JPA：映射多对多关系指的是不正确的列名答案

【问题标题】：JPA: Mapping ManyToMany relationship refers to incorrect column nameJPA：映射多对多关系指的是不正确的列名
【发布时间】：2016-03-04 23:13:22
【问题描述】：

这个问题与我昨天问的previous question 有关。我在 Employee 和 SkillSet 表之间有多对多的关系，每个关系都有额外的列 numberOfYears

employeeId  skillSetId  numberOfYears 
10          101         2

由于 EmployeeSkillSet 表中没有 ID 列，我使用@IdClass 来定义复合键

@Entity
class Employee {
    private @Id Long id;
    @OneToMany(mappedBy="employeeId")
    private List<EmployeeSkillSet> skillSets;
}
class SkillSet {
    private @Id Long id;
}

@IdClass(EmpSkillKey.class)
@Entity
class EmployeeSkillSet {
    @Id 
    @Column("employee_id")
    private Long employeeId;
    @Id
    @Column("skill_id")
    private @Id Long skillId;

    @ManyToOne  
    private Employee employee;
    private int numberOfYears;
}

class EmpSkillKey{
   private int employeeId;
   private int skillId;
}


interface EmployeeRepository extends JPARepository{
   List<Employee> getEmployeesBySkillSetSkillId(long id);
}

上述 JPA 存储库方法工作正常，并根据技能集 ID 为我提供了员工列表。但是，当我尝试遍历列表并获取 EmployeeSkillSet 对象时，它会引发错误，因为它试图映射到不正确的列 employee 而不是 employeeId。

List<Employee> emps = employeeRepository.getEmployeesBySkillSetSkillId(101);
for(Employee e: emps){  // this line throws error
  EmployeeSkillSet  ess = e.getEmployeeSkillSet();
  int n = ess.getNumberOfYears();
}

生成的查询是这样的。（我已将其转换为 Employee 用例，无法共享实际查询）

select ud.employee_id , ud.employee_id , ud.employee , ud.employee_value , rd.employee_id 
 from employee_skill_set ud left outer join employee rd 
 on ud.employee=rd.employee_id 
 where ud.employee_id=?

例外

WARN  - SqlExceptionHelper         - SQL Error: 207, SQLState: ZZZZZ
ERROR - SqlExceptionHelper         - Invalid column name 'employee'.
org.hibernate.exception.GenericJDBCException: could not extract ResultSet
    at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:54)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:126)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:112)
    at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:91)
    at org.hibernate.loader.plan.exec.internal.AbstractLoadPlanBasedLoader.getResultSet(AbstractLoadPlanBasedLoader.java:449)
    at org.hibernate.loader.plan.exec.internal.AbstractLoadPlanBasedLoader.executeQueryStatement(AbstractLoadPlanBasedLoader.java:202)
    at org.hibernate.loader.plan.exec.internal.AbstractLoadPlanBasedLoader.executeLoad(AbstractLoadPlanBasedLoader.java:137)
    at org.hibernate.loader.plan.exec.internal.AbstractLoadPlanBasedLoader.executeLoad(AbstractLoadPlanBasedLoader.java:102)
    at org.hibernate.loader.collection.plan.AbstractLoadPlanBasedCollectionInitializer.initialize(AbstractLoadPlanBasedCollectionInitializer.java:100)
    at org.hibernate.persister.collection.AbstractCollectionPersister.initialize(AbstractCollectionPersister.java:693)
    at org.hibernate.event.internal.DefaultInitializeCollectionEventListener.onInitializeCollection(DefaultInitializeCollectionEventListener.java:92)
    at org.hibernate.internal.SessionImpl.initializeCollection(SessionImpl.java:1933)
    at org.hibernate.collection.internal.AbstractPersistentCollection$4.doWork(AbstractPersistentCollection.java:559)
    at org.hibernate.collection.internal.AbstractPersistentCollection.withTemporarySessionIfNeeded(AbstractPersistentCollection.java:261)
    at org.hibernate.collection.internal.AbstractPersistentCollection.initialize(AbstractPersistentCollection.java:555)
    at org.hibernate.collection.internal.AbstractPersistentCollection.read(AbstractPersistentCollection.java:143)
    at org.hibernate.collection.internal.PersistentBag.iterator(PersistentBag.java:294)

可能我无法在同一个班级中定义@Id employeeId 和@ManyToOne employee。但是那怎么解决呢？

【问题讨论】：

标签： java jpa orm spring-data-jpa

【解决方案1】：

没关系，我找到了解决方案。用@JoinColumn 和实际的列名注释了@ManyToOne 关系。不知道为什么它要求将updatable 和insertable 设为false。必须弄清楚我的基础知识:)

@IdClass(EmpSkillKey.class)
@Entity
class EmployeeSkillSet {
    @Id 
    @Column("employee_id")
    private Long employeeId;
    @Id
    @Column("skill_id")
    private @Id Long skillId;

    @JoinColumn(name="employee_id", insertable=false, updatable=false)
    @ManyToOne  
    private Employee employee;
    private int numberOfYears;
}

【讨论】：