me on my SpringBoot integration Test case. I marked this anotation on the Entity class where the error occurs when loading child objects method on the toString(). – Jay Aug 05 '19 at 15:15

score 104 · Answer 2 · edited May 23 '17 at 12:02

104

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

edited May 23 '17 at 12:02

Community

1
1

answered Dec 04 '14 at 20:33

Wilianto Indrawan

2,220
3
18
25

8

Can you explain its meaning as well? – Mohit Kanwar Jul 13 '15 at 11:26
2

I'm also curious what this does. It did fix the issue I was having, but I'd like to understand why. – Hassan Oct 01 '15 at 22:46
After thousands of "just put lazy=false in your configuration file" and "just initialize every object with Hibernate.initialize()" finally a concrete and viable solution for me. This option should be made standard in hibernate! – saimiris_devel Jan 21 '16 at 09:23
3

for persistence.xml: `` – ACV Aug 24 '16 at 19:15
10

DO NOT USE THIS PROPERTY, IF SPRING MANAGES YOUR TRANSACTIONS THIS PROPERTY WILL LEAD TO TRANSACTIONS EXPLOSION, SIMPLY SPRING WILL SHUTDOWN THE APPLICATION – Youans Oct 18 '17 at 12:09
To doing that you will use an anti-pattern ... so better don't do that. – Rubén Fanjul Estrada Dec 24 '19 at 19:16

score 99 · Accepted Answer · edited Jun 04 '19 at 08:45

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

Thank you! I solved my problem using openSession() instead of getCurrentSession() as one of the links you gave me suggested this, but now I'm affraid if it is wrong to do so — Blerta Dhimitri, Feb 05 '14 at 11:39
No it is probably fine. But read some more to be able to fully control your sessions and transactions. It is realy important to know the basics because all higher level technologies like Spring, Hibernate and more operate on the very same concept. — goroncy, Feb 05 '14 at 12:08

Vlad Mihalcea · Answer 4 · 2021-01-08T21:07:24.883

76

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.

edited Jan 08 '21 at 21:07

answered Sep 13 '16 at 08:10

Vlad Mihalcea

103,297
39
432
788

How Can I Identify which call is having problem ? I am finding it as hard to identify the call. Is there any way ? For testing purpose I used `FetchType=EAGER`, but this is not correct solution, right ? – Shantaram Tupe Nov 22 '17 at 07:25
Just use logging. And EAGER is bad, yes. – Vlad Mihalcea Nov 22 '17 at 11:56
Then, you should use either DTOs or initialize all associations before leaving the `@Transactional` service. – Vlad Mihalcea Nov 22 '17 at 12:26
2

We should advocate the best enterprise practice but not the quick fix. – etlds Feb 22 '18 at 17:08
1

This is the best fix – phamductri Jul 27 '20 at 02:53
Why is EAGER bad? In @OneToMany too? – stephan f Nov 11 '20 at 17:50
[Here's why](https://vladmihalcea.com/eager-fetching-is-a-code-smell/). – Vlad Mihalcea Nov 11 '20 at 22:33

score 24 · Answer 5 · edited Nov 30 '17 at 18:18

24

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)

edited Nov 30 '17 at 18:18

Ahmed Ashour

4,209
10
29
46

answered Sep 20 '17 at 10:21

Smruti R Tripathy

609
1
9
14

36

Yes it may fix it but now you are loading the whole tree of data. This will have negative performance impacts in most cases – astro8891 Jun 11 '18 at 01:39

score 17 · Answer 6 · answered Jun 15 '18 at 21:24

17

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

answered Jun 15 '18 at 21:24

Shaaban Ebrahim

7,015
1
14
19

10

This is considered as an anti-pattern. https://vladmihalcea.com/the-hibernate-enable_lazy_load_no_trans-anti-pattern/ – Sudip Bhandari Aug 24 '18 at 11:11
If it's an antipattern, could you please suggest an alternative? I'm also working with Spring Data and the other answers I've looked at don't help me. – MiguelMunoz Jan 19 '21 at 12:42

score 8 · Answer 7 · edited Sep 07 '16 at 14:37

8

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.

edited Sep 07 '16 at 14:37

Garf365

3,503
5
27
37

answered Sep 07 '16 at 14:17

R Pidugu

370
3
6

How do you re-attach the entity to the session? – Pere Jan 12 '21 at 02:12

score 5 · Answer 8 · answered Apr 13 '15 at 08:44

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")

score 5 · Answer 9 · answered Apr 23 '18 at 11:36

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

score 3 · Answer 10 · answered Mar 15 '16 at 01:55

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.

score 3 · Answer 11 · answered Jul 14 '19 at 19:18

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

score 2 · Answer 12 · answered Mar 23 '17 at 21:55

2

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_

answered Mar 23 '17 at 21:55

xihui

113
1
7

score 1 · Answer 13 · answered Oct 27 '16 at 12:38

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

score 1 · Answer 14 · edited Mar 31 '20 at 04:05

1

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

edited Mar 31 '20 at 04:05

giannis christofakis

7,151
4
48
63

answered Oct 26 '18 at 09:47

Aman Goel

2,535
1
16
17

score 1 · Answer 15 · answered Oct 30 '18 at 14:04

1

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

answered Oct 30 '18 at 14:04

absin

900
7
16

score 0 · Answer 16 · answered Nov 14 '20 at 00:37

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.

score -1 · Answer 17 · answered Aug 25 '15 at 21:21

-1

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

answered Aug 25 '15 at 21:21

Artavazd Manukyan

73
1
2

3

can you please explain this? – newday Oct 14 '15 at 20:42

score -2 · Answer 18 · answered Jul 31 '14 at 10:12

-2

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

answered Jul 31 '14 at 10:12

Sandeep Roniyaar

206
1
8
26

10

generally adding lazy=false is not a good idea. that is why lazy is true by default – MoienGK Nov 06 '14 at 11:51
The OP clearly said beforehand that he's not allowed to do that. – GingerHead May 19 '15 at 16:09

score -2 · Answer 19 · answered Apr 09 '16 at 00:29

-2

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>

answered Apr 09 '16 at 00:29

ArunDhwaj IIITH

3,730
1
20
13

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

19 Answers19

Linked

Related