我有以下两个实体，当我尝试将项目添加到我的汽车表时，它会显示以下错误消息；因此，它不允许我拥有多于一辆带“自动”传输的汽车。

错误:

 #1062 - Duplicate entry 'Auto' for key 'UK_bca5dfkfd4fjdhfh4ddirfhdhesr' 

实体:

汽车

@Entity
public class Car  implements java.io.Serializable {


    @Id
    @GeneratedValue
    long id;
    @Column(name="transmission", nullable = false)
    String transmission;
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "car")
    Set<CarFactory> factories;
    ...
}

汽车表的示例值:

10 Auto
12 Auto
43 Manual
54 Manual
65 Normal
68 Standard
90 Normal
99 NoGear

汽车工厂

@Entity
public class CarFactory implements java.io.Serializable {

   @Id
   @JoinColumn(name="transmission",referencedColumnName = "transmission")
   @ManyToOne
   Car car;

   @Id
   @JoinColumn(name="factory_id", referencedColumnName= "id")
   @ManyToOne
   Factory factory;

   ...
}

CarFactory 表的预期值

Auto Fac1
Auto Fac2
Manual Fac1
Auto Fac5
Standard Fac6
Normal Fac3
NoGear Fac1

我已关注此 question 的答案也一样，但是没有用。

长话短说，我需要一个表，其中包含来自其他表的两个外键，以及组合的主键。它不应该强制参与表中的唯一外键。

最佳答案

我模拟了您的用例，您可以在 GitHub 上找到测试。 .

这些是映射:

@Entity(name = "Car")
public static class Car implements Serializable {

    @Id
    @GeneratedValue
    long id;

    @Column(name="transmission", nullable = false)
    String transmission;
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "car")
    Set<CarFactory> factories;
}

@Entity(name = "Factory")
public static class Factory  implements Serializable {

    @Id
    @GeneratedValue
    long id;
}

@Entity(name = "CarFactory")
public static class CarFactory implements Serializable {

    @Id
    @ManyToOne
    @JoinColumn(name = "transmission", referencedColumnName = "transmission")
    Car car;

    @ManyToOne
    @Id
    Factory factory;

    public void setCar(Car car) {
        this.car = car;
    }

    public void setFactory(Factory factory) {
        this.factory = factory;
    }
}

这是添加一些测试数据的方式:

doInTransaction(session -> {
    Car car = new Car();
    car.transmission = "Auto";

    Car car1 = new Car();
    car1.transmission = "Manual";

    Factory factory = new Factory();
    session.persist(factory);
    session.persist(car);
    session.persist(car1);

    CarFactory carFactory = new CarFactory();
    carFactory.setCar(car);
    carFactory.setFactory(factory);

    CarFactory carFactory1 = new CarFactory();
    carFactory1.setCar(car1);
    carFactory1.setFactory(factory);

    session.persist(carFactory);
    session.persist(carFactory1);
});

并且测试工作正常:

@Test
public void test() {
    doInTransaction(session -> {
        List<CarFactory> carFactoryList = session.createQuery("from CarFactory").list();
        assertEquals(2, carFactoryList.size());
    });
}

更新

由于以下唯一约束，您会遇到异常:

alter table Car add constraint UK_iufgc8so6uw3pnyih5s6lawiv  unique (transmission)

这是正常行为，因为 FK 必须唯一标识 PK 行。就像您不能有更多具有相同 PK 的行一样，您的 FK 标识符引用也不能超过一行。

您的映射是问题所在。您需要引用其他内容，而不是 transmision。您需要一个唯一的汽车标识符，例如 VIN(车辆识别号)，因此您的映射变为:

@Entity(name = "Car")
public static class Car implements Serializable {

    @Id
    @GeneratedValue
    long id;

    @Column(name="vin", nullable = false)
    String vin;

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "car")
    Set<CarFactory> factories;
}

@Entity(name = "CarFactory")
public static class CarFactory implements Serializable {

    @Id
    @ManyToOne
    @JoinColumn(name = "vin", referencedColumnName = "vin")
    Car car;

    @ManyToOne
    @Id
    Factory factory;

    public void setCar(Car car) {
        this.car = car;
    }

    public void setFactory(Factory factory) {
        this.factory = factory;
    }
}

这样，vin 是唯一的，并且 Child 关联可以引用 唯一 一个父级。

关于java - 如何建立基于非关键字段的关系？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29998660/