使用Criteria and Restrictions进行查询时,Hibernate异常"无法解析属性"

Question

我在hibernate中有一个OneToMany关系定义如下:

@Entity
@Table(name = "groups")
public class Group extends BaseModel {// BaseModel defines id as @Id and @GeneratedValue

    @OneToMany
    @JoinColumn(name = "group_id")
    private List<User> users;

    // other fields, getters and setters omitted 
}


@Entity
@Table(name = "users")
public class User extends BaseModel {

    @ManyToOne
    @JoinColumn(name = "group_id")
    private Group group;

    // other fields, getters and setters omitted 
}

列group_id位于users表中.
调用方法Group.getUsers()并User.getGroup()正常工作.但我还需要在列之后进行查询group_id:

Criteria criteria = Activator.getDefault().getSQLSession().createCriteria(User.class);
Criterion c = Restrictions.eq("group_id", 1); // an id of a group
criteria.add(c);

该Criterion对象是在一个方法中创建的,它可以用于其他一对多表,也可以包含其他列,因此我不能使用方法getUsers().

不幸的是,上面的代码给出了以下异常:

org.hibernate.QueryException: could not resolve property: group_id of: com.example.User
    at org.hibernate.persister.entity.AbstractPropertyMapping.propertyException(AbstractPropertyMapping.java:81)
    at org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:75)
    at org.hibernate.persister.entity.AbstractEntityPersister.getSubclassPropertyTableNumber(AbstractEntityPersister.java:1482)
    at org.hibernate.persister.entity.BasicEntityPropertyMapping.toColumns(BasicEntityPropertyMapping.java:62)
    and so on ...

可能是什么问题呢？

---

编辑:

在用户759837建议更改(Criterion c = Restrictions.eq("group", 1);)之后,当我调用时criteria.list(),收到以下错误消息:could not get a field value by reflection getter of com.example.Group.id

java.lang.IllegalArgumentException: Can not set java.lang.Long field com.example.BaseModel.id to java.lang.Long
    at sun.reflect.UnsafeFieldAccessorImpl.throwSetIllegalArgumentException(Unknown Source)
    at sun.reflect.UnsafeFieldAccessorImpl.throwSetIllegalArgumentException(Unknown Source)
    at sun.reflect.UnsafeFieldAccessorImpl.ensureObj(Unknown Source)
    at sun.reflect.UnsafeObjectFieldAccessorImpl.get(Unknown Source)
    at java.lang.reflect.Field.get(Unknown Source)
    at org.hibernate.property.DirectPropertyAccessor$DirectGetter.get(DirectPropertyAccessor.java:59)
    at org.hibernate.tuple.entity.AbstractEntityTuplizer.getIdentifier(AbstractEntityTuplizer.java:227)
    at org.hibernate.persister.entity.AbstractEntityPersister.getIdentifier(AbstractEntityPersister.java:3875)
    at org.hibernate.persister.entity.AbstractEntityPersister.isTransient(AbstractEntityPersister.java:3583)
    at org.hibernate.engine.ForeignKeys.isTransient(ForeignKeys.java:203)
    at org.hibernate.engine.ForeignKeys.getEntityIdentifierIfNotUnsaved(ForeignKeys.java:242)
    at org.hibernate.type.EntityType.getIdentifier(EntityType.java:456)
    at org.hibernate.type.ManyToOneType.nullSafeSet(ManyToOneType.java:130)
    ...

BaseModel类是

@MappedSuperclass
public abstract class BaseModel {

    @Id
    @GeneratedValue
    private Long id;

    public Long getId() {
        return id;
    }

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

我也尝试long id过,但这是同样的错误.

---

编辑2:

经过大量挖掘后,它看起来该Criterion对象应该接收一个组对象作为参数,而不是一个id:Restrictions.eq("group", {A_GROUP_OBJECT});

我可以发送身份证吗？

Answer 1

这似乎有效：

Criterion c = Restrictions.eq("group.id", 1); // an id of a group