2010-12-06 60 views
3

我正在開發一個拍賣網站。問題就出在3個實體使用:Hibernate查詢導致大量額外的不必要的查詢

  • 產品(具有零個或多個ProductBid)
  • ProductBid(有零或一個ProductBidRejection)
  • ProductBidRejection

我使用Hibernate查詢獲得投標:

select pb from ProductBid pb left join pb.rejection pbr where pbr is null and pb.product = :product order by pb.amount desc

This generates this query (via console):

select 
    productbid0_.id as id4_, 
    productbid0_.amount as amount4_, 
    productbid0_.bid_by as bid4_4_, 
    productbid0_.date as date4_, 
    productbid0_.product_id as product5_4_ 
from 
    product_bids productbid0_ 
left outer join 
    product_bid_rejections productbid1_ 
     on productbid0_.id=productbid1_.product_bid_id 
where 
(
    productbid1_.id is null 
) 
and productbid0_.product_id=?

But for each bid it gets it also generates:

select 
    productbid0_.id as id3_1_, 
    productbid0_.date_rejected as date2_3_1_, 
    productbid0_.product_bid_id as product4_3_1_, 
    productbid0_.reason as reason3_1_, 
    productbid0_.rejected_by as rejected5_3_1_, 
    productbid1_.id as id4_0_, 
    productbid1_.amount as amount4_0_, 
    productbid1_.bid_by as bid4_4_0_, 
    productbid1_.date as date4_0_, 
    productbid1_.product_id as product5_4_0_ 
from 
    product_bid_rejections productbid0_ 
inner join 
    product_bids productbid1_ 
     on productbid0_.product_bid_id=productbid1_.id 
where 
    productbid0_.product_bid_id=?

These are my entities:

ProductBid


@Entity 
@Table(name = "product_bids") 
public class ProductBid 
{ 
    @Column(name = "id", nullable = false) 
    @Id 
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private int id; 

    @JoinColumn(name = "product_id", nullable = false) 
    @Index(name="product") 
    @ManyToOne(fetch = FetchType.LAZY) 
    private Product product; 

    @Column(name = "amount", nullable = false) 
    private BigDecimal amount; 

    @JoinColumn(name = "bid_by", nullable = false) 
    @Index(name="bidBy") 
    @ManyToOne(fetch = FetchType.LAZY) 
    @Fetch(FetchMode.JOIN) 
    private User bidBy; 

    @Column(name = "date", nullable = false) 
    @Type(type = "org.joda.time.contrib.hibernate.PersistentDateTime") 
    private DateTime date; 

    @OneToOne(fetch = FetchType.LAZY, mappedBy = "productBid") 
    private ProductBidRejection rejection; 
} 

ProductBidRejection

 
@Entity 
@Table(name = "product_bid_rejections") 
public class ProductBidRejection 
{ 
    @Id 
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    @Column(name = "id", nullable = false) 
    private long id;

@Column(name = "reason", nullable = false, columnDefinition = "TEXT") 
private String reason; 

@Column(name = "date_rejected", nullable = false) 
@Type(type = "org.joda.time.contrib.hibernate.PersistentDateTime") 
private DateTime dateRejected; 

@ManyToOne(fetch = FetchType.LAZY) 
@JoinColumn(name = "rejected_by", nullable = false) 
private User rejectedBy; 

@OneToOne(fetch = FetchType.LAZY) 
@JoinColumn(name = "product_bid_id", nullable = false) 
@Fetch(FetchMode.JOIN) 
private ProductBid productBid; 

}

回答

2

因爲您在ProductBid上有@Fetch(FetchMode.JOIN)。 因此,對於您檢索的每個ProductBidRejections,它還會加載一個ProductBid。

UPDATE

嘗試此查詢。它會得到不同的PB和預先抓取PBR

select distinct pb from ProductBid pb left join fetch pb.rejection pbr where pbr is null and pb.product = :product order by pb.amount desc

+0

刪除ProductBid上的@Fetch(FetchMode.JOIN)將從查詢中刪除'productbid1'。但它仍然使用每個出價的選擇。 – 2010-12-06 13:48:43

0

的使用標準,而不是HQL你的問題將得到解決

session.createCriteria(ProductBid.class).add(Restrictions.eq("product",yourproduct)).list(); 

和ProductBid實體類使用註釋LY加入渴望ProductBidRejection