Question

我正在使用命名查询（hibernate 4）.Entity定义如下。

    @Entity
       @NamedQuery(
        name = "findAllProduct",
        query = "SELECT PC.pincode,PO.description"
        +" FROM PRODUCT_VENDOR_PAYMENT_OPTION_LOCATION PVPOL" 
        +" INNER JOIN PVPOL.paymentId PID" 
        +" INNER JOIN PVPOL.pincode PC"
        +" INNER JOIN PVPOL.paymentOptions PO"
        +" where PVPOL.id = :id"

        )


    public class PRODUCT_VENDOR_PAYMENT_OPTION_LOCATION extends baseEntity.Entity {

        @Id
        @GeneratedValue(strategy=GenerationType.AUTO)
        private int id;

        @Column(name="Payment_Id")
        @OneToMany(cascade=CascadeType.ALL,fetch=FetchType.EAGER,mappedBy="id")
        private Set<Product_Catalog_Vendor> paymentId; 

        @Column(name="pincode_id")
        @OneToMany(cascade=CascadeType.ALL,fetch=FetchType.EAGER,mappedBy="pincode_id")
        private Set<Pincodes> pincode;      

        @Column(name = "payment_options")

        @OneToMany(cascade=CascadeType.ALL,fetch=FetchType.EAGER,mappedBy="paymentOptions") 
        private Set<Payment_Options> paymentOptions;

//Protected setter getter here

}

在sql下生成Hibernate： -

    select  pincode2_.pincode as col_0_0_,    paymentopt3_.Description as col_1_0_ 
from    PRODUCT_VENDOR_PAYMENT_OPTION_LOCATION product_ve0_ 
inner join    Product_Catalog_Vendor paymentid1_ 
        on product_ve0_.id=paymentid1_.Id 
inner join    Pincodes pincode2_ 
        on product_ve0_.id=pincode2_.pincode_id 
inner join    payement_options paymentopt3_ 
        on product_ve0_.id=paymentopt3_.payment_options 
where    product_ve0_.id=?

而不是

select pincode2_.pincode as col_0_0_,  paymentopt3_.Description as col_1_0_ 
from PRODUCT_VENDOR_PAYMENT_OPTION_LOCATION product_ve0_
INNER JOIN product_catalog_vendor paymentid1_ 
        ON **product_ve0_.payment_id = paymentid1_.id** 
INNER JOIN PINCODES pincode2_ 
        ON **product_ve0_.pincode_id = pincode2_.pincode_id**
INNER JOIN payement_options paymentopt3_ 
        ON **product_ve0_.payment_options=paymentopt3_.payment_options**
where product_ve0_.id=1;

Product_catalog_vendor类：

@Entity
    public class Product_Catalog_Vendor extends baseEntity.Entity {

            @Id
            @Column(name="Id")
            private int id ;



//Setters and getters here

}

Pincodes实体：

@Entity
public class Pincodes extends baseEntity.Entity {



    @Id
    private int pincode_id;

    @Column(name="pincode")
    private int pincode;


//Setters and getters here
}

payment_options以下实体：

@Entity
@Table(name="payement_options")     
public class Payment_Options extends baseEntity.Entity {

    @Id
    @Column(name="payment_options")
    private int paymentOptions;


//Setter getter

}

我在很多网站上搜索过但无法找到幕后的原因。如果我做错了，请给我建议。一些好的参考将不胜感激。感谢

Answer 1

只是为了让你的问题正确，你的查询加入paymentid1_.Id而不是paymentid1_.id？或者我错过了预期和真实查询之间的差异？

我不是专业人士，但只是猜测我会说你的查询加入了Product_Catalog_Vendor的id：

@Id
@Column(name="Id")
private int id ;

因为这就是为什么它的 Id 而不是id ...

Answer 2

我认为由于单向映射你根本不需要mappedBy，或者无论如何，你是以一种奇怪的方式使用它们。只有当关联是双向的（不是你的情况）时才需要mappedBy，并且应该引用一个字段，该字段的类型与声明mappedBy的实体的类型相同（而不是类似的String类型）在您的情况下。）您以mappedBy属性

的方式使用了referencedColumnName

在你的例子中：

@Entity
public class Product_Catalog_Vendor extends baseEntity.Entity {
  @ManyToOne
  private PRODUCT_VENDOR_PAYMENT_OPTION_LOCATION pvpol;
}

和

public class PRODUCT_VENDOR_PAYMENT_OPTION_LOCATION extends baseEntity.Entity {
  @Column(name="Payment_Id")
  @OneToMany(cascade=CascadeType.ALL,fetch=FetchType.EAGER,mappedBy="pvpol")
  private Set<Product_Catalog_Vendor> paymentId;
}

Hibernate从Named Query生成一个连接错误列的查询？

2 个答案: