JPA/Hibernate tries to store too many parameters in entity with composite Id (and doubling properties)

Question

I have entity with composite Id

@Entity
@IdClass(value = BorrowId.class)
@Table(name = "BORROW")
public class Borrow {

    @Id
    @Column(name = "BOOK_ID", insertable = false, updatable = false)
    private long bookId;

    @Id
    @Column(name = "BORROWER_ID", insertable = false, updatable = false)
    private long borrowerId;

    @Id
    @ManyToOne(fetch = FetchType.LAZY, cascade = {CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REFRESH})
    @JoinColumn(name = "BOOK_ID")
    private Book book;

    @Id
    @ManyToOne(fetch = FetchType.EAGER, cascade = {CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REFRESH})
    @JoinColumn(name = "BORROWER_ID")
    private Borrower borrower;

    @Id
    @Column(name = "BORROW_DATE")
    private Date borrowDate;

    @Column(name = "RETURN_DATE")
    private Date returnDate;

BorrowId has properties (everything with getters/setters)

    private long bookId;
    private long borrowerId;

    private Date borrowDate;

and equals,hashCode methods

when I'm trying to persist Borrow entity (book and borrower properties set to corresponding entities, then calling entityManager.persist(borrow); ) I've got in my logs:

Hibernate: insert into BORROW (BORROWER_ID, BOOK_ID, RETURN_DATE, BORROW_DATE) values (?, ?, ?, ?) // this insert statement is correct (that's a look of my table) Invalid value "5" for parameter "parameterIndex" [90008-174]

So it looks like someone is doing something wrong :) How to solve this? is my entity declaration faulty? (I wanted to use CompositeId to learn something about using it)

The mapping is wrong, and you're making your own life difficult by having such complex composite IDs. Follow the best practices and use a single-column autogenerated ID. Everything will be much easier, in all layers of the app, and also faster and cleaner. — JB Nizet, Nov 22 '13 at 13:40
As i said, i want to use composite Id to know something about it (it's not a production app). What's wrong with mappings ? this mapping creates correct tables in DB (borrow with only 4 columns), reads data from it correctly too, but it can't persist a new Borrow — razor, Nov 22 '13 at 14:07
The properties annotated with `@Id` must match with the properties in the ID class. book and borrower should not be annotated with `@Id`, but with `@MapsId`. See http://docs.jboss.org/hibernate/core/4.2/manual/en-US/html_single/#d5e2309 — JB Nizet, Nov 22 '13 at 14:13
Thanks, `@MapsId` on Borrower and Book entites solves the problem. I didn't tried it, because it's nowhere used in examples with `@IdClass` (only with `@EmbeddedId`). If these examples even work then ? — razor, Nov 22 '13 at 14:27

score 2 · Accepted Answer · edited Jun 08 '15 at 23:12

Thanks to @JB nizet

@Id
@Column(name = "BOOK_ID", insertable = false, updatable = false)
private long bookId;

@Id
@Column(name = "BORROWER_ID", insertable = false, updatable = false)
private long borrowerId;

@MapsId("bookId")
@ManyToOne(fetch = FetchType.LAZY, cascade = {CascadeType.PERSIST,
                                              CascadeType.MERGE,
                                              CascadeType.REFRESH})
@JoinColumn(name = "BOOK_ID")
private Book book;

@MapsId("borrowerId")
@ManyToOne(fetch = FetchType.EAGER, cascade = {CascadeType.PERSIST, 
                                               CascadeType.MERGE,
                                               CascadeType.REFRESH})
@JoinColumn(name = "BORROWER_ID")
private Borrower borrower;

@Id
@Column(name = "BORROW_DATE")
private Date borrowDate;

Solves the problem, strange, because nowhere, in any example/tutorial there was no @MapsId used with @IdClass. I'm not sure if now @JoinColumn is needed

JPA/Hibernate tries to store too many parameters in entity with composite Id (and doubling properties)

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