无法将 java.lang.Integer 字段设置为 java.lang.ZA0FAEF0851B4194C46F2B2B

Question

User declaration:

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

Pattern declaration:

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

UserPatternDeclaration:

public class UserPattern {
    @Id
    @GeneratedValue
    Integer id;

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

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

request to database:

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

I got following exception:

 ...
    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)
    ....

Please help to fix this issue.

error message looks very very strange.

I have read related topic click but I don't found out answer.

PS

hibernate log(before exception):

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=?

In browser I see following message:

HTTP Status 500....could not get a field value by reflection getter of...model.User.id

Answer 1

What happens if you change your HQL query to from UserPattern where user.id = :user_id and pattern.id = :pattern_id ?

I think Hibernate is confusing objects and ID fields.

Answer 2

You need to modify your query as follows:

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

In your query, you are trying to match a User object with an Integer object.

Answer 3

If your field name is "id", your getter and setter methods should be named

public Integer getId(){return id;}
public void setId(Integer id){this.id = id};

If your are using Eclipse, generate the getter/setter by right click -> Source -> Generate Getters and Setters...

Make sure your getters and setter are public. Also you should add @Table -Annotation to all your Entities

Answer 4

i think maybe your annotation should be

@ManyToOne(TargetEntity=....class)

Answer 5

I struggled with this for days and have finally found the answer.
You used to be able to tell Hibernate what type the parameter was by using alternatives to setParameter() such as setInteger etc. But these are now deprecated.

The problem is Hibernate is failing to interpret the type correctly for some reason. But there is an overloaded version of setParameter which allows you to set the type directly.

Firstly import the types: import org.hibernate.type.StandardBasicTypes;

Then use the following syntax to set the Parameter: q.setParameter("name", variable containing the value, StandardBasicTypes.INTEGER);

So in your case this would look like q.setParameter("user_id", userid, StandardBasicTypes.INTEGER);

Hope this helps!