org.hibernate.MappingException:无法确定:java.util.List的类型,在表:College,对于列:[org.hibernate.mapping.Column(students)]

Question

现在,我正在学习hibernate,并开始在我的项目中使用它.这是一个CRUD应用程序.我使用hibernate进行所有的crud操作.它适用于所有人.但是,One-To-Many和Many-To-One,我厌倦了尝试它.最后它给了我以下错误.

org.hibernate.MappingException: Could not determine type for: java.util.List, at table: College, for columns: [org.hibernate.mapping.Column(students)]

然后我再次浏览了这个视频教程.一开始对我来说非常简单.但是,我不能让它发挥作用.现在,它说

org.hibernate.MappingException: Could not determine type for: java.util.List, at table: College, for columns: [org.hibernate.mapping.Column(students)]

我在互联网上运行了一些搜索,有人告诉它在Hibernate中有一个错误,有些人说,通过添加@GenereatedValue,这个错误将被清除.但是,nothings对我有用,

我希望我能得到一些解决方案!

谢谢!

在这里我的代码:

College.java

@Entity
public class College {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private int collegeId;
private String collegeName;


private List<Student> students;

@OneToMany(targetEntity=Student.class, mappedBy="college", fetch=FetchType.EAGER)
public List<Student> getStudents() {
    return students;
}
public void setStudents(List<Student> students) {
    this.students = students;
}//Other gettters & setters omitted

Student.java

@Entity
public class Student {


@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private int studentId;
private String studentName;


private College college;

@ManyToOne
@JoinColumn(name="collegeId")
public College getCollege() {
    return college;
}
public void setCollege(College college) {
    this.college = college;
}//Other gettters & setters omitted

Main.java:

public class Main {

private static org.hibernate.SessionFactory sessionFactory;

  public static SessionFactory getSessionFactory() {
    if (sessionFactory == null) {
      initSessionFactory();
    }
    return sessionFactory;
  }

  private static synchronized void initSessionFactory() {
    sessionFactory = new AnnotationConfiguration().configure().buildSessionFactory();

  }

  public static Session getSession() {
    return getSessionFactory().openSession();
  }

  public static void main (String[] args) {
                Session session = getSession();
        Transaction transaction = session.beginTransaction();
        College college = new College();
        college.setCollegeName("Dr.MCET");

        Student student1 = new Student();
        student1.setStudentName("Peter");

        Student student2 = new Student();
        student2.setStudentName("John");

        student1.setCollege(college);
        student2.setCollege(college);



        session.save(student1);
        session.save(student2);
        transaction.commit();
  }


}

安慰:

 Exception in thread "main" org.hibernate.MappingException: Could not determine type  for: java.util.List, at table: College, for columns:  [org.hibernate.mapping.Column(students)]
at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:306)
at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:290)
at org.hibernate.mapping.Property.isValid(Property.java:217)
at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:463)
at org.hibernate.mapping.RootClass.validate(RootClass.java:235)
at org.hibernate.cfg.Configuration.validate(Configuration.java:1330)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1833)
at test.hibernate.Main.initSessionFactory(Main.java:22)
at test.hibernate.Main.getSessionFactory(Main.java:16)
at test.hibernate.Main.getSession(Main.java:27)
at test.hibernate.Main.main(Main.java:43)

XML:

<?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>
    <!-- Database connection settings -->
    <property name="connection.driver_class">com.mysql.jdbc.Driver</property>
    <property name="connection.url">jdbc:mysql://localhost:3306/dummy</property>
    <property name="connection.username">root</property>
    <property name="connection.password">1234</property>
    <!-- JDBC connection pool (use the built-in) -->
    <property name="connection.pool_size">1</property>
    <!-- SQL dialect -->
    <property name="dialect">org.hibernate.dialect.MySQLDialect</property>
    <!-- Enable Hibernate's automatic session context management -->
    <property name="current_session_context_class">thread</property>
    <!-- Disable the second-level cache -->
    <property name="cache.provider_class">org.hibernate.cache.NoCacheProvider</property>
    <!-- Echo all executed SQL to stdout -->
    <property name="show_sql">true</property>
    <!-- Drop and re-create the database schema on startup -->
    <property name="hbm2ddl.auto">update</property>

    <mapping class="test.hibernate.Student" />
    <mapping class="test.hibernate.College" />
</session-factory>

Answer 1

您正在使用字段访问策略(由@Id注释确定).将任何JPA相关注释放在每个字段的正上方而不是getter属性

@OneToMany(targetEntity=Student.class, mappedBy="college", fetch=FetchType.EAGER)
private List<Student> students;

Answer 2

添加@ElementCollection到列表字段解决了此问题:

    @Column
    @ElementCollection(targetClass=Integer.class)
    private List<Integer> countries;

Answer 3

访问策略的问题

作为JPA提供者,Hibernate可以内省实体属性(实例字段)或访问者(实例属性).默认情况下,@Id注释的放置提供默认访问策略.当放置在字段上时,Hibernate将假定基于字段的访问.Hibernate放置在标识符getter上,将使用基于属性的访问.

基于现场的访问

使用基于字段的访问时,添加其他实体级方法要灵活得多,因为Hibernate不会考虑持久化状态的那些部分

@Entity
public class Simple {

@Id
private Integer id;

@OneToMany(targetEntity=Student.class, mappedBy="college", 
fetch=FetchType.EAGER)
private List<Student> students;

//getter +setter
}

基于财产的访问

使用基于属性的访问时,Hibernate使用访问器来读取和写入实体状态

@Entity
public class Simple {

private Integer id;
private List<Student> students;

@Id
public Integer getId() {
    return id;
}

public void setId( Integer id ) {
    this.id = id;
}
@OneToMany(targetEntity=Student.class, mappedBy="college", 
fetch=FetchType.EAGER)
public List<Student> getStudents() {
   return students;
}
public void setStudents(List<Student> students) {
    this.students = students;
}

}

但是,您不能同时使用基于字段和基于属性的访问.它会为你显示那样的错误

更多想法请关注此事

Answer 4

只需在数组列表变量上插入 @ElementCollection 注释，如下所示：

@ElementCollection
private List<Price> prices = new ArrayList<Price>();

我希望这有帮助

Answer 5

@Access(AccessType.PROPERTY)
@OneToOne(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
@JoinColumn(name="userId")
public User getUser() {
    return user;
}

我有同样的问题,我通过添加解决了它 @Access(AccessType.PROPERTY)