Question

我有两个实体foo和bar有多对多关系。我为它们创建了manyToMany实体FooBar，将其主键用作带有@IdClass的复合主键ID：

Foo类：

@Entity
@Table(name = "foos")
public class Foo implements Serializable {
    private static final long serialVersionUID = -1L;

    @Id
    @Column(name="name")
    private String name;

    @OneToMany(fetch = FetchType.EAGER, mappedBy = "bar", orphanRemoval = true)
    @NotFound(action = NotFoundAction.IGNORE)
    @JsonIgnoreProperties
    @JsonIgnore
    private List<Foo_bar> fooBars;

    // constructors, getters, setters, ..
}

班级酒吧：

@Entity
@Table(name = "bars")
public class Bar implements Serializable {
    private static final long serialVersionUID = -1L;

    @Id
    @Column(name="name")
    private String name;

    @OneToMany(fetch = FetchType.EAGER, mappedBy = "foo", orphanRemoval = true)
    @NotFound(action = NotFoundAction.IGNORE)
    @JsonIgnoreProperties
    @JsonIgnore
    private List<Foo_bar> fooBars;

    // constructors, getters, setters, ..
}

FooBar类：

@Entity
@IdClass(FooBarID.class)
@Table(name = "foo_bars")
public class Foo_bar implements Serializable {

    @Id
    @ManyToOne(cascade=CascadeType.MERGE, fetch=FetchType.EAGER)
    private Foo foo;

    @Id
    @ManyToOne(cascade=CascadeType.MERGE, fetch=FetchType.EAGER)
    private Bar bar;

    // constructors, getters, setters, ..
}

FooBarId类：

public class FooBarID implements Serializable {

    Foo foo;
    Bar bar;

    public FooBarID() {
    }

    public FooBarID(Foo foo, Bar bar) {
        this.foo = foo;
        this.bar = bar;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((foo == null) ? 0 :     foo.hashCode());
        result = prime * result + ((bar == null) ? 0 :     bar.hashCode());
        return result;
    }

    @Override
    public boolean equals(Object obj) {
        return
                (this.foo == ((FooBarID) obj).foo)
                && (this.bar == ((FooBarID) obj).bar);

    }
}

如果我现在尝试使用jpa和以下代码创建并保存新的FooBar：

Foo foo = this.fooRepository.getByName('foo123').get();
Bar bar = this.barRepository.getByName('bar123').get();

FooBar fooBar = new FooBar();
fooBar.setFoo(foo);
fooBar.setBar(bar);
this.fooBarRepository.save(fooBar); <-- exception

抛出InvalidDataAccessApiUsageException：Can not set package.Foo field package.FooBarID.foo to java.lang.String; nested exception is java.lang.IllegalArgumentException: Can not set package.Foo field package.FooBarID.foo to java.lang.String

有人知道如何解决此问题吗？：/

Spring JPA ManyToMany复合@IdClass

0 个答案: