JPA：在 fied 和 getter 上设置 @Id 的区别

Question

我使用 EclipseLink，但得到了非常奇怪的结果。请考虑以下代码：

此代码有效：

@Entity
@Table(name = "someTable")
public class SomeClass{
    @Id// PAY ATTENTION TO THIS LINE
    private String id;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    @Column (name = "somecol")// PAY ATTENTION TO THIS LINE
    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

此代码也有效：

@Entity
@Table(name = "someTable")
public class SomeClass{
    @Id// PAY ATTENTION TO THIS LINE
    private String id;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    //@Column (name = "somecol")// PAY ATTENTION TO THIS LINE
    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

此代码也有效：

@Entity
@Table(name = "someTable")
public class SomeClass{

    private String id;

    @Id// PAY ATTENTION TO THIS LINE
    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    //@Column (name = "somecol")// PAY ATTENTION TO THIS LINE
    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

此代码无效：

@Entity
@Table(name = "someTable")
public class SomeClass{

    private String id;

    @Id // PAY ATTENTION TO THIS LINE
    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    @Column (name = "somecol")// PAY ATTENTION TO THIS LINE
    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

我得到以下异常：

Exception Description: Entity class [class SomeClass] has no primary key specified. It should define either an @Id, @EmbeddedId or an @IdClass. If you have defined PK using any of these annotations then make sure that you do not have mixed access-type (both fields and properties annotated) in your entity class hierarchy.
    at org.eclipse.persistence.exceptions.ValidationException.noPrimaryKeyAnnotationsFound(ValidationException.java:1425)
    at org.eclipse.persistence.internal.jpa.metadata.accessors.classes.EntityAccessor.validatePrimaryKey(EntityAccessor.java:1542)
    at org.eclipse.persistence.internal.jpa.metadata.accessors.classes.EntityAccessor.processMappingAccessors(EntityAccessor.java:1249)
    at org.eclipse.persistence.internal.jpa.metadata.accessors.classes.EntityAccessor.process(EntityAccessor.java:699)
    at org.eclipse.persistence.internal.jpa.metadata.MetadataProject.processStage2(MetadataProject.java:1808)
    at org.eclipse.persistence.internal.jpa.metadata.MetadataProcessor.processORMMetadata(MetadataProcessor.java:573)
    at org.eclipse.persistence.internal.jpa.deployment.PersistenceUnitProcessor.processORMetadata(PersistenceUnitProcessor.java:607)
    at org.eclipse.persistence.internal.jpa.EntityManagerSetupImpl.predeploy(EntityManagerSetupImpl.java:1948)
    at org.eclipse.persistence.internal.jpa.deployment.JPAInitializer.callPredeploy(JPAInitializer.java:100)
    at org.eclipse.persistence.jpa.PersistenceProvider.createEntityManagerFactoryImpl(PersistenceProvider.java:104)
    at org.eclipse.persistence.jpa.PersistenceProvider.createEntityManagerFactory(PersistenceProvider.java:188)
    at org.eclipse.gemini.jpa.ProviderWrapper.createEntityManagerFactory(ProviderWrapper.java:128)
    at org.eclipse.gemini.jpa.proxy.EMFServiceProxyHandler.createEMF(EMFServiceProxyHandler.java:151)
    at org.eclipse.gemini.jpa.proxy.EMFServiceProxyHandler.syncGetEMFAndSetIfAbsent(EMFServiceProxyHandler.java:127)
    at org.eclipse.gemini.jpa.proxy.EMFServiceProxyHandler.invoke(EMFServiceProxyHandler.java:73)
    at com.sun.proxy.$Proxy8.createEntityManager(Unknown Source)

为什么最后的代码不起作用？怎么解释？

Answer 1

那是因为有类似 @Access 的内容，如果您想使用混合模式，则必须在实体和字段级别指定。有两个值 AccessType.PROPERTY 和 AccesType.FIELD.

~~默认访问类型由您放置标识符注释的位置定义 (@Id)。如果你把它放在字段上 - 它将是 AccessType.FIELD，如果你把它放在 getter - 它将是 AccessType.PROPERTY。~~ - 编辑，不是由 JPA 定义.

如果你想注释的不是字段而是属性（字段上仍然有@Id）你必须定义一个getter并将它注释为AccessType.PROPERTY。（或者 getter 上的 @Id 反之亦然）。

JPA：在 fied 和 getter 上设置 @Id 的区别

JPA: difference between setting @Id on fied and on getter

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