Java JPA：在fied和getter上设置@Id的区别

我使用日食，我得到了非常奇怪的结果。请考虑下面的代码：

此代码适用于：

@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;
    }
}

此代码也适用于：

@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;
    }
}

@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;
    }
}

此代码也适用于：

@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;
    }
}

@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;
    }
}

此代码不起作用：

@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;
    }
}

我得到以下例外情况：

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？如何解释？

这是因为如果要使用混合模式，必须在实体和字段级别指定类似于

@Access

的内容。有两个值

AccessType.PROPERTY

和

AccessType.FIELD

默认访问类型由您放置标识符注释的位置定义（

@Id

）。如果您将它放在字段上-它将是

AccessType.field

，如果您将它放在getter上-它将是

AccessType.PROPERTY

-已编辑，未由JPA定义

---

如果您不想注释字段而是属性（字段上仍有@Id），则必须定义一个getter并将其注释为

AccessType.PROPERTY

。（对于getter上的

@Id

，反之亦然）。

我感到困惑：如果放置

@Id

的位置决定了

访问类型，为什么它会抱怨“没有指定主键”？根据您的答案，最后一个示例将默认为
AccessType.PROPERTY
，并且
@Id
注释放在了正确的位置，不是吗？只需进一步搜索，就可以得到以下答案：您所说的与答案相同，但结果是，在答案的注释部分，提示了此类行为（
@Id
决定
AccessType
）不是JPA标准。（它也不能解释OP面临的行为）我刚刚读了你发布的内容。老实说，我记得关于AccessType的这一点，只是因为我在JPA培训中有一个关于它的练习（但我使用Hibernate作为提供者）.我现在理解的是EclipseLink的处理方式不同。不过，你是对的。我的回答没有回答实际问题，这就是为什么没有PK例外：）。但肯定有助于避免这些症状。我想JPA的标准坚持访问类型在整个类中都必须是标准的（即使在我相信的整个层次结构中也是如此）。供应商可自行决定如何处理不一致的情况。答案不错，因为它为其他人（包括我）找到更多问题提供了正确的方向：）