问题描述
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我在休眠中定义了一个OnetoMany关系,如下所示:
@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
}
table1列在users表中。
调用方法Group.getUsers()
和User.getGroup()
可以正常工作。但是我还需要在列“ 1”之后进行查询:
Criteria criteria = Activator.getDefault().getsqlSession().createCriteria(User.class);
Criterion c = Restrictions.eq(\"group_id\",1); // an id of a group
criteria.add(c);
Criterion
对象是在方法中创建的,它可以用于其他一对多表或可以包含其他列,因此我无法使用方法method7ѭ。
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user759837建议更改(Criterion c = Restrictions.eq(\"group\",1);
)之后,当我调用criteria.list()
时,出现以下错误消息: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;
}
}
我也尝试过用ѭ14but,但这是相同的错误。
编辑2:
经过大量挖掘,看起来ѭ6对象应该接收一个组对象作为参数,而不是id :: 16
我可以在那里发送一个ID吗?
解决方法
您的列是group_id,您应该使用属性group
...
标准c = Restrictions.eq(\“ group \”,1); //群组的ID
...
,这似乎可行:
Criterion c = Restrictions.eq(\"group.id\",1); // an id of a group
,如果您使用oracle作为数据库,则可能是group_id是其中的关键字,将名称更改为其他名称然后尝试。