如何在运行时通过表名在JPA中检索实体？ [英] How to retrieve an entity in JPA by table name at runtime?

问题描述

是否可以通过实体的本地表名来确定实体类名？

旧代码的某些部分 （在此示例中为CUSM_LANGUAGE），我需要 找到实体类名 >（LanguageEntity）

Some part of legacy code does only have a String value of the table (CUSM_LANGUAGE in this example), and I need to find the entity class name (LanguageEntity)

@Entity
@Table(name = "CUSM_LANGUAGE")
public class LanguageEntity implements java.io.Serializable {
    private static final long serialVersionUID = 1L;
    @Column(name = "LANG_DESCRIPTION")
    private String description = null;
    ....

推荐答案

在JPA 2.0中，您有类： http://docs.oracle.com/javaee/6/api/javax/persistence/EntityManagerFactory.html

In JPA 2.0 you have class: http://docs.oracle.com/javaee/6/api/javax/persistence/EntityManagerFactory.html

这个类有getMetamodel（）方法，我希望包含你的数据库的所有信息。

This class has getMetamodel() method which I expect to contains all information about your database.

在Metamodel（ http://docs.oracle.com/javaee/6/api/javax/持久化/元模型/ Metamodel.html ）你有getEntities（），这是我们得到的：EntityType

In Metamodel (http://docs.oracle.com/javaee/6/api/javax/persistence/metamodel/Metamodel.html) you have getEntities() that's how we get: EntityType

我不知道如何从EntityType获得有关表它正在使用。但是，即使它不是直接的可能（我怀疑），你可以做如下所述。

I'm not sure how get from EntityType information about table it is using. But even it's not directly possible (which I doubt) you can do it as described below.

因为你有遗留代码，我希望你有JPA< 2.0，它没有这个getMetamodel方法。这就是为什么你可以得到实体的信息强烈取决于你使用的特定JPA。

Because you have legacy code I expect you have JPA < 2.0 which doesn't have this getMetamodel method. That's why informations you can get about entities strongly depends on specific JPA you're using.

但是如果你有很多类，你关心（你可以创建这个列表你可以自己创建索引：

However if you have bunch of classes you care about (and you can create this list easily) you can create index on your own:

String tableNameOfEntityClass = LanguageEntity.class.getAnnotation(Table.class).name();

所以你可以把所有你关心的类列出，然后创建关键tableName和值引用类。

So you can put all classes you care to list and then create map with key tableName and value refering class.

当然，如果没有我上面描述的Metamodel，它也不是很有用，因为你可以创建准备好的映射，而不是List和实体类。

Of course it's not very useful without Metamodel which I described above because as well you can create ready map instead of List with entity classes.

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