ClassCastException：整数无法转换为 Long，同时尝试迭代实体 ID答案

【问题标题】：ClassCastException: Integer cannot be cast to Long, while trying to iterate over entity IDsClassCastException：整数无法转换为 Long，同时尝试迭代实体 ID
【发布时间】：2023-03-28 04:06:01
【问题描述】：

我的服务中有以下方法：

public Set<BoardCard> moveHoldedCardsToNewBoard(Board newBoard, Board prevBoard) {
        Set<BoardCard> boardCards = new HashSet<>();
        if (prevBoard != null) {
            List<Long> holdedCardIds = getExcludedCardIds(prevBoard);
            for (Long item: holdedCardIds) {

            }
    }

当我想循环holdedCardIds 列表时，我收到：java.lang.ClassCastException: java.lang.Integer cannot be cast to java.lang.Long in this place -> @ 987654324@

我的 getExcludedCardIds() 看起来像：

@Override
public List<Long> getExcludedCardIds(Board board) {

        return boardCardRepository.getExcludedCardIds(board.getId());
    }

存储库：

@Repository
public interface BoardCardRepository extends JpaRepository<BoardCard, Long>, QueryDslPredicateExecutor<BoardCard> {

    @Query(value = "SELECT bc.card_id FROM boards_cards bc WHERE bc.board_id =:boardId AND bc.on_hold=true", nativeQuery = true)
    List<Long> getExcludedCardIds(@Param("boardId") Long boardId);
}

实体：

@Entity
@NamedEntityGraph(name = "graph.BoardCard", attributeNodes = {})
@Table(name = "boards_cards")
public class BoardCard implements Serializable {

    private static final long serialVersionUID = -9019060375256960701L;

    @EmbeddedId
    private BoardCardId id = new BoardCardId();

}
@Embeddable
public class BoardCardId implements Serializable {

    private static final long serialVersionUID = -3630484760647866357L;

    @ManyToOne
    private Board board;

    @ManyToOne
    private Card card;
}

    @Entity
    @Table(name = "boards")
    public class Board extends BaseEntity implements Serializable {
        @Id
        @SequenceGenerator(name = "boards_id_seq", sequenceName = "boards_id_seq", allocationSize = 1)
        @GeneratedValue(strategy = GenerationType.AUTO, generator = "boards_id_seq")
        private Long id;
}




 @Entity    
   @Table(name = "cards")     
   public class Card extends BaseEntity implements Serializable {

                @Id
                @SequenceGenerator(name = "cards_id_seq", sequenceName = "cards_id_seq", allocationSize = 1)
                @GeneratedValue(strategy = GenerationType.AUTO, generator = "cards_id_seq")
                private Long id;
}

在我的 POSTGRES schema.sql 中，BoardCard 实体定义如下：

CREATE TABLE IF NOT EXISTS boards_cards(
                board_id INTEGER,
                card_id INTEGER,
                on_hold BOOLEAN DEFAULT FALSE,
                CONSTRAINT pk_user_card PRIMARY KEY (board_id, card_id),
                FOREIGN KEY(board_id) REFERENCES boards(id),
                FOREIGN KEY(card_id) REFERENCES cards(id)
            );

我发现 here ，在 postgresql 中相当于 LONG 类型的是 bigint 。但是，如果我尝试使用它，它将如何影响我的应用程序的性能方面？

那么，告诉我如何解决这个问题？

【问题讨论】：

标签： java postgresql spring-data spring-data-jpa

【解决方案1】：

我找到了解决方案here。解决方案是使用 JPQL 查询 而不是 SQL 查询。

重构的存储库：

@Repository
public interface BoardCardRepository extends JpaRepository<BoardCard, Long>, QueryDslPredicateExecutor<BoardCard> {

    @Query(value = "SELECT id.card.id FROM BoardCard WHERE id.board.id = :boardId AND onHold = true")
    List<Long> getExcludedCardIds(@Param("boardId") Long boardId);
}

【讨论】：

【解决方案2】：

(long) can be casted into (int), and (int) can be casted to (long)

然而，

(Long) **cannot** be casted into (Integer)

反之亦然，因为它们不是原语。 bigint 也是如此

这是您的根本问题，尽管我不确定它在您的程序中的哪个位置导致了演员表。

【讨论】：

【解决方案3】：

如果您使用java-8，只需将Integer 对象列表转换为包含相同值的对象列表即可解决此问题，在long 中，然后将由java 自动装箱。

getExcludedCardIds(prevBoard).stream
                             .mapToLong(x -> x).collect(Collectors.toList);

【讨论】：