創建一個表，如果不在休眠中退出

的hibernate.cfg.xml

<hibernate-configuration> 
<session-factory name=""> 
    <property name="hibernate.connection.driver_class">org.postgresql.Driver</property> 
    <property name="hibernate.connection.password">toor</property> 
    <property name="hibernate.connection.url">jdbc:postgresql://localhost:5432/hibernatedb</property> 
    <property name="hibernate.connection.username">postgres</property> 
    <property name="hibernate.dialect">org.hibernate.dialect.PostgreSQLDialect</property> 
    <property name="current_session_context_class">thread</property> 
    <property name="hbm2dll.auto">update</property> 
    <property name="show_sql">true</property> 
    <mapping class="org.firsthibernatproject.dto.Person"/> 
</session-factory> 
</hibernate-configuration>

Enity：

@Entity 
@Table (name="PERSON_DETAILS") 
public class Person { 
    @Id 
    @Column (name="id") 
    private int id; 
    @Column (name="first_name") 
    private String firstName; 
    @Column (name="last_name") 
    private String lastName; 
    private Date joinedDate; 
    private String adresse; 
    private String description; 



    public String getAdresse() { 
     return adresse; 
    } 
    public void setAdresse(String adresse) { 
     this.adresse = adresse; 
    } 
    public String getDescription() { 
     return description; 
    } 
    public void setDescription(String description) { 
     this.description = description; 
    } 
    public Date getJoinedDate() { 
     return joinedDate; 
    } 
    public void setJoinedDate(Date joinedDate) { 
     this.joinedDate = joinedDate; 
    } 
    public int getId() { 
     return id; 
    } 
    public void setId(int id) { 
     this.id = id; 
    } 
    public String getFirstName() { 
     return firstName; 
    } 
    public void setFirstName(String firstName) { 
     this.firstName = firstName; 
    } 
    public String getLastName() { 
     return lastName; 
    } 
    public void setLastName(String lastName) { 
     this.lastName = lastName; 
    } 


}

這是我試圖創建新表的代碼：

public class HibernateTest { 
    public static void main (String args[]) { 
     Person person = new Person(); 
     person.setId(3); 
     person.setFirstName("Orihime"); 
     person.setLastName("Inoue"); 
     person.setAdresse("Karakura Town"); 
     person.setJoinedDate(new Date()); 
     person.setDescription("She has a unique power."); 

     SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory(); 
     Session session = sessionFactory.openSession(); 
     session.beginTransaction(); 
     session.save(person); 
     session.getTransaction().commit(); 

    } 
}

由於你可以在看到屬性我使用的值update這將創建表，如果不存在於數據庫中，而是我得到了日食的控制檯此錯誤消息：

Hibernate: insert into PERSON_DETAILS (adresse, description, first_name, joinedDate, last_name, id) values (?, ?, ?, ?, ?, ?) 
mai 24, 2014 2:49:06 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions 
WARN: SQL Error: 0, SQLState: 42P01 
mai 24, 2014 2:49:06 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions 
ERROR: ERREUR: la relation « person_details » n'existe pas 
    Position : 13 
mai 24, 2014 2:49:06 PM org.hibernate.engine.jdbc.batch.internal.AbstractBatchImpl release 
INFO: HHH000010: On release of batch it still contained JDBC statements 
Exception in thread "main" org.hibernate.exception.SQLGrammarException: could not execute statement

它說，我嘗試添加表沒有存在。

如何讓我的程序創建新表格。

我還遇到了另一個問題，即向已創建的表中添加新列，它向我顯示該列不存在，但如果表中不存在該列，則應創建該列。

來源

2014-05-24 Ichigo Kurosaki

你會嘗試改變屬性名稱hibernate.hbm2ddl.auto的 – Kalyan

這是問題，謝謝:) 你能發佈此評論作爲一個答案，讓我可以使這個話題解決。 –

發表回覆:) – Kalyan

請更改屬性名稱

hibernate.hbm2ddl.auto

來源

2014-05-24 13:30:24 Kalyan

創建一個表，如果不在休眠中退出

回答

相關問題