[已解决] org.hibernate.AnnotationException: 未为实体类指定标识符

最近我正在处理一个Hibernate项目，我添加了一些实体bean，当执行时，我得到了下面的异常堆栈跟踪。

Initial SessionFactory creation failed.org.hibernate.AnnotationException: No identifier specified for entity: com.journaldev.hibernate.model.Address
org.hibernate.AnnotationException: No identifier specified for entity: com.journaldev.hibernate.model.Address
	at org.hibernate.cfg.InheritanceState.determineDefaultAccessType(InheritanceState.java:277)
	at org.hibernate.cfg.InheritanceState.getElementsToProcess(InheritanceState.java:224)
	at org.hibernate.cfg.AnnotationBinder.bindClass(AnnotationBinder.java:775)
	at org.hibernate.cfg.Configuration$MetadataSourceQueue.processAnnotatedClassesQueue(Configuration.java:3788)
	at org.hibernate.cfg.Configuration$MetadataSourceQueue.processMetadata(Configuration.java:3742)
	at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1410)
	at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1844)
	at com.journaldev.hibernate.util.HibernateUtil.buildSessionFactory(HibernateUtil.java:22)
	at com.journaldev.hibernate.util.HibernateUtil.getSessionFactory(HibernateUtil.java:34)
	at com.journaldev.hibernate.main.HibernateExample.main(HibernateExample.java:15)
Exception in thread "main" java.lang.ExceptionInInitializerError
	at com.journaldev.hibernate.util.HibernateUtil.buildSessionFactory(HibernateUtil.java:29)
	at com.journaldev.hibernate.util.HibernateUtil.getSessionFactory(HibernateUtil.java:34)
	at com.journaldev.hibernate.main.HibernateExample.main(HibernateExample.java:15)

问题出在我忘记在实体bean中指定主键，我的bean定义如下；Address.java

package com.journaldev.hibernate.model;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.OneToOne;
import javax.persistence.PrimaryKeyJoinColumn;
import javax.persistence.Table;

import org.hibernate.annotations.GenericGenerator;
import org.hibernate.annotations.Parameter;

@Entity
@Table(name = "ADDRESS")
@Access(value=AccessType.FIELD)
public class Address {

	@Column(name = "emp_id", unique = true, nullable = false)
	@GeneratedValue(generator = "gen")
	@GenericGenerator(name = "gen", strategy = "foreign", parameters = { @Parameter(name = "property", value = "employee") })
	private long id;

	@Column(name = "address_line1")
	private String addressLine1;

	@Column(name = "zipcode")
	private String zipcode;

	@Column(name = "city")
	private String city;

	@OneToOne
	@PrimaryKeyJoinColumn
	private Employee employee;

	public long getId() {
		return id;
	}

	public void setId(long id) {
		this.id = id;
	}

	public String getAddressLine1() {
		return addressLine1;
	}

	public void setAddressLine1(String addressLine1) {
		this.addressLine1 = addressLine1;
	}

	public String getZipcode() {
		return zipcode;
	}

	public void setZipcode(String zipcode) {
		this.zipcode = zipcode;
	}

	public String getCity() {
		return city;
	}

	public void setCity(String city) {
		this.city = city;
	}

	public Employee getEmployee() {
		return employee;
	}

	public void setEmployee(Employee employee) {
		this.employee = employee;
	}

	@Override
	public String toString() {
		return "AddressLine1= " + addressLine1 + ", City=" + city
				+ ", Zipcode=" + zipcode;
	}
}

解决问题所需的全部就是用@Id注解标注主键字段。我将我的id字段声明更改为以下内容，问题得以解决。

@Id
@Column(name = "emp_id", unique = true, nullable = false)
@GeneratedValue(generator = "gen")
@GenericGenerator(name = "gen", strategy = "foreign", parameters = { @Parameter(name = "property", value = "employee") })
private long id;

这是一个简单的修复，但有一种情况会让人更加困惑。通常，Hibernate会根据在变量或getter-setter上使用的注解自动确定访问bean属性的方式。然而，我们可以显式地为我们的实体bean定义访问类型。有两种访问类型：

1. Field：在这种情况下，Hibernate将查找变量上的注解，就像我们上面为Address类定义的那样@Access(value=AccessType.FIELD)。
2. Property：在这种情况下，Hibernate将查找getter-setter方法上的注解，语法如下@Access(value=AccessType.PROPERTY)

如果注解未根据访问类型定义，我们仍然会遇到这个异常。例如，如果访问类型是property，而我们已将所有注解添加到bean变量上，我们将遇到此异常。下面是一个会抛出此异常的示例类。Employee.java

package com.journaldev.hibernate.model;

import javax.persistence.Access;
import javax.persistence.AccessType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToOne;
import javax.persistence.Table;

import org.hibernate.annotations.Cascade;

@Entity
@Table(name = "EMPLOYEE")
@Access(value=AccessType.FIELD)
public class Employee {

	private long id;

	private String name;

	private double salary;

	private Address address;

	@Id
	@GeneratedValue(strategy = GenerationType.IDENTITY)
	@Column(name = "emp_id")
	public long getId() {
		return id;
	}

	public void setId(long id) {
		this.id = id;
	}

	@OneToOne(mappedBy = "employee")
	@Cascade(value = org.hibernate.annotations.CascadeType.ALL)
	public Address getAddress() {
		return address;
	}

	public void setAddress(Address address) {
		this.address = address;
	}

	@Column(name = "emp_name")
	public String getName() {
		System.out.println("Employee getName called");
		return name;
	}

	public void setName(String name) {
		System.out.println("Employee setName called");
		this.name = name;
	}

	@Column(name = "emp_salary")
	public double getSalary() {
		return salary;
	}

	public void setSalary(double salary) {
		this.salary = salary;
	}

	@Override
	public String toString() {
		return "Id= " + id + ", Name= " + name + ", Salary= " + salary
				+ ", {Address= " + address + "}";
	}

}

请注意，所有JPA注解都与getter方法一起使用，而访问类型被定义为Field，只需将访问类型更改为property即可解决此问题。