一尘不染

Hibernate无法从资源解析映射文档

hibernate

我知道这已经被问过十二遍了。今天,我一定很沮丧,因为我尝试了一百种方法来修改xml文件,但我无法弄清楚。

运行驱动程序时出现错误堆栈:

16 [main] INFO org.hibernate.cfg.Environment - Hibernate 3.3.2.GA
31 [main] INFO org.hibernate.cfg.Environment - hibernate.properties not found
31 [main] INFO org.hibernate.cfg.Environment - Bytecode provider name : javassist
31 [main] INFO org.hibernate.cfg.Environment - using JDK 1.4 java.sql.Timestamp handling
78 [main] INFO org.hibernate.cfg.Configuration - configuring from resource: test_hibernate/hibernate.cfg.xml
78 [main] INFO org.hibernate.cfg.Configuration - Configuration resource: test_hibernate/hibernate.cfg.xml
125 [main] INFO org.hibernate.cfg.Configuration - Reading mappings from resource : test_hibernate/contact.hbm.xml
Exception in thread "main" Could not parse mapping document from resource            test_hibernate/contact.hbm.xml
java.lang.NullPointerException
at test_hibernate.MyExample.main(MyExample.java:29)

这是我如何建立表格的MySQL代码:

create database HibernateTest;
drop table contact;
Create table CONTACT(
ID int(15),
FIRSTNAME varchar(50) ,
LASTNAME varchar(50),
EMAIL varchar(150)); 
describe contact;

这是我的映射文件:

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping>
<class name="test_hibernate.Contact" table="CONTACT">
<id name="id" column="ID" >
<generator class="increment" />
</id>
<property name="firstName" column="FIRSTNAME" />
<property name="lastName" column="LASTNAME" />
<property name="email" column="EMAIL" />
</class>
</hibernate-mapping>

这是我的配置文件:

<?xml version='1.0' encoding='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD//EN"
"http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">

<hibernate-configuration>
<session-factory>
  <property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>
  <property name="hibernate.connection.url">jdbc:mysql://localhost/hibernatetest</property>
  <property name="hibernate.connection.username">###</property>
  <property name="hibernate.connection.password">###</property>
  <property name="hibernate.connection.pool_size">10</property>
  <property name="show_sql">true</property>
  <property name="dialect">org.hibernate.dialect.MySQLDialect</property>
  <property name="hibernate.hbm2ddl.auto">update</property>
  <!-- Mapping files -->
  <mapping resource="test_hibernate/contact.hbm.xml"/>
</session-factory>
</hibernate-configuration>

任何帮助表示赞赏,谢谢!

一些其他信息:安装程序包test_hibernate和这是我的持久性文件:

package test_hibernate;
public class Contact {
    private String firstName;
    private String lastName;
    private String email;
    private long id;

    public String getEmail() {
        return email;
    }

    public String getFirstName() {
        return firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public void setEmail(String string) {
        email = string;
    }

    public void setFirstName(String string) {
        firstName = string;
    }

    public void setLastName(String string) {
        lastName = string;
    }

    public long getId() {
        return id;
    }

    public void setId(long l) {
        id = l;
    }
}

这是我的司机:

package test_hibernate;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;
import org.hibernate.Transaction;

public class MyExample {

public static void main(String[] args) {
Session session = null;

try{
    // This step will read hibernate.cfg.xml and prepare hibernate for use
    SessionFactory sessionFactory = new Configuration().configure("hibernate.cfg.xml").buildSessionFactory();
    session=sessionFactory.openSession();
    Transaction tx = session.beginTransaction();

    //Create new instance of Contact and set values in it by reading them from form object
    System.out.println("Inserting Record");

    Contact contact = new Contact();
    contact.setId(3);
    contact.setFirstName("Smitha");
    contact.setLastName("Rao");
    contact.setEmail("smithaxxx@yahoo.com");
    session.save(contact);
    tx.commit();
    System.out.println("Done");
}catch(Exception e){
    System.out.println(e.getMessage());
}finally{
    // Actual contact insertion will happen at this step

    session.flush();        
    session.close();
}
}
}

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2020-06-20

共1个答案

一尘不染

您的映射文件看起来不错。我认为这可能是由于您在双引号中使用了 而不是 。”请尝试在所有双引号中使用

2020-06-20