Question

我正在尝试学习hibernate框架，创建一个简单的程序，将类 Cliente 推送到postgres上的表中，返回的错误如下：

Exception in thread "main" org.hibernate.exception.SQLGrammarException: could not extract ResultSet

Caused by: org.postgresql.util.PSQLException: ERROR: the relation "clienti_id_seq" does not exist

这是PGAdmin的数据库（很抱歉，我无法直接上传图片）http://i.imgur.com/Fz9o1fR.png?1

这是 Cliente

类

public class Cliente {

private long clienteId;
private String clienteNome;
private String clienteCognome;
private String clienteTelefono;
private String clienteMail;
private String clientePermesso;
private long clienteCommessa;

public long getClienteId() {
    return clienteId;
}

public void setClienteId(long clienteId) {
    this.clienteId = clienteId;
}

public String getClienteNome() {
    return clienteNome;
}

public void setClienteNome(String clienteNome) {
    this.clienteNome = clienteNome;
}
public String getClienteCognome() {
    return clienteCognome;
}

public void setClienteCognome(String clienteCognome) {
    this.clienteCognome = clienteCognome;
}

 public String getClienteTelefono() {
    return clienteTelefono;
}

public void setClienteTelefono(String clienteTelefono) {
    this.clienteTelefono = clienteTelefono;
}

 public String getClienteMail() {
    return clienteMail;
}

public void setClienteMail(String clienteMail) {
    this.clienteMail = clienteMail;
}

 public String getClientePermesso() {
    return clientePermesso;
}

public void setClientePermesso(String clientePermesso) {
    this.clientePermesso = clientePermesso;
}

 public long getClienteCommessa() {
    return clienteCommessa;
}

public void setClienteCommessa(long clienteNome) {
    this.clienteCommessa = clienteCommessa;
}

}

这是我的映射文件

<hibernate-mapping>
<class name="beans.Cliente" table="Clienti">
    <id name="clienteId" type="integer" column="id" >

        <generator class="sequence"> 
            <param name="sequence">CLIENTI_ID_seq</param>            
        </generator>
    </id>
    <property name="clienteNome" column="nome" type="string">            
    </property>
    <property name="clienteCognome" column="cognome" type="string">            
    </property>
        <property name="clienteTelefono" column="telefono" type="string">            
    </property>
    <property name="clienteMail" column="mail" type="string">            
 </property>
        <property name="clientePermesso" column="permesso" type="string">            
    </property>
    <property name="clienteCommessa" column="commessa" type="string">            
</property>
</class>
</hibernate-mapping>

这是 hibernate cfg.xml 文件，关于url pass和user的信息是正确的

<?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.PostgreSQLDialect</property>
<property name="hibernate.connection.driver_class">org.postgresql.Driver</property>
<property name="hibernate.connection.url">jdbc:postgresql:postgres</property>

<property name="hibernate.connection.url">jdbc:postgresql://localhost:5432/Georilievi</property>
<property name="hibernate.connection.username">postgres</property>
<property name="hibernate.connection.password">Fabio1990</property>
<property name="hibernate.current_session_context_class">thread</property>


<mapping resource="Clienti.hbm.xml"/>
</session-factory>

</hibernate-configuration>

这是主要文件，在这里我创建了Cliente，并试图推进postgres数据库

package main;

import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.boot.registry.StandardServiceRegistryBuilder;
import org.hibernate.cfg.Configuration;
import beans.Cliente;

public class Main {
public static void main(String [] args){
    // Create a configuration instance
    Configuration configuration = new Configuration();
    // Provide configuration file
    configuration.configure("hibernate.cfg.xml");
    // Build a SessionFactory
    SessionFactory factory = configuration.buildSessionFactory(new StandardServiceRegistryBuilder().configure().build());
    // Get current session, current session is already associated with Thread
    Session session = factory.getCurrentSession();

    // Begin transaction

    session.getTransaction().begin();

    Cliente cliente = new Cliente();
    cliente.setClienteNome("Fabio");
    cliente.setClienteCognome("Tramontana");
    cliente.setClienteTelefono("3343052346");
    cliente.setClienteMail("info.tramontanafabio@gmail.com");
    cliente.setClientePermesso("admin");
    cliente.setClienteCommessa(0);
    // Save*/
    session.save(cliente);
    // Commit, calling of commit will cause save an instance of employee
    session.getTransaction().commit();
    }
}

谢谢你们所有人帮助我，我不明白错误，我认为这是在发电机的声明中。

Answer 1

请尝试将生成器类从序列更改为标识并查看。我想问题是自动生成ID。

请参阅以下链接： http://www.roseindia.net/hibernate/hibernateidgeneratorelement.shtml http://docs.jboss.org/hibernate/core/3.3/reference/en/html/mapping.html#mapping-declaration-property

org.postgresql.util.PSQLException：错误：关系＆＃34; clienti_id_seq＆＃34;不存在

1 个答案: