JPA + Hibernate + PostgreSQL工程示例。
注意 user 在 postgresql 中为保留关键字,所以如果你persist一个名为User的类的时候就会报语法错误。
<dependencies>
<!-- JPA -->
<dependency>
<groupId>org.hibernate.javax.persistence</groupId>
<artifactId>hibernate-jpa-2.0-api</artifactId>
<version>1.0.1.Final</version>
</dependency>
<!-- hibernate -->
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<version>4.0.1.Final</version>
</dependency>
<!-- log4j -->
<dependency>
<groupId>org.slf4j</groupId>
<artifactId>slf4j-log4j12</artifactId>
<version>1.6.1</version>
</dependency>
<!-- JDBC Driver -->
<dependency>
<groupId>postgresql</groupId>
<artifactId>postgresql</artifactId>
<version>9.1-901.jdbc4</version>
</dependency>
</dependencies><?xml version="1.0" encoding="UTF-8"?> <persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"> <persistence-unit name="test" transaction-type="RESOURCE_LOCAL"> <provider>org.hibernate.ejb.HibernatePersistence</provider> <properties> <!-- standard properties --> <property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost:5432/jpa-test" /> <property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver" /> <property name="javax.persistence.jdbc.user" value="postgres" /> <property name="javax.persistence.jdbc.password" value="111111" /> <!-- hibernate-specific properties --> <property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQLDialect" /> <property name="hibernate.hbm2ddl.auto" value="create-drop"/> <property name="hibernate.show_sql" value="true"/> <property name="hibernate.format_sql" value="true"/> </properties> </persistence-unit> </persistence>
@Entity
@Table(name = "USER_")
public class User {
@Id @GeneratedValue
private long id;
private String name;
private String password;
// getters and setters ...
}public static void main(String[] args) {
// obtain factory
EntityManagerFactory emf = Persistence.createEntityManagerFactory("test");
// obtain EntityManager
EntityManager em = emf.createEntityManager();
// start transaction
em.getTransaction().begin();
User u = new User();
u.setName("bruce");
em.persist(u);
em.getTransaction().commit();
}JPA + Hibernate + PostgreSQL + Maven基本配置示例
原文:http://blog.csdn.net/neosmith/article/details/18866027