Hibernate table not mapped error in HQL query

Question

I have a web application that use Hibernate to make CRUD operations over a database. I got an error saying that the table is not mapped. See the Java files:

Error message:

org.springframework.orm.hibernate3.HibernateQueryException: Books is not mapped [SELECT COUNT(*) FROM Books]; nested exception is org.hibernate.hql.ast.QuerySyntaxException: Books is not mapped [SELECT COUNT(*) FROM Books]
at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:660)
at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:411)
...
Caused by: org.hibernate.hql.ast.QuerySyntaxException: Books is not mapped [SELECT COUNT(*) FROM Books]
at org.hibernate.hql.ast.util.SessionFactoryHelper.requireClassPersister(SessionFactoryHelper.java:181)
at org.hibernate.hql.ast.tree.FromElementFactory.addFromElement(FromElementFactory.java:111)
at org.hibernate.hql.ast.tree.FromClause.addFromElement(FromClause.java:93)
...

Here's my DAO.java method:

public int getTotalBooks(){
    return DataAccessUtils.intResult(hibernateTemplate.find(
          "SELECT COUNT(*) FROM Books"));
}

Book.java:

@Entity
@Table(name="Books")
public class Book {

    @Id
    @GeneratedValue
    @Column(name="id")
    private int id;

    @Column(name="title", nullable=false)
    private String title;
    ...
}

How should I modify it in order to work?

Please name of the class. Issue might be with definition of table name. Please check if the same name is used in both the places. (Case sensitive) — FullStackDev, Jun 02 '17 at 14:35

score 159 · Accepted Answer · edited Jan 14 '21 at 10:44

159

The exception message says:

Books is not mapped [SELECT COUNT(*) FROM Books]; nested exception is org.hibernate.hql.ast.QuerySyntaxException: Books is not mapped [SELECT COUNT(*) FROM Books]

Books is not mapped. That is, that there is no mapped type called Books.

And indeed, there isn't. Your mapped type is called Book. It's mapped to a table called Books, but the type is called Book. When you write HQL (or JPQL) queries, you use the names of the types, not the tables.

So, change your query to:

select count(*) from Book

Although I think it may need to be

select count(b) from Book b

If HQL doesn't support the * notation.

edited Jan 14 '21 at 10:44

E_net4

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answered Jan 21 '13 at 20:34

Tom Anderson

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this is not working, see the error: Caused by: org.hibernate.hql.ast.QuerySyntaxException: Books is not mapped [select count(*) from Books] – Pascut Jan 21 '13 at 20:37
now I understand what you're saying, I'm in a rush so I just read quickly all the answers..my bad. I'll try now to change from Books to Book. I think you're right – Pascut Jan 21 '13 at 20:50
50

The key takeaway here for me is: In your HQL, you must use Entity names and not table names. That made all the difference. – Michael Plautz Sep 04 '14 at 19:08
3

Sure it says it's not mapped but the error and you don't say how to do the mapping. – Philip Rego Feb 21 '18 at 16:42
Dont mix between the table name and the repository name. Your hibernate query is against the repository – Java Main Feb 25 '20 at 15:21
1

Be aware: This error appears also if you use `SELECT b FROM book` because it is case-sensitive. Instead you should the exact classname (`SELECT b FROM Book`) – phil Jun 02 '20 at 07:47

Gray · Answer 2 · 2021-11-13T16:13:38.743

hibernate3.HibernateQueryException: Books is not mapped [SELECT COUNT(*) FROM Books];

Hibernate is trying to say that it does not know an entity named "Books". Let's look at your entity:

@javax.persistence.Entity
@javax.persistence.Table(name = "Books")
public class Book {

Right. The table name for Book has been renamed to "Books" but the entity name is still "Book" from the class name. If you want to set the entity name, you should use the @Entity annotation's name instead:

// this allows you to use the entity Books in HQL queries
@javax.persistence.Entity(name = "Books")
public class Book {

That sets both the entity name and the table name.

The opposite problem happened to me when I was migrating from the Person.hbm.xml file to using the Java annotations to describe the hibernate fields. My old XML file had:

<hibernate-mapping package="...">
    <class name="Person" table="persons" lazy="true">
       ...
</hibernate-mapping>

And my new entity had a @Entity(name=...) which I needed to set the name of the table.

// this renames the entity and sets the table name
@javax.persistence.Entity(name = "persons")
public class Person {
    ...

What I then was seeing was HQL errors like:

QuerySyntaxException: Person is not mapped
     [SELECT id FROM Person WHERE id in (:ids)]

The problem with this was that the entity name was being renamed to persons as well. I should have set the table name using:

// no name = here so the entity can be used as Person
@javax.persistence.Entity
// table name specified here
@javax.persistence.Table(name = "persons")
public class Person extends BaseGeneratedId {

score 19 · Answer 3 · answered Apr 04 '17 at 16:45

This answer comes late but summarizes the concept involved in the "table not mapped" exception(in order to help those who come across this problem since its very common for hibernate newbies). This error can appear due to many reasons but the target is to address the most common one that is faced by a number of novice hibernate developers to save them hours of research. I am using my own example for a simple demonstration below.

The exception:

org.hibernate.hql.internal.ast.QuerySyntaxException: subscriber is not mapped [ from subscriber]

In simple words, this very usual exception only tells that the query is wrong in the below code.

Session session = this.sessionFactory.getCurrentSession();
List<Subscriber> personsList = session.createQuery(" from subscriber").list();

This is how my POJO class is declared:

@Entity
@Table(name = "subscriber")
public class Subscriber

But the query syntax "from subscriber" is correct and the table subscriber exists. Which brings me to a key point:

It is an HQL query not SQL.

and how its explained here

HQL works with persistent objects and their properties not with the database tables and columns.

Since the above query is an HQL one, the subscriber is supposed to be an entity name not a table name. Since I have my table subscriber mapped with the entity Subscriber. My problem solves if I change the code to this:

Session session = this.sessionFactory.getCurrentSession();
List<Subscriber> personsList = session.createQuery(" from Subscriber").list();

Just to keep you from getting confused. Please note that HQL is case sensitive in a number of cases. Otherwise it would have worked in my case.

Keywords like SELECT , FROM and WHERE etc. are not case sensitive but properties like table and column names are case sensitive in HQL.

https://www.tutorialspoint.com/hibernate/hibernate_query_language.htm

To further understand how hibernate mapping works, please read this

Victor Mwenda · Answer 4 · 2021-03-02T06:14:58.733

4

Ensure your have the Entity annotation and Table annotation and set the table name on the table annotation.

@Entity
@Table(name = "table_name")
public class TableName {

}

edited Mar 02 '21 at 06:14

answered Apr 13 '20 at 09:12

Victor Mwenda

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Exactly my case: I had `@Entity(name = "table_name")` but didn't have `@Table` annotation. After I changed it to `@Entity @Table(name = "table_name")`, it worked. Note: didn't have to specify `(name = "table_name")` for `@Entity` annotation. – izogfif Oct 21 '20 at 10:51

score 3 · Answer 5 · edited May 12 '15 at 11:30

3

Thanks everybody. Lots of good ideas. This is my first application in Spring and Hibernate.. so a little more patience when dealing with "novices" like me..

Please read Tom Anderson and Roman C.'s answers. They explained very well the problem. And all of you helped me.I replaced

SELECT COUNT(*) FROM Books

with

select count(book.id) from Book book

And of course, I have this Spring config:

<bean id="sessionFactory" class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean">
<property name="packagesToScan" value="extjs.model"/>

Thank you all again!

edited May 12 '15 at 11:30

Tim

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answered Jan 21 '13 at 21:04

Pascut

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How does this _add_ to the answers that you reference? Duplicating their information doesn't really help the page as a reference. – Gray Jan 18 '19 at 14:30
I don't understand what's your point here, I haven't created this answer to get more upvotes. I created a synthesis of all the answers. This is not the accepted answer as you can well see. My answer is from 2013, your answer is from 2016. So you say that my answer is extra? Sorry but you are looking for upvotes. – Pascut Jan 21 '19 at 10:28
1

Nope, not looking for upvotes @pascut. Just looking to make sure there isn't duplicate information. Just about every question has multiple answers but I tend to ask folks to think about removal if their's doesn't add anything new and comes later. – Gray Jan 21 '19 at 22:32
Thanks for your good intentions. I do believe that my answer also adds good value to the question. – Pascut Jan 22 '19 at 10:25

score 2 · Answer 6 · answered May 12 '19 at 00:09

2

In addition to the accepted answer, one other check is to make sure that you have the right reference to your entity package in sessionFactory.setPackagesToScan(...) while setting up your session factory.

answered May 12 '19 at 00:09

Durja

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Roman C · Answer 7 · 2013-01-21T20:38:17.600

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In the Spring configuration typo applicationContext.xml where the sessionFactory configured put this property

<bean id="sessionFactory" class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean">
  <property name="packagesToScan" value="${package.name}"/>

edited Jan 21 '13 at 20:38

answered Jan 21 '13 at 20:29

Roman C

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I don't know what is the package name so I've put context property for its name – Roman C Jan 21 '13 at 20:35
right now it is: But it's not working.. – Pascut Jan 21 '13 at 20:40
extjs.model.Books should be there, not in the property but in the package – Roman C Jan 21 '13 at 20:40
not working: Caused by: org.hibernate.hql.ast.QuerySyntaxException: Books is not mapped [select count(*) from Books] and now it is: – Pascut Jan 21 '13 at 20:48
Because your class named Book instead of Books – Roman C Jan 21 '13 at 20:49
in the property should be the name of the package not the name of the class – Roman C Jan 21 '13 at 20:50
you should either change the class name to plural or change the query to singular name – Roman C Jan 21 '13 at 20:53
initially it was: I changed it because you said so.. – Pascut Jan 21 '13 at 20:54
I said to change it to the package name not the class name as you did – Roman C Jan 21 '13 at 20:54
you were right with your first answer.. the package should be there. But the problem was that I was using Books instead of Book. Thanks anyway.. – Pascut Jan 21 '13 at 20:57
along as I read the comments from answers you wanted to create hibernare.cfg.xml for this should be another property `` – Roman C Jan 21 '13 at 21:13
these properties are mutual exclusive – Roman C Jan 21 '13 at 21:13
in the `hibernate.cfg.xml` should be ``, this still useful to IDE. – Roman C Jan 21 '13 at 21:16
these settings are very important, your answers will help lots of peoples :D – Pascut Jan 21 '13 at 21:18

score 1 · Answer 8 · answered Oct 30 '22 at 17:09

1

I had the same issue with the native query and what was missing is nativeQuery = true

@Query(
  value = "SELECT * FROM USERS u WHERE u.status = 1", 
  nativeQuery = true)
Collection<User> findAllActiveUsersNative();

answered Oct 30 '22 at 17:09

jfk

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score 0 · Answer 9 · answered Apr 16 '18 at 14:20

Hibernate also is picky about the capitalization. By default it's going to be the class name with the First letter Capitalized. So if your class is called FooBar, don't pass "foobar". You have to pass "FooBar" with that exact capitalization for it to work.

score 0 · Answer 10 · answered Oct 19 '20 at 09:40

I had same problem , instead @Entity I used following code for getting records

    List<Map<String, Object>> list = null;
            list = incidentHibernateTemplate.execute(new HibernateCallback<List<Map<String, Object>>>() {

            @Override
            public List<Map<String, Object>> doInHibernate(Session session) throws HibernateException {
                Query query = session.createSQLQuery("SELECT * from table where appcode = :app");
                    query.setParameter("app", apptype);
                query.setResultTransformer(Transformers.ALIAS_TO_ENTITY_MAP);
                return query.list();

                }
            });

I used following code for update

private @Autowired HibernateTemplate incidentHibernateTemplate;
Integer updateCount = 0;

    updateCount = incidentHibernateTemplate.execute((Session session) -> {
        Query<?> query = session
                    .createSQLQuery("UPDATE  tablename SET key = :apiurl, data_mode = :mode WHERE apiname= :api ");
          query.setParameter("apiurl", url);
          query.setParameter("api", api);
          query.setParameter("mode", mode);
            return query.executeUpdate();
        }
    );

score 0 · Answer 11 · answered Jan 12 '23 at 18:09

Had this issue with an old HQL app, turns out that the mapping of a certain class was not inside the hibernate config file hibernate.cfg.xml

<mapping class="com.your.package.MyClass" />

Then, I was able to use the mapping correctly using

FROM MyClass WHERE prop LIKE ":myId"

Hibernate table not mapped error in HQL query

11 Answers11

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