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

2022-09-04 01:58:02

用户声明:

@Entity
public class User {
    @Id
    @GeneratedValue
    private Integer id;
    ....

模式声明:

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

用户模式声明:

public class UserPattern {
    @Id
    @GeneratedValue
    Integer id;

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

    @ManyToOne
    @JoinColumn(name = "pattern_id")
    Pattern pattern;
    ...

对数据库的请求:

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

我得到了以下例外:

 ...
    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)
    ....

请帮助解决此问题。

错误消息看起来非常非常奇怪。

我已经阅读了相关主题点击,但我没有找到答案。

附言

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=?

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

答案 1

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

我认为Hibernate混淆了对象和ID字段。


答案 2

您需要修改查询,如下所示:

from UserPattern where user.id = :user_id and pattern.id = :pattern_id

在查询中,您尝试将对象与对象进行匹配。UserInteger


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