Question

我正在研究遗留系统，它没有正确的单元测试，也没有任何一个，所以我需要非常小心我在系统中做的任何更改。

我的基本情景是：我有一个与地址有ManyToOne关系的人，我不能改变这种关系。但是知道有些类型的地址，例如：账单地址和邮寄地址。

public class Person {

@Id
@Column(name = "PERSON_ID")
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;

@Column(name = "PERSON_NAME")
private String personName;

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "ADDRESS_ID")
private Address address;

getters and setters

地址类

public class Address {

private static final long serialVersionUID = 1L;

@Id
@Column(name = "ADDRESS_ID")
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;

@Column(name = "STREET_1_NAME")
private String street1Name;

@Column(name = "STREET_2_NAME")
private String street2Name;

getters and setters

我正在考虑将ADDRESS_TYPE列添加到ADDRESS表中，但我正在努力实现它。我只想拥有一个类型为Address的新属性postalAddress，并使postalAddress像getPostalAddress一样简单。

Answer 1

感谢Alan，您的建议有所帮助，但由于与遗留代码的兼容性，我最终使用了InheritanceType。

我的地址类现在看起来像：

@Entity
@Table(name = "ADDRESS")
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(
    name="ADDRESS_TYPE",
    discriminatorType=DiscriminatorType.INTEGER
)
@DiscriminatorValue("null")

public class Address  {

我的邮政地址如下：

@Entity
@Table(name = "ADDRESS")
@DiscriminatorValue(value="2")
public class PostalAddress extends Address {

    private static final long serialVersionUID = 1L;

    @ManyToOne
    @JoinColumn(name = "INSURED_ID")
    private Insured insured;

    public void setInsured(Insured insured) {
        this.insured = insured;
    }

    public Insured getInsured() {
        return insured;
    }

}

因此，对于过去插入的地址没有类型我假设为null，为了兼容性，新的地址也将为'null'，只有PostalAddress将为'2'。

我的Person类看起来像

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "ADDRESS_ID")
private Address address;

@Transient
private PostalAddress postalAddress;

我知道这听起来很奇怪，但我被要求有一个关系Person 1 - N PostalAddress。

Answer 2

只需添加：

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "POSTAL_ADDRESS_ID")
private Address postalAddress;

这会在管理关系的Person表中添加一个新列。无需将ADDRESS_TYPE添加到ADDRESS表，因为它将是多余的。

Answer 3

继我之前发表的评论之后，听起来就像你想要使用地图从人到地址一对多。

按照建议添加address_type列并创建枚举地址类型（POSTAL，BILLING）：

亲自：

//there are various MapKey annotations however, without testing I think these are what you need
@OneToMany
@MapKeyEnumerated(EnumType.String) //tells Hibernate we want the type to contain BILLING/POSTAL etc rather than the ENUM ordinals (0,1 etc)
@MapKeyColumn(name ="address_type")
private Map<AddressType, Address) addresses;

public Address getPostalAddress(){
return addresses.get(AddressType.POSTAL);
}

public Address getBillingAddress(){
return addresses.get(AddressType.Billing);
}

//or even
public Address getAddress(AddressType type){
return addresses.get(type);
}

我假设地址表对人有FK。

Hibernate：为两个不同的类映射相同的列

3 个答案: