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org.hibernate.hql.internal.ast.QuerySyntaxException: table is not mapped

由 感情迁移 提交于 2019-11-28 17:22:39
Vlad Dobrydin

Finally I found a mistake! Hope this is useful to someone. When doing a request to the database(in my case it Apache Derby), name of base need write the first letter upper case other in lower case.

This is wrong query:

session.createQuery("select first_name from CUSTOMERV").

This is valid query

session.createQuery("select first_name from Customerv").

And class entity must be same name as database, but I'm not sure.

Sanjeev Kumar

in HQL query, Don't write the Table name, write your Entity class name in your query like 

String s = "from Entity_class name";
query qry = session.createUqery(s);

hibernate.cfg.xml file should have the mapping for the tables like below. Check if it is missing in your file.

......
<hibernate-configuration>
......
......
  <session-factory>
......
<mapping class="com.test.bean.dbBean.testTableHibernate"/>
......
 </session-factory>

</hibernate-configuration>
.....

If you are using the JPA annotations to create the entities and then make sure that the table name is mapped along with @Table annotation instead of @Entity.

Incorrectly mapped :

@Entity(name="DB_TABLE_NAME")
public class DbTableName implements Serializable {
   ....
   ....
}

Correctly mapped entity :

@Entity
@Table(name="DB_TABLE_NAME")
public class DbTableName implements Serializable {
   ....
   ....
}
connectoram

May be this will make it more clear, and of course makes sense too.

@Entity
@Table(name = "users")

/**
 * 
 * @author Ram Srinvasan
 * Use class name in NamedQuery
 * Use table name in NamedNativeQuery
 */
@NamedQueries({ @NamedQuery(name = "findUserByName", query = "from User u where u.name= :name") })

@NamedNativeQueries({ @NamedNativeQuery(name = "findUserByNameNativeSQL", query = "select * from users u where u.name= :name", resultClass = User.class) })
public class User implements Principal {
...
}

There is one more chance to get this exception even we used class name i.e., if we have two classes with same name in different packages. we'll get this problem.

I think hibernate may get ambiguity and throws this exception, so the solution is to use complete qualified name(like com.test.Customerv)

I added this answer that will help in scenario as I mentioned. I got the same scenario got stuck for some time.

None of the other solution worked for me.

Even if I don't think its the best practice, I Had to add it into the code like this

configuration.addAnnotatedClass(com.myOrg.entities.Person.class);

here

public static SessionFactory getSessionFactory() {
    Configuration configuration = new Configuration().configure();

    configuration.addAnnotatedClass(com.myOrg.entities.Person.class);

    StandardServiceRegistryBuilder builder = new StandardServiceRegistryBuilder()
            .applySettings(configuration.getProperties());
    SessionFactory sessionFactory = configuration.buildSessionFactory(builder.build());
    return sessionFactory;
}

I too have faced similar issue when i started to work on Hibernate. All in all i can say is in the createQuery one needs to pass the name of the entity class not the table name to which the entity is mapped to.

If you by any chance using java for configuration, you may need to check the below bean declaration if you have package level changes. Eg: com.abc.spring package changed to com.bbc.spring 

@Bean
    public SessionFactory sessionFactory() {

        LocalSessionFactoryBuilder builder = new LocalSessionFactoryBuilder(dataSource());
        //builder.scanPackages("com.abc.spring");    //Comment this line as this package no longer valid.
        builder.scanPackages("com.bbc.spring");
        builder.addProperties(getHibernationProperties());

        return builder.buildSessionFactory();
    }

It means your table is not mapped to the JPA. Either Name of the table is wrong (Maybe case sensitive), or you need to put an entry in the XML file.

Happy Coding :)

Other persons that are using mapping classes for Hibernate, make sure that have addressed correctly to model package in sessionFactory bean declaration in the following part:

<property name="packagesToScan" value="com.mblog.model"></property>

Problem partially was solved. Besides creating jdbc/resource(DB Derby) had to create JDBC Connection Pool for db resource in Glassfish admin console, and check it on pinging. Now all CRUD operation work just fine. I check, object Customer in database adding properly, update and delete too. But in Glassfish output log have same exception: 

SEVERE:   org.hibernate.hql.internal.ast.QuerySyntaxException: CUSTOMERV is not mapped [select concat(first_name, ' ', last_name) as name from CUSTOMERV]
    at org.hibernate.hql.internal.ast.QuerySyntaxException.generateQueryException(QuerySyntaxException.java:96)
    at org.hibernate.QueryException.wrapWithQueryString(QueryException.java:120)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:234)
    .......

Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: CUSTOMERV is not mapped
    at org.hibernate.hql.internal.ast.util.SessionFactoryHelper.requireClassPersister(SessionFactoryHelper.java:189)
    at org.hibernate.hql.internal.ast.tree.FromElementFactory.addFromElement(FromElementFactory.java:109)

Should use Entity class name for em.createQuery method or Should use em.createNativeQuery method for native query without entity class

With Entity class:

em.createQuery("select first_name from CUSTOMERV")

Without Entity class or Native query:

em.createNativeQuery("select c.first_name from CUSTOMERV c")

In my case: spring boot 2 ,multiple datasource(default and custom). entityManager.createQuery go wrong: 'entity is not mapped'

while debug, i find out that the entityManager's unitName is wrong(should be custom,but the fact is default) the right way:

@PersistenceContext(unitName = "customer1") // !important, 
private EntityManager em;

the customer1 is from the second datasource config class:

@Bean(name = "customer1EntityManagerFactory")
public LocalContainerEntityManagerFactoryBean entityManagerFactory(EntityManagerFactoryBuilder builder,
        @Qualifier("customer1DataSource") DataSource dataSource) {
    return builder.dataSource(dataSource).packages("com.xxx.customer1Datasource.model")
            .persistenceUnit("customer1")
            // PersistenceUnit injects an EntityManagerFactory, and PersistenceContext
            // injects an EntityManager.
            // It's generally better to use PersistenceContext unless you really need to
            // manage the EntityManager lifecycle manually.
            // 【4】
            .properties(jpaProperties.getHibernateProperties(new HibernateSettings())).build();
}

Then,the entityManager is right.

But, em.persist(entity) doesn't work,and the transaction doesn't work.

Another important point is:

@Transactional("customer1TransactionManager") // !important
public Trade findNewestByJdpModified() {
    //test persist,working right!
    Trade t = new Trade();
    em.persist(t);
    log.info("t.id" + t.getSysTradeId());

    //test transactional, working right!
    int a = 3/0;
}

customer1TransactionManager is from the second datasource config class:

@Bean(name = "customer1TransactionManager")
public PlatformTransactionManager transactionManager(
        @Qualifier("customer1EntityManagerFactory") EntityManagerFactory entityManagerFactory) {
    return new JpaTransactionManager(entityManagerFactory);
}

The whole second datasource config class is :

package com.lichendt.shops.sync;

import javax.persistence.EntityManagerFactory;
import javax.sql.DataSource;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.beans.factory.annotation.Qualifier;
import org.springframework.boot.autoconfigure.jdbc.DataSourceProperties;
import org.springframework.boot.autoconfigure.orm.jpa.HibernateSettings;
import org.springframework.boot.autoconfigure.orm.jpa.JpaProperties;
import org.springframework.boot.context.properties.ConfigurationProperties;
import org.springframework.boot.orm.jpa.EntityManagerFactoryBuilder;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.data.jpa.repository.config.EnableJpaRepositories;
import org.springframework.orm.jpa.JpaTransactionManager;
import org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean;
import org.springframework.transaction.PlatformTransactionManager;
import org.springframework.transaction.annotation.EnableTransactionManagement;

@Configuration
@EnableTransactionManagement
@EnableJpaRepositories(entityManagerFactoryRef = "customer1EntityManagerFactory", transactionManagerRef = "customer1TransactionManager",
        // 【1】这里写的是DAO层的路径 ，如果你的DAO放在 com.xx.DAO下面，则这里写成 com.xx.DAO
        basePackages = { "com.lichendt.customer1Datasource.dao" })
public class Custom1DBConfig {

    @Autowired
    private JpaProperties jpaProperties;

    @Bean(name = "customer1DatasourceProperties")
    @Qualifier("customer1DatasourceProperties")
    @ConfigurationProperties(prefix = "customer1.datasource")
    public DataSourceProperties customer1DataSourceProperties() {
        return new DataSourceProperties();
    }

    @Bean(name = "customer1DataSource")
    @Qualifier("customer1DatasourceProperties")
    @ConfigurationProperties(prefix = "customer1.datasource") //
    // 【2】datasource配置的前缀，对应上面 【mysql的yaml配置】
    public DataSource dataSource() {
        // return DataSourceBuilder.create().build();
        return customer1DataSourceProperties().initializeDataSourceBuilder().build();
    }

    @Bean(name = "customer1EntityManagerFactory")
    public LocalContainerEntityManagerFactoryBean entityManagerFactory(EntityManagerFactoryBuilder builder,
            @Qualifier("customer1DataSource") DataSource dataSource) {
        return builder.dataSource(dataSource).packages("com.lichendt.customer1Datasource.model") // 【3】这里是实体类的包路径
                .persistenceUnit("customer1")
                // PersistenceUnit injects an EntityManagerFactory, and PersistenceContext
                // injects an EntityManager.
                // It's generally better to use PersistenceContext unless you really need to
                // manage the EntityManager lifecycle manually.
                // 【4】
                .properties(jpaProperties.getHibernateProperties(new HibernateSettings())).build();
    }

    @Bean(name = "customer1TransactionManager")
    public PlatformTransactionManager transactionManager(
            @Qualifier("customer1EntityManagerFactory") EntityManagerFactory entityManagerFactory) {
        return new JpaTransactionManager(entityManagerFactory);
    }
}

In Apache Derby DB, refrain from using table names as "user" or so because they are reserved keywords on Apache Derby but will work fine on MySql.

In the Query, you must specify the name of the Entity class that you want to fetch the data from in the FROM clause of the Query.

List<User> users=session.createQuery("from User").list();

Here, User is the name of my Java Entity class(Consider the casing of the name as in Java it matters.)

标签
java
hibernate
derby
glassfish-4
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