Question

我已为我的实体Employee配置了复合主键，如下所示

Employee.java：

@Entity
@Table(name="employee")
@Proxy(lazy=false)
@IdClass(EmployeeId.class)
public class Employee implements Serializable {
    private static final long serialVersionUID = 1L;

    private EmployeeId employeeId;
    private Person person;
    private Branch branch;
    private boolean isActive;

    public Employee() {

    }    

    @EmbeddedId
    @AttributeOverrides({
        @AttributeOverride(name="person", column = @Column(name="person_id")),
        @AttributeOverride(name="branch", column = @Column(name="branch_id"))})

    public EmployeeId getEmployeeId() {
        return employeeId;
    }

    public void setEmployeeId(EmployeeId employeeId) {
        this.employeeId = employeeId;
    }

    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="person_id")
    public Person getPerson() {
        return person;
    }

    public void setPerson(Person person) {
        this.person = person;
    }


    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="branch_id")
    public Branch getBranch() {
        return branch;
    }

    public void setBranch(Branch branch) {
        this.branch = branch;
    }

    @Column(name="is_active")
    public boolean getIsActive() {
        return isActive;
    }

    public void setIsActive(boolean isActive) {
        this.isActive = isActive;
    }

}

EmployeeId.java：

@Embeddable
public class EmployeeId implements Serializable {
    private static final long serialVersionUID = 1L;

    private Person person;
    private Branch branch;

    public EmployeeId() {

    }

    public EmployeeId(Person argPerson, Branch argbranch) {
        this.person = argPerson;
        this.branch = argbranch;
    }


    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="person_id", insertable=false, updatable=false)
    public Person getPerson() {
        return person;
    }
    public void setPerson(Person person) {
        this.person = person;
    }

    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="branch_id", insertable=false, updatable=false)
    public Branch getBranch() {
        return branch;
    }
    public void setBranch(Branch branch) {
        this.branch = branch;
    }
}

我使用类SessionFactory创建了一个org.springframework.orm.hibernate5.LocalSessionFactoryBean bean，并将所有hbm.xml映射为MappingLocations。

我的代码抛出以下错误：

Caused by: java.lang.IllegalArgumentException: expecting IdClass mapping
at org.hibernate.metamodel.internal.AttributeFactory$3.resolveMember(AttributeFactory.java:971)
at org.hibernate.metamodel.internal.AttributeFactory$5.resolveMember(AttributeFactory.java:1029)
at org.hibernate.metamodel.internal.AttributeFactory.determineAttributeMetadata(AttributeFactory.java:451)
at org.hibernate.metamodel.internal.AttributeFactory.buildIdAttribute(AttributeFactory.java:128)
at org.hibernate.metamodel.internal.MetadataContext.buildIdClassAttributes(MetadataContext.java:337)
at org.hibernate.metamodel.internal.MetadataContext.applyIdMetadata(MetadataContext.java:269)
at org.hibernate.metamodel.internal.MetadataContext.wrapUp(MetadataContext.java:190)
at org.hibernate.metamodel.internal.MetamodelImpl.initialize(MetamodelImpl.java:219)
at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:296)
at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:476)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:707)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:723)
at org.springframework.orm.hibernate5.LocalSessionFactoryBean.buildSessionFactory(LocalSessionFactoryBean.java:504)
at org.springframework.orm.hibernate5.LocalSessionFactoryBean.afterPropertiesSet(LocalSessionFactoryBean.java:488)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFac

如何避免此错误？我使用的是spring-orm-4.3.1-RELEASE和hibernate-core-5.2.0.Final。

更新

我创建了一个示例项目，运行时出现以下错误...

Caused by: org.hibernate.AnnotationException: Property of @IdClass not found in entity sample.domain.Employee: employee

参考代码：https://www.dropbox.com/s/axr8l01iqh0qr29/idclass-using-hibernate5.tar.gz?dl=0

我做错了什么？请在此提供您的意见

Answer 1

复合键映射可以使用IdClass或Embeddable完成。如果要使用IdClass，则必须使用@Id在Employee中注释您的字段。

@IdClass(EmployeeId.class)
    class Person{
    @Id
     private Person person;
    @Id   
     private Branch branch;
    }

如果要将Embedded用作复合键，请从Person中删除@IdClass（EmployeeId.class）注释。您也不需要Person类中的person和branch字段，因为它们是在Embedded类中定义的。

Answer 2

您的情况对应于JPA 2.1 Specification的 2.4.1对应于派生身份的主键一章。

Employee的身份来自Person和Branch的身份。你还没有显示其中任何一个的代码，所以我假设他们有简单的主键。在该关系中，Person和Branch是“父实体”，Employee是“依赖”实体。

Employee的ID可以使用IdClass或EmbeddedId映射，而不是同时使用。{/ p>

参见 2.4.1.1派生身份规范章节。

如果您想使用IdClass，那么：

id类的属性名称和依赖实体类的Id属性必须如下对应：


实体类中的Id属性和id类中的相应属性必须具有相同的名称。


...


如果实体中的Id属性是与父实体的多对一或一对一关系，则id类中的对应属性必须是（...）类型父实体的Id属性。

所以你的类看起来像这样（getters，setter，多余的注释等省略）

@Entity
@IdClass(EmployeeId.class)
public class Employee {
   @Id
   @ManyToOne
   private Person person;
   @Id
   @ManyToOne
   private Branch branch;
}

public class EmployeeId {
    private Long person; // Corresponds to the type of Person ID, name matches the name of Employee.person
    private Long branch; // Corresponds to the type of Branch ID, name matches the name of Employee.branch
}

如果您使用EmbeddedId，则：

如果从属实体使用嵌入式id表示其主键，则对应于relationship属性的嵌入式id中的属性必须与父实体的主键具有相同的类型，并且必须由{{指定1}}注释应用于关系属性。必须使用MapsId注释的value元素来指定关系属性所对应的嵌入式ID中的属性名称。

代码看起来像这样：

MapsId

Answer 3

更改为：

@Entity
@Table(name = "employee")
@Proxy(lazy = false)
@IdClass(EmployeeId.class)
public class Employee implements Serializable {
private static final long serialVersionUID = 1L;

private EmployeeId id;
private Person person;
private Branch branch;
private boolean isActive;

public Employee() {

}

@EmbeddedId
@AttributeOverrides({@AttributeOverride(name = "person", column = @Column(name = "person_id") ),
    @AttributeOverride(name = "branch", column = @Column(name = "branch_id") )})

public EmployeeId getId() {
return id;
}

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

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "person_id")
public Person getPerson() {
return person;
}

public void setPerson(Person person) {
this.person = person;
}

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "branch_id")
public Branch getBranch() {
return branch;
}

public void setBranch(Branch branch) {
this.branch = branch;
}

@Column(name = "is_active")
public boolean getIsActive() {
return isActive;
}

public void setIsActive(boolean isActive) {
this.isActive = isActive;
}

}

Answer 4

IdClass不应该被定义为可嵌入 -

@Entity
@Table(name="employee")
@IdClass(EmployeeId.class)
public class Employee implements Serializable {
   private static final long serialVersionUID = 1L;

   @Id   
   @ManyToOne
   private Person person;
   @Id
   @ManyToOne
   private Branch branch;

   private boolean isActive;

   public Employee() { }
   //....
}

和 -

public class EmployeeId implements Serializable {
    private static final long serialVersionUID = 1L;

    private Person person;
    private Branch branch;

    public EmployeeId() {}

    public EmployeeId(Person argPerson, Branch argbranch) {
        this.person = argPerson;
        this.branch = argbranch;
    }
}

阅读您的评论 - 我是否可以建议您将Employee映射到person_id和branch_id，而不是JPA对象Person and Branch？这将让我们测试您的hbm配置是否正确。我还建议发布您的hbm配置，因为我认为此问题中缺少信息

因此该表类似于 -

@Entity
@Table(name="employee")
@IdClass(EmployeeId.class)
public class Employee implements Serializable {
   private static final long serialVersionUID = 1L;

   @Id
   private Long personId;
   @Id
   private Long branchId;

   private boolean isActive;

   public Employee() { }
   //....
}

和 -

public class EmployeeId implements Serializable {
    private static final long serialVersionUID = 1L;

    private Long personId;
    private Long branchId;

    public EmployeeId() {}

    public EmployeeId(Person argPerson, Branch argbranch) {
        this.person = argPerson;
        this.branch = argbranch;
    }
}

Answer 5

使用包含@IdClass的类提及ID注释。在this帖子

上查看答案

Answer 6

此链接可以帮助您 JPA - EmbeddedId with @ManytoOne

不支持嵌入式id类中定义的关系映射。然后你需要像这样更改embeddedId类

@Embeddable
public class EmployeeId implements Serializable {
    private static final long serialVersionUID = 1L;

    private Long personId;
    private Long branchId;

    public EmployeeId() {

    }

    public EmployeeId(Long argPerson, Long argbranch) {
        this.personId = argPerson;
        this.branchId = argbranch;
    }


     @Column(name = "person_id")
    public Long getPersonId() {
        return personId;
    }
    public void setPersonId(Long personId) {
        this.personId = personId;
    }

    @Column(name = "branch_id")
    public Long getBranchId() {
        return branchId;
    }
    public void setBranchId(Long branchId) {
        this.branchId = branchId;
    }
}

Answer 7

JPA复合主键

指定映射到实体的多个字段或属性的复合主键类。

主键类和字段中的字段或属性的名称   主要字段或实体的属性必须对应和   他们的类型必须相同。

答案就在这里。为您阅读说明。 enter link description here

（示例代码）

@Entity
@Table(name = "EMP_PROJECT")
@IdClass(ProjectAssignmentId.class)
public class ProjectAssignment {
   @Id
   @Column(name = "EMP_ID", insertable = false, updatable = false)
   private int empId;

   @Id
   @Column(name = "PROJECT_ID", insertable = false, updatable = false)
   private int projectId;

   @ManyToOne
   @JoinColumn(name = "EMP_ID")
   Professor employee;

   @ManyToOne
   @JoinColumn(name = "PROJECT_ID")
   Project project;
   ....
}

public class ProjectAssignmentId implements Serializable {
   private int empId;
   private int projectId;
  ...
}

java.lang.IllegalArgumentException：期待IdClass映射

7 个答案: