无法将java.lang.Integer字段设置为java.lang.Integer

gst*_*low 14 java persistence hibernate hql

用户声明:

@Entity
public class User {
    @Id
    @GeneratedValue
    private Integer id;
    ....
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模式声明:

@Entity
public class Pattern {
    @Id
    @GeneratedValue
    Integer id;
    ...
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UserPatternDeclaration:

public class UserPattern {
    @Id
    @GeneratedValue
    Integer id;

    @ManyToOne
    @JoinColumn(name = "user_id")
    User user;

    @ManyToOne
    @JoinColumn(name = "pattern_id")
    Pattern pattern;
    ...
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请求数据库:

Session session = sessionFactory.getCurrentSession();
Query query = session.createQuery("from UserPattern where user = :user_id and pattern = :pattern_id ");
query.setParameter("user_id", userId);
query.setParameter("pattern_id", pattern_id);
List<UserPattern> list = query.list();//exception throws here
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我有以下异常:

 ...
    java.lang.IllegalArgumentException: Can not set java.lang.Integer field 
    com.....s.model.User.id to java.lang.Integer
        at sun.reflect.UnsafeFieldAccessorImpl.throwSetIllegalArgumentException(UnsafeFieldAccessorImpl.java:164)
        at sun.reflect.UnsafeFieldAccessorImpl.throwSetIllegalArgumentException(UnsafeFieldAccessorImpl.java:168)
        at sun.reflect.UnsafeFieldAccessorImpl.ensureObj(UnsafeFieldAccessorImpl.java:55)
        at sun.reflect.UnsafeObjectFieldAccessorImpl.get(UnsafeObjectFieldAccessorImpl.java:36)
        at java.lang.reflect.Field.get(Field.java:379)
    ....
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请帮助解决此问题.

错误信息看起来很奇怪.

我已阅读相关主题点击但我没有找到答案.

PS

休眠日志(异常之前):

Hibernate: 
    select
        userpatter0_.id as id1_2_,
        userpatter0_.amountSearched as amountSe2_2_,
        userpatter0_.amountplayed as amountpl3_2_,
        userpatter0_.pattern_id as pattern_4_2_,
        userpatter0_.user_id as user_id5_2_ 
    from
        UserPattern userpatter0_ 
    where
        userpatter0_.user_id=? 
        and userpatter0_.pattern_id=?
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在浏览器中,我看到以下消息:

HTTP Status 500....could not get a field value by reflection getter of...model.User.id
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Pim*_*mgd 15

如果将HQL查询更改为from UserPattern where user.id = :user_id and pattern.id = :pattern_id?会发生什么?

我认为Hibernate会混淆对象和ID字段.

  • 不,但想法是这样的:如果一切都是公开的...您需要调用什么来获取用户的ID?UserPattern.user.id!这就是为什么它是user.id。 (2认同)