JPA OneToMany映射异常：字段XXX无法声明它被另一个字段映射

Question

在我的项目中，我有下面报告的类，具有双向关系映射。当我尝试通过 JPA 读取 DataProviderImpl 类型的对象时，出现以下异常：

org.apache.openjpa.persistence.ArgumentException: Field "persistence.entity.DataProviderImpl.methods" cannot declare that it is mapped by another field. Its mapping strategy( org.apache.openjpa.jdbc.meta.strats.HandlerCollectionTableFieldStrategy) does not support mapping by another field.

我已经检查了其他问题，但似乎没有任何问题适用于我的情况。有什么提示吗？提前致谢！

<persistence
xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence
http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
version="2.0">

<persistence-unit name="DataProvider" transaction-type="JTA"> <!-- transaction-type="RESOURCE_LOCAL"  info at:  http://tomee.apache.org/jpa-concepts.html -->
    <jta-data-source>
        osgi:service/jdbc/mysqlPoolXAds
    </jta-data-source>
    <non-jta-data-source>
        osgi:service/jdbc/mysqlds
    </non-jta-data-source>      
    <class>persistence.entity.DataProviderImpl</class>
    <exclude-unlisted-classes>true</exclude-unlisted-classes>
    <properties>
    </properties>
</persistence-unit>
<persistence-unit name="RestMethod" transaction-type="JTA"> <!-- transaction-type="RESOURCE_LOCAL"  info at:    http://tomee.apache.org/jpa-concepts.html -->
    <jta-data-source>
        osgi:service/jdbc/mysqlPoolXAds
    </jta-data-source>
    <non-jta-data-source>
        osgi:service/jdbc/mysqlds
    </non-jta-data-source>      
    <class>persistence.entity.RestMethodImpl</class>
    <exclude-unlisted-classes>true</exclude-unlisted-classes>
    <properties>
    </properties>
</persistence-unit>

@Entity(name = "DataProvider")
public class DataProviderImpl implements DataProvider {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;
    private String name;
    // ...

    @OneToMany(fetch = FetchType.LAZY, mappedBy="dataProvider")
    private List<RestMethod> methods;


    public DataProviderImpl() {
        super();
        this.id = 0;
    }

    @Override
    public int getId() {
        return id;
    }
    @Override
    public void setId(int id) {
        this.id = id;
    }
    @Override
    public String getName() {
        return name;
    }
    @Override
    public void setName(String name) {
        this.name = name;
    }
    @Override
    public List<RestMethod> getMethods() {
        return methods;
    }
    @Override
    public void setMethods(List<RestMethod> methods) {
        this.methods = methods;
    }
}

    public interface DataProvider {
    public int getId();
    public void setId(int id);
    public String getName();
    public void setName(String name);
    public List<RestMethod> getMethods();
    public void setMethods(List<RestMethod> methods);
}

    @Entity(name = "RestMethod")
public class RestMethodImpl implements RestMethod {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;
    private String nickname;
    // ...

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="dataProviderId")
    private DataProvider dataProvider;

    public RestMethodImpl() {
        super();
        this.id = 0;
    }

    @Override
    public int getId() {
        return id;
    }
    @Override
    public void setId(int id) {
        this.id = id;
    }
    @Override
    public String getNickname() {
        return nickname;
    }
    @Override
    public void setNickname(String nickname) {
        this.nickname = nickname;
    }
    @Override
    public DataProvider getDataProvider() {
        return dataProvider;
    }
    @Override
    public void setDataProvider(DataProvider dataProvider) {
        this.dataProvider = dataProvider;
    }
}

    public interface RestMethod  {
    public int getId();
    public void setId(int id);
    public String getNickname();
    public void setNickname(String nickname);
    public DataProvider getDataProvider();
    public void setDataProvider(DataProvider dataProvider);
}

Answer 1

假设您使用了来自 javax.persistence 的导入语句，您的映射注释绝对应该使用额外的参数 targetEntity=...，否则对象关系映射器无法准确确定哪个 class 是为给定的接口类型实例化，例如 RestMethod。所以你的映射将是：

@Entity(name = "DataProvider")
public class DataProviderImpl implements DataProvider {
    // ...
    @OneToMany(targetEntity=RestMethodImpl.class, fetch = FetchType.LAZY, mappedBy="dataProvider")
    private List<RestMethod> methods;

因此：

@Entity(name = "RestMethod")
public class RestMethodImpl implements RestMethod {

    // ...  
    @ManyToOne(targetEntity=DataProviderImpl.class, fetch = FetchType.LAZY)
    @JoinColumn(name="dataProviderId")
    private DataProvider dataProvider;

我不知道这对你的情况是否有帮助，但你当然应该尝试使用 JPA 映射器的这些额外信息来调整你的代码。

我刚发现您给定的 persistence.xml 的另一个问题。您应该在同一个 persistence-unit 块中同时拥有 classes（DataProviderImpl 和 RestMethodImpl），因为根据您的配置构建的实体上下文必须了解这两个 class es 施加正确的映射策略。

JPA OneToMany映射异常：字段XXX无法声明它被另一个字段映射

JPA OneToMany mapping exception: Field XXX cannot declare that it is mapped by another field

java

jpa