Question

我正在使用PostgreSQL并创建了称为hib的表和架构。当我在db中创建外来表并运行应用程序时，它成功地将数据持久保存到数据库中。但是如果数据库中没有该表，则Hibernate不会在hib模式中创建表，我认为hibernate不会读取hibernate.cfg.xml中提到的hdm2ddl.auto属性。根据日志，它不提供类似创建或更新表（如果存在）的日志。它直接尝试插入值。尝试了许多方法，但无法像创建，更新，创建-删除那样工作。每次相同的错误ERROR：关系“ hib.alien”不存在。这是持久性类。

@Entity()
@Table(name="hib.alien")
public class Alien {
    @Id
    private int aid;
    private String aname;
    private String acolor;

    public Alien() {}

    public Alien(int aid, String aname, String acolor) {
        this.aid = aid;
        this.aname = aname;;
        this.acolor = acolor;
    }

    public int getAid() {
        return aid;
    }

    public void setAid(int aid) {
        this.aid = aid;
    }

    public String getAname() {
        return aname;
    }

    public void setAname(String aname) {
        this.aname = aname;
    }

    public String getAcolor() {
        return acolor;
    }

    public void setAcolor(String acolor) {
        this.acolor = acolor;
    }

    @Override
    public String toString() {
        return "Alien [aid=" + aid + ", aname=" + aname + ", acolor=" + acolor + "]";
    }
}

hibernate.cfg.xml文件

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
        "-//Hibernate/Hibernate Configuration DTD 3.0//EN"
        "http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
    <session-factory>
        <property name="hbm2ddl.auto">update</property>
        <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://localhost:5432/hib</property>
        <property name="hibernate.connection.username">postgres</property>
        <property name="hibernate.connection.password">123</property>
        <property name="show_sql">true</property>
    </session-factory>
</hibernate-configuration>

和主班

public class App 
{
    public static void main( String[] args )
    {
        Alien alien = new Alien(1, "Nagaraj Kharvi", "Red");
        System.out.println(alien);

        Configuration con = new Configuration().configure().addAnnotatedClass(Alien.class);
        SessionFactory sf = con.buildSessionFactory();
        Session s = sf.openSession();
        Transaction tx = s.beginTransaction();
        s.save(alien);
        tx.commit();
    }
}

这是日志：

INFO: HHH000400: Using dialect: org.hibernate.dialect.PostgreSQLDialect
Mar 03, 2019 10:28:08 AM org.hibernate.engine.transaction.jta.platform.internal.JtaPlatformInitiator initiateService
INFO: HHH000490: Using JtaPlatform implementation: [org.hibernate.engine.transaction.jta.platform.internal.NoJtaPlatform]
Hibernate: insert into hib.alien (acolor, aname, aid) values (?, ?, ?)
Mar 03, 2019 10:28:09 AM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 0, SQLState: 42P01
Mar 03, 2019 10:28:09 AM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: ERROR: relation "hib.alien" does not exist
  Position: 13
Mar 03, 2019 10:28:09 AM org.hibernate.internal.ExceptionMapperStandardImpl mapManagedFlushFailure
ERROR: HHH000346: Error during managed flush [org.hibernate.exception.SQLGrammarException: could not execute statement]
Exception in thread "main" javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not execute statement

Answer 1

对于PostgreSQL实体应如下所示，这样该表将创建在那种模式下。

 @Entity()
 @Table(name = "alien"  schema = "\"hib\"")
 public class Alien {
    @Id
    private int aid;
    public int getAid() {
        return aid;
    }
    public void setAid(int aid) {
        this.aid = aid;
    }
  }

Hibernate不在给定模式下在PostgreSQL中创建表

1 个答案: