Reputation: 2893

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

I have example web application Hibernate 4.3.5 + Derby database 10.10.1.1+ Glassfish4.0 with IDE NetBeans 8.0Beta.

I have the next exception:

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)
at org.hibernate.hql.internal.ast.tree.FromClause.addFromElement(FromClause.java:95)
at org.hibernate.hql.internal.ast.HqlSqlWalker.createFromElement(HqlSqlWalker.java:331)
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElement(HqlSqlBaseWalker.java:3633)
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElementList(HqlSqlBaseWalker.java:3522)
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromClause(HqlSqlBaseWalker.java:706)
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:562)
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:299)
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:247)
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:278)
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:206)
... 72 more

Form from index.xhtml

<h:panelGrid id="panel1" columns="2" border="1"
                 cellpadding="5" cellspacing="1">
        <f:facet name="header">
            <h:outputText value="Add Customer Information"/>
        </f:facet>
        <h:outputLabel value="First Name:"/>
        <h:inputText value="#{customer.firstName}" id="fn"/>
        <h:outputLabel value="Last Name:"/>
        <h:inputText value="#{customer.lastName}" id="ln"/>
        <h:outputLabel value="Email:"/>
        <h:inputText value="#{customer.email}" id="eml"/>
        <h:outputLabel value="Date of Birth:"/>
        <h:inputText value="#{customer.sd}" id="s"/>
        <f:facet name="footer">
            <h:outputLabel value="#{customer.msg}" id="msg" styleClass="msg"/>
            <h:commandButton value="Save" action="#{customer.saveCustomer}">
            </h:commandButton>
        </f:facet>
    </h:panelGrid>

Customer.java

    package com.javaknowledge.entity;

    import com.javaknowledge.dao.CustomerDao;
    import java.text.ParseException;
    import java.text.SimpleDateFormat;
    import java.util.ArrayList;
    import java.util.Date;
    import java.util.List;
    import javax.faces.bean.ManagedBean;
    import javax.faces.bean.SessionScoped;
    import javax.persistence.*;    

    @ManagedBean
    @SessionScoped

    public class Customer implements java.io.Serializable {

    private Integer custId;
    private String firstName;
    private String lastName;
    private String email;
    private Date dob;
    private String sd, msg, selectedname;
    SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");


    public Customer() {
    }

    public Customer(String firstName, String lastName, String email, Date dob) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.email = email;
        this.dob = dob;
    }

    public String getSd() {
        return sd;
    }

    public void setSd(String sd) {
        this.sd = sd;
    }

    public Integer getCustId() {
        return this.custId;
    }

    public void setCustId(Integer custId) {
        this.custId = custId;
    }

    public String getFirstName() {
        return this.firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return this.lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }
    @Column(name = "EMAIL")
    public String getEmail() {
        return this.email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    @Column(name = "DOB")
    public Date getDob() {
        return this.dob;
    }

    public void setDob(Date dob) {
        this.dob = dob;
    }

    public String getMsg() {
        return msg;
    }

    public void setMsg(String msg) {
        this.msg = msg;
    }

    public String getSelectedname() {
        return selectedname;
    }

    public void setSelectedname(String selectedname) {
        this.selectedname = selectedname;
    }

    public void saveCustomer() {
        try {
            Date d = sdf.parse(sd);
            System.out.println(d);
            this.dob = d;
        } catch (ParseException e) {
            e.printStackTrace();
        }
        CustomerDao dao = new CustomerDao();
        dao.addCustomer(this);
        this.msg = "Member Info Saved Successfull!";
        clearAll();
    }
    public void updateCustomer() {
        try {
            Date d = sdf.parse(sd);
            System.out.println(d);
            this.dob = d;
        } catch (ParseException e) {
            e.printStackTrace();
        }
        CustomerDao dao = new CustomerDao();
        dao.updateCustomer(this);
        this.msg = "Member Info Update Successfull!";
        clearAll();
    }
    public void deleteCustomer() {
        CustomerDao dao = new CustomerDao();
        dao.deleteCustomer(custId);
        this.msg = "Member Info Delete Successfull!";
        clearAll();
    }

    public List<Customer> getAllCustomers() {
        List<Customer> users = new ArrayList<Customer>();
        CustomerDao dao = new CustomerDao();
        users = dao.getAllCustomers();
        return users;
    }

    public void fullInfo() {
        CustomerDao dao = new CustomerDao();
        List<Customer> lc = dao.getCustomerById(selectedname);
        System.out.println(lc.get(0).firstName);
        this.custId = lc.get(0).custId;
        this.firstName = lc.get(0).firstName;
        this.lastName = lc.get(0).lastName;
        this.email = lc.get(0).email;
        this.dob = lc.get(0).dob;
        this.sd = sdf.format(dob);
    }

    private void clearAll() {
        this.firstName = "";
        this.lastName = "";
        this.sd = "";
        this.email = "";
        this.custId=0;
    }

   }

hibernate.cfg.xml

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC "-//Hibernate/Hibernate Configuration DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
  <session-factory>
    <property name="hibernate.dialect">org.hibernate.dialect.DerbyDialect</property>
    <property name="hibernate.connection.driver_class">org.apache.derby.jdbc.ClientDriver</property>
    <property name="hibernate.connection.url">jdbc:derby://localhost:1527/derbyDB</property>
    <property name="hibernate.connection.username">user1</property>
    <property name="hibernate.connection.password">user1</property>
    <property name="hibernate.hbm2ddl.auto">create</property>

    <property name="c3p0.min_size">1</property>
    <property name="c3p0.max_size">5</property>
    <property name="c3p0.timeout">300</property>
    <property name="c3p0.max_statements">50</property>
    <property name="c3p0.idle_test_period">300</property>

    <mapping class="com.javaknowledge.entity.Customer" resource="com/javaknowledge/entity/Customer.hbm.xml"/>
  </session-factory>
</hibernate-configuration>

Customer.hbm.xml

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
  <class name="com.javaknowledge.entity.Customer" table="CUSTOMERV" schema="APP">
        <id name="custId" type="java.lang.Integer">
            <column name="cust_id" />
            <generator class="increment" />
        </id>
        <property name="firstName" type="string">
            <column name="first_name" length="45" not-null="true" />
        </property>
        <property name="lastName" type="string">
            <column name="last_name" length="45" not-null="true" />
        </property>
        <property name="email" type="string">
            <column name="email" length="45" not-null="true" />
        </property>
        <property name="dob" type="date">
            <column name="dob" length="10" not-null="true" />
        </property>
   </class>
</hibernate-mapping>

Upvotes: 129

Answers (22)

Spilvergo

Reputation: 76

in my case was that i forgot the "nativeQuery = true"

Upvotes: 2

Shiwantha Viraj

Reputation: 331

add parameter nativeQuery = true

ex:

@Query(value="Update user set user_name =:user_name,password =:password where user_id =:user_id",nativeQuery = true)

Upvotes: 2

G. Ciardini

Reputation: 1307

The mistake in my case is that I used session.createQuery() instead of session.createSQLQuery()

Upvotes: 1

saneryee

Reputation: 3405

In Hibernate,

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

The Customerv is your Entity Name, not your Table Name

Upvotes: 6

ITisha

Reputation: 964

In my case I just forgot to add nativeQuery = true

@Query( value = "some sql query ...", nativeQuery = true)

For Spring Boot with Spring Data JPA

Upvotes: 45

Madhusanka Edirimanna

Reputation: 1428

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:

public List<Book> list() {
    List<Book> list=SessionFactory.getCurrentSession().createQuery("from book").list();
    return list;
}

The mistake I did in the above snippet is that I have used the table name foo inside createQuery. Instead, I got to use Foo, the actual class name.

public List<Book> list() {                                                                               
    List<Book> list=SessionFactory.getCurrentSession().createQuery("from Book").list();
    return list;
}

Upvotes: 2

Barani r

Reputation: 2367

Another solution that worked:

The data access object that actually throwed this exception is

public List<Foo> findAll() {
    return sessionFactory.getCurrentSession().createQuery("from foo").list();
}

The mistake I did in the above snippet is that I have used the table name foo inside createQuery. Instead, I got to use Foo, the actual class name.

public List<Foo> findAll() {
    return sessionFactory.getCurrentSession().createQuery("from Foo").list();

Thanks to this blog: https://www.arundhaj.com/blog/querysyntaxexception-not-mapped.html

Upvotes: 2

Sanjeev Kumar

Reputation: 997

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);

Upvotes: 89

raj240

Reputation: 666

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.

Upvotes: 5

Majid

Reputation: 14253

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>

Upvotes: 4

Kshitiz Bathwal

Reputation: 87

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.)

Upvotes: 1

Cjo

Reputation: 1351

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 {
   ....
   ....
}

Upvotes: 25

Debiprasad Behera

Reputation: 63

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 :)

Upvotes: 5

hatanooh

Reputation: 4287

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);
    }
}

Upvotes: 3

Radhakrishnan

Reputation: 1040

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")

Upvotes: 2

Suresh

Reputation: 1601

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();
    }

Upvotes: 2

Pipo

Reputation: 5093

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;
}

Upvotes: 8

Barani r

Reputation: 2367

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>
.....

Upvotes: 16

Siva kumar

Reputation: 235

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.

Upvotes: 6

connectoram

Reputation: 81

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 {
...
}

Upvotes: 5

Vlad Dobrydin

Reputation: 2893

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.

Upvotes: 157

Vlad Dobrydin

Reputation: 2893

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)

Upvotes: 0

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

Answers (22)

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