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主页 / 问题 / 984546
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YuriiS
YuriiS
Asked:2020-05-23 14:14:51 +0000 UTC2020-05-23 14:14:51 +0000 UTC 2020-05-23 14:14:51 +0000 UTC

为什么交叉连接在使用 count() 函数的查询中起作用?

  • 772

我正在spring mvc使用Oracle. hibernate当我对数据库进行查询并选择所有数据时,得到一个对象列表,然后我获取这个列表的大小,然后一切正常。但是从数据库中提取所有数据是不好的。我需要发出一个请求,该请求将返回满足传递给请求的参数的行数。为此,我正在尝试使用Criteria. 我创建了一个构建器、一个谓词并执行了一个查询,但结果不是我所期望的。查询中的 Hibernate 进行交叉连接,然后将表中的所有行乘以包含所需参数的行。表中只有7条记录,其中只有4条满足参数,查询结果是28条,而不是预期的4条。下面是代码片段。实体类:

@Entity(name = "Relationship")
@Table(name = "RELATIONSHIP")
public class Relationship extends IdEntity{
    private Long id;
    private User userFrom;
    private User userTo;
    private Date acceptedFriends;
    private RelationshipStatusType statusType;
    private Set<User> users = new HashSet<>();

    public Relationship() {
    }

    @Id
    @SequenceGenerator(name = "R_SHIP_SQ", sequenceName = "RELATIONSHIP_SEQ", allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "R_SHIP_SQ")
    @Column(name = "RELATIONSHIP_ID")
    @Override
    public Long getId() {
        return id;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "ID_USER_FROM")
    public User getUserFrom() {
        return userFrom;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "ID_USER_TO")
    public User getUserTo() {
        return userTo;
    }

    @Enumerated(EnumType.STRING)
    @Column(name = "STATUS_TYPE")
    public RelationshipStatusType getStatusType() {
        return statusType;
    }

    @JsonIgnore
    @ManyToMany(mappedBy = "statuses")
    Set<User> getUsers() {
        return users;
    }

枚举类:

public enum RelationshipStatusType {
    REQUESTED("REQUESTED"),
    CANCELED("CANCELED"),
    DECLINED("DECLINED"),
    DELETED("DELETED"),
    ACCEPTED("ACCEPTED");

    private String value;

    RelationshipStatusType(final String value){
        this.value = value;
    }

    public String getValue(){
        return value;
    }

    @Override
    public String toString(){
        return this.getValue();
    }
}

DAO:

@Repository("relationshipDAO")
@Transactional
public class RelationshipDAO extends GeneralDAO<Relationship> {

    private static final String GET_RELATIONSHIP = "SELECT * FROM RELATIONSHIP WHERE ID_USER_FROM = ? AND ID_USER_TO = ?";

    @SuppressWarnings("unchecked")
    public Relationship getRelationship(Long idUserFrom, Long idUserTo)throws InternalServerError {
        Relationship relationship;
        NativeQuery<Relationship> query = (NativeQuery<Relationship>) getEntityManager().createNativeQuery(GET_RELATIONSHIP, Relationship.class);
        try {
            relationship = query.setParameter(1, idUserFrom).setParameter(2, idUserTo).uniqueResult();
        }catch (NoResultException e){
            System.err.println(e.getMessage());
            throw e;
        }
        return relationship;
    }

    @SuppressWarnings("unchecked")
    public long getQuantityFriends(Long idUser, String status) throws InternalServerError{

        CriteriaBuilder criteriaBuilder = getEntityManager().getCriteriaBuilder();
        CriteriaQuery<Long> criteriaQuery = criteriaBuilder.createQuery(Long.class);
        Root<Relationship> relationshipRoot = criteriaQuery.from(Relationship.class);
        Predicate idUserPredicate = criteriaBuilder.equal(relationshipRoot.get("userTo"), idUser);
        Predicate statusPredicate = criteriaBuilder.equal(relationshipRoot.get("statusType"), RelationshipStatusType.valueOf(status));
        criteriaQuery.select(criteriaBuilder.count(criteriaQuery.from(Relationship.class)))
                .where(criteriaBuilder.and(idUserPredicate, statusPredicate));
        TypedQuery<Long> query = getEntityManager().createQuery(criteriaQuery);
        System.out.println("Quantity rows = " + query.getSingleResult());
        return query.getSingleResult();
    }
}

作为请求的结果,我在控制台中看到了这个:

23-May-2019 01:01:22.049 INFO [http-nio-8080-exec-9] org.hibernate.hql.internal.QueryTranslatorFactoryInitiator.initiateService HHH000397: Using ASTQueryTranslatorFactory
Hibernate: select count(relationsh1_.RELATIONSHIP_ID) as col_0_0_ from RELATIONSHIP relationsh0_ cross join RELATIONSHIP relationsh1_ where relationsh0_.ID_USER_TO=90 and relationsh0_.STATUS_TYPE=?
Quantity rows = 28

为什么会弹出以cross join​​及如何解决?

java
  • 1 1 个回答
  • 10 Views

1 个回答

  • Voted
  1. Best Answer
    default locale
    2020-05-23T14:37:02Z2020-05-23T14:37:02Z

    试试这样:

    criteriaQuery.select(criteriaBuilder.count(relationshipRoot))
                 .where(criteriaBuilder.and(idUserPredicate, statusPredicate));
    

    该调用criteriaQuery.from(Relationship.class)创建了一个新的根,它被添加到 c 请求cross join中。这是写在文档中的AbstractQuery.from:

    创建并添加与给定实体对应的查询根,形成具有任何现有根的笛卡尔积。

    • 1

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