如何映射一个组合键与JPA和Hibernate?

在这段代码中，如何为组合键生成一个Java类(如何在hibernate中组合键):

create table Time (
     levelStation int(15) not null,
     src varchar(100) not null,
     dst varchar(100) not null,
     distance int(15) not null,
     price int(15) not null,
     confPathID int(15) not null,
     constraint ConfPath_fk foreign key(confPathID) references ConfPath(confPathID),
     primary key (levelStation, confPathID)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

当前回答

假设你有以下数据库表:

首先，你需要创建包含复合标识符的@Embeddable:

@Embeddable
public class EmployeeId implements Serializable {
 
    @Column(name = "company_id")
    private Long companyId;
 
    @Column(name = "employee_number")
    private Long employeeNumber;
 
    public EmployeeId() {
    }
 
    public EmployeeId(Long companyId, Long employeeId) {
        this.companyId = companyId;
        this.employeeNumber = employeeId;
    }
 
    public Long getCompanyId() {
        return companyId;
    }
 
    public Long getEmployeeNumber() {
        return employeeNumber;
    }
 
    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof EmployeeId)) return false;
        EmployeeId that = (EmployeeId) o;
        return Objects.equals(getCompanyId(), that.getCompanyId()) &&
                Objects.equals(getEmployeeNumber(), that.getEmployeeNumber());
    }
 
    @Override
    public int hashCode() {
        return Objects.hash(getCompanyId(), getEmployeeNumber());
    }
}

有了这些，我们可以映射使用复合标识符的Employee实体，方法是用@EmbeddedId注释它:

@Entity(name = "Employee")
@Table(name = "employee")
public class Employee {
 
    @EmbeddedId
    private EmployeeId id;
 
    private String name;
 
    public EmployeeId getId() {
        return id;
    }
 
    public void setId(EmployeeId id) {
        this.id = id;
    }
 
    public String getName() {
        return name;
    }
 
    public void setName(String name) {
        this.name = name;
    }
}

Phone实体与Employee有@ManyToOne关联，需要通过两个@JoinColumnmappings从父类引用复合标识符:

@Entity(name = "Phone")
@Table(name = "phone")
public class Phone {
 
    @Id
    @Column(name = "`number`")
    private String number;
 
    @ManyToOne
    @JoinColumns({
        @JoinColumn(
            name = "company_id",
            referencedColumnName = "company_id"),
        @JoinColumn(
            name = "employee_number",
            referencedColumnName = "employee_number")
    })
    private Employee employee;
 
    public Employee getEmployee() {
        return employee;
    }
 
    public void setEmployee(Employee employee) {
        this.employee = employee;
    }
 
    public String getNumber() {
        return number;
    }
 
    public void setNumber(String number) {
        this.number = number;
    }
}

2018-01-15 19:17:50

其他回答

让我们举一个简单的例子。假设有两个表，分别叫test和customer，描述如下:

create table test(
  test_id int(11) not null auto_increment,
  primary key(test_id));

create table customer(
  customer_id int(11) not null auto_increment,
  name varchar(50) not null,
  primary key(customer_id));

还有一个表用于跟踪测试和客户:

create table tests_purchased(
  customer_id int(11) not null,
  test_id int(11) not null,
  created_date datetime not null,
  primary key(customer_id, test_id));

我们可以看到，在表tests_bought中，主键是一个复合键，因此我们将使用<composite-id…>…</composite-id hbm.xml映射文件中的>标签。因此PurchasedTest.hbm.xml看起来像这样:

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
  "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
  "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping>
  <class name="entities.PurchasedTest" table="tests_purchased">

    <composite-id name="purchasedTestId">
      <key-property name="testId" column="TEST_ID" />
      <key-property name="customerId" column="CUSTOMER_ID" />
    </composite-id>

    <property name="purchaseDate" type="timestamp">
      <column name="created_date" />
    </property>

  </class>
</hibernate-mapping>

但这并没有结束。在Hibernate中我们使用session。load (entityClass, id_type_object)使用主键查找并加载实体。在组合键的情况下，ID对象应该是一个单独的ID类(在上面的情况下是purchasedtestd类)，它只声明主键属性如下:

import java.io.Serializable;

public class PurchasedTestId implements Serializable {
  private Long testId;
  private Long customerId;

  // an easy initializing constructor
  public PurchasedTestId(Long testId, Long customerId) {
    this.testId = testId;
    this.customerId = customerId;
  }

  public Long getTestId() {
    return testId;
  }

  public void setTestId(Long testId) {
    this.testId = testId;
  }

  public Long getCustomerId() {
    return customerId;
  }

  public void setCustomerId(Long customerId) {
    this.customerId = customerId;
  }

  @Override
  public boolean equals(Object arg0) {
    if(arg0 == null) return false;
    if(!(arg0 instanceof PurchasedTestId)) return false;
    PurchasedTestId arg1 = (PurchasedTestId) arg0;
    return (this.testId.longValue() == arg1.getTestId().longValue()) &&
           (this.customerId.longValue() == arg1.getCustomerId().longValue());
  }

  @Override
  public int hashCode() {
    int hsCode;
    hsCode = testId.hashCode();
    hsCode = 19 * hsCode+ customerId.hashCode();
    return hsCode;
  }
}

重要的一点是，我们还实现了hashCode()和equals()两个函数，因为Hibernate依赖于它们。

2017-03-09 07:07:41

你需要使用@EmbeddedId:

@Entity
class Time {
    @EmbeddedId
    TimeId id;

    String src;
    String dst;
    Integer distance;
    Integer price;
}

@Embeddable
class TimeId implements Serializable {
    Integer levelStation;
    Integer confPathID;
}

2010-08-27 14:24:21

使用hbm.xml

    <composite-id>

        <!--<key-many-to-one name="productId" class="databaselayer.users.UserDB" column="user_name"/>-->
        <key-property name="productId" column="PRODUCT_Product_ID" type="int"/>
        <key-property name="categoryId" column="categories_id" type="int" />
    </composite-id>

使用注释

复合键类

public  class PK implements Serializable{
    private int PRODUCT_Product_ID ;    
    private int categories_id ;

    public PK(int productId, int categoryId) {
        this.PRODUCT_Product_ID = productId;
        this.categories_id = categoryId;
    }

    public int getPRODUCT_Product_ID() {
        return PRODUCT_Product_ID;
    }

    public void setPRODUCT_Product_ID(int PRODUCT_Product_ID) {
        this.PRODUCT_Product_ID = PRODUCT_Product_ID;
    }

    public int getCategories_id() {
        return categories_id;
    }

    public void setCategories_id(int categories_id) {
        this.categories_id = categories_id;
    }

    private PK() { }

    @Override
    public boolean equals(Object o) {
        if ( this == o ) {
            return true;
        }

        if ( o == null || getClass() != o.getClass() ) {
            return false;
        }

        PK pk = (PK) o;
        return Objects.equals(PRODUCT_Product_ID, pk.PRODUCT_Product_ID ) &&
                Objects.equals(categories_id, pk.categories_id );
    }

    @Override
    public int hashCode() {
        return Objects.hash(PRODUCT_Product_ID, categories_id );
    }
}

实体类

@Entity(name = "product_category")
@IdClass( PK.class )
public  class ProductCategory implements Serializable {
    @Id    
    private int PRODUCT_Product_ID ;   

    @Id 
    private int categories_id ;

    public ProductCategory(int productId, int categoryId) {
        this.PRODUCT_Product_ID = productId ;
        this.categories_id = categoryId;
    }

    public ProductCategory() { }

    public int getPRODUCT_Product_ID() {
        return PRODUCT_Product_ID;
    }

    public void setPRODUCT_Product_ID(int PRODUCT_Product_ID) {
        this.PRODUCT_Product_ID = PRODUCT_Product_ID;
    }

    public int getCategories_id() {
        return categories_id;
    }

    public void setCategories_id(int categories_id) {
        this.categories_id = categories_id;
    }

    public void setId(PK id) {
        this.PRODUCT_Product_ID = id.getPRODUCT_Product_ID();
        this.categories_id = id.getCategories_id();
    }

    public PK getId() {
        return new PK(
            PRODUCT_Product_ID,
            categories_id
        );
    }    
}

2017-12-14 01:56:06

主键类必须定义equals和hashCode方法

When implementing equals you should use instanceof to allow comparing with subclasses. If Hibernate lazy loads a one to one or many to one relation, you will have a proxy for the class instead of the plain class. A proxy is a subclass. Comparing the class names would fail. More technically: You should follow the Liskows Substitution Principle and ignore symmetricity. The next pitfall is using something like name.equals(that.name) instead of name.equals(that.getName()). The first will fail, if that is a proxy.

http://www.laliluna.de/jpa-hibernate-guide/ch06s06.html

2013-01-18 16:28:46