Question

我使用EclipseLink，结果非常奇怪。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)

为什么上次代码不起作用？怎么解释呢？

Answer 1

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

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

如果要注释不是字段而是属性（在字段上仍然有@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。）

JPA：在fied和getter上设置@Id之间的区别

1 个答案: