Question

我有3张桌子：人物 - ＆gt; Person_Address_Type {person_address_type_id（PK），person_id，address_id，type} - ＆gt;地址。 Person_Address_Type用于指示Person和地址之间关系的类型。例如：

Mr. A has unique HOME addresses.
Mr. A has one or many OFFICE addresses.
Mr. A has one or many RELAXING addresses.
... And other types of address may be added later.

如何使Hibernate的映射具有以下java实体：

Person {
     getHomeAddress(): Address;
     getOfficeAddresses(): Address*;
     getRelaxingAddresses(): Address*;
}

我知道有短暂的领域。但是，我们不能用瞬态做HQL查询。我需要一个映射。我该怎么做？

由于

Answer 1

您的映射应该是这样的

表格结构

CREATE TABLE `person` (
    `id_person` BIGINT(20) NOT NULL AUTO_INCREMENT,
    `firstname` VARCHAR(50) NULL DEFAULT NULL,
    `lastname` VARCHAR(50) NULL DEFAULT NULL,
    `birth_date` DATE NULL DEFAULT NULL,
    `cell_phone` VARCHAR(15) NULL DEFAULT NULL,
    PRIMARY KEY (`id_person`)
)ENGINE=InnoDB;

CREATE TABLE `address_type` (
    `id_address_type` BIGINT(10) NOT NULL AUTO_INCREMENT,
    `name` VARCHAR(50) NULL DEFAULT NULL,   
    PRIMARY KEY (`id_address_type`)
)ENGINE=InnoDB;

CREATE TABLE `address` (
    `id_address` BIGINT(10) NOT NULL AUTO_INCREMENT,
    `fk_address_type`BIGINT(20) NULL DEFAULT NULL,
    PRIMARY KEY (`id_address`),
    CONSTRAINT `FK_ADDRESS_TYPE` FOREIGN KEY (`fk_address_type`) REFERENCES `address_type` (`id_address_type`),
)ENGINE=InnoDB;

CREATE TABLE `person_address` (
    `id_person_address` BIGINT(10) NOT NULL AUTO_INCREMENT, 
    `fk_person` BIGINT(20) NULL DEFAULT NULL,
    `fk_address`BIGINT(20) NULL DEFAULT NULL,       
    PRIMARY KEY (`id_person_address`),
    CONSTRAINT `FK_PERSON` FOREIGN KEY (`fk_person`) REFERENCES `person` (`id_person`),
    CONSTRAINT `FK_ADDRESS` FOREIGN KEY (`fk_address`) REFERENCES `address` (`id_address`)  
)ENGINE=InnoDB;

实体类

<强> Person.java

@Entity
@Table(name = "person",uniqueConstraints = {
@UniqueConstraint(columnNames = "id_person")})
public class Person implements Serializable {
    private static final long serialVersionUID = -1798070786993154676L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id_person", unique = true, nullable = false)
    private Integer id;

    @OneToMany(cascade=CascadeType.ALL)
    @JoinTable(name="person_address", joinColumns={@JoinColumn(name="fk_person", referencedColumnName="id_person")}
    , inverseJoinColumns={@JoinColumn(name="fk_address", referencedColumnName="id_address")})
    private Set<Address> address;

   // Getter and Setter methods 
}

<强> Address.java

@Entity
@Table(name = "address", uniqueConstraints = {
@UniqueConstraint(columnNames = "id_address")})
public class Address implements Serializable {

    private static final long serialVersionUID = -6790693372846798580L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id_address", unique = true, nullable = false)
    private Integer addressId;

    @OneToOne(mappedBy="address", cascade=CascadeType.ALL)
    private AddressType addressType;    

   // Getter and Setter methods
}

的 AddressType.java 的

@Entity @Table(name="address_type",uniqueConstraints = { @UniqueConstraint(columnNames = "id_address_type")}) public class AddressType { @Id @Column(name="id_address_type", unique=true, nullable=false) @GeneratedValue(strategy = GenerationType.IDENTITY) private Long addressTypeId; @OneToOne @PrimaryKeyJoinColumn private Address address; // Getter and Setter methods }

现在当你拨打getAddress()时，这将返回与Person相关的所有地址，然后根据地址类型对该地址进行分组，即通过调用address.getAddressType().getName()方法。

您还可以在Address和AddressType实体之间进行@ManyToMany映射

Answer 2

从我对Hibernate的简单了解，似乎Hibernate不直接支持关系表的样式。

最简单的方法是创建Person_Address_Type的单独视图，这样您就可以拥有3个关系表：PERSON_HOME_ADDRESS，PERSON_OFFICE_ADDRESS，PERSON_RELAX_ADDRESS等。

每个视图都包含PERSON_ID和ADDRESS_ID

使用这些视图，您可以使用ManyToMany/ManyToOne执行正常的JoinTable。

但是，如果要直接使用这些实体更新关系，则此方法可能无效（取决于DBMS，某些DBMS允许通过视图更新/插入）。

另一种方法是创建一个新实体，如PersonAddressRelationship，其中包含对Person，Address和RelationshipType枚举的引用。

然后，Person to Address中的三个关系不会直接映射。相反，您可以根据映射的homeAddress在relaxAddresses方法中填充officeAddresses，@PostLoad和addressRelationships的值。（并在[{1}}方法中更新addressRelationships基础homeAddress，relaxAddresses和officeAddresses

Answer 3

我刚刚发现@WhereJoinTable是为了这个（理论上，我没有在实践中尝试过）

与连接表中的类型的单向关联

3 个答案: