Can not set java.lang.Integer field to java.lang.Integer

java hibernate persistence hql

25,501

Solution 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.

Solution 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.

Solution 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

25,501

Author by

gstackoverflow

Updated on September 14, 2020

Comments

gstackoverflow over 3 years

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.

P.S.

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

Pimgd almost 10 years

Do you have getters and setters for that id field?
gstackoverflow almost 10 years

Yes, I have getters and setters for all fields
Pimgd almost 10 years

What's up with the difference in "patern_id" and "pattern_id"?
gstackoverflow almost 10 years

@Pimgd - fixed, recreate database tables(drope and then create) but I see old result
Pimgd almost 10 years

What kind of database are you using? Can you retrieve an empty result set without problems using that code? Can you retrieve a User object by doing a from user where id = ? query?
gstackoverflow almost 10 years

1.mysql 2.don't understand what do you want
gstackoverflow almost 10 years

3.**Query query1 = session.createQuery("from User where id= 1"); System.out.println(query1.list());** - this code works good

gstackoverflow almost 10 years

It is enough strange because I know that HQL uses class name instead of table name, and property names instead of column name. UserPattern hasn't user.id property.
gstackoverflow almost 10 years

It is enough strange because I know that HQL uses class name instead of table name, and property names instead of column name. UserPattern hasn't user.id property.
Pimgd almost 10 years

No but the idea is this: If everything was public... what would you need to call to get to the user's id? UserPattern.user.id! So that's why it's user.id.
dieter almost 10 years

Oh. It's true. @Table annotation allows to specify the name of the table and some other details. docs.jboss.org/hibernate/orm/3.5/javadocs/org/hibernate/…