在我的Spring Data项目中,我有一个以下实体:
@Entity
@NamedEntityGraph(name = "graph.CardPair", attributeNodes = {})
@Table(name = "card_pairs")
public class CardPair extends BaseEntity implements Serializable {
private static final long serialVersionUID = 7571004010972840728L;
@Id
@SequenceGenerator(name = "card_pairs_id_seq", sequenceName = "card_pairs_id_seq", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.AUTO, generator = "card_pairs_id_seq")
private Long id;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "card1_id")
private Card card1;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "card2_id")
private Card card2;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "selected_card_id")
private Card selectedCard;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "board_id")
private Board board;
....
}
我需要实现一个Spring Data存储库方法,该方法将根据CardPair
,cardPairId
和board
查找card
。对我来说更复杂的情况是Card
,因为我需要构建一个card1 = card or card2 = card
是否可以通过Spring Data存储库方法名称提供此条件?
例如
cardPairRepository.findByIdAndBoardAnd[card1 = card or card2 = card]AndSelectedCardIsNull(cardPairId, board, card);
是否可以在方法名称中翻译此条件card1 = card or card2 = card
?
答案 0 :(得分:7)
使用方法名称是不可能实现这一点的,因为您需要在括号中使用最后一个条件:id = ? AND board = ? AND (card1 = ? OR card2 = ?)
。无法从方法名称推断出这一点,因为它仅用于涵盖基本情况。
您的案例有三种方式:
答案 1 :(得分:1)
您可以在此处找到所有操作:http://docs.spring.io/spring-data/jpa/docs/1.3.0.RELEASE/reference/html/jpa.repositories.html第2.2.2节
findByIdAndBoardAndCard1OrCard2(cardPairId, board, card, card);
我没有考虑过这个的顺序,但它应该是这样的