Question

我尝试以编程方式配置我的EntityManager。我的供应商是OpenJPA，我编写了一个简单的控制台应用程序。这是我的代码

public static void main(String[] args) {
    Map<String, String> properties = new HashMap<>();

    properties.put("openjpa.ConnectionDriverName", "org.postgresql.Driver");
    properties.put("openjpa.ConnectionURL", "jdbc:postgresql://localhost:5432/shop");

    properties.put("openjpa.ConnectionUserName", "bob");
    properties.put("openjpa.ConnectionPassword", "secret");

    properties.put("openjpa.RuntimeUnenhancedClasses", "supported");
    properties.put("openjpa.MetaDataFactory", "jpa(Types=de.jpa.demo.dto.Category;de.jpa.demo.dto.User;de.jpa.demo.dto.Order;de.jpa.demo.dto.Product)");
    properties.put("openjpa.jdbc.SynchronizeMappings", "buildSchema(foreignKeys=true)");

    EntityManagerFactory factory = Persistence.createEntityManagerFactory("test", properties);

    OpenJPAConfiguration configuration = ((EntityManagerFactoryImpl) factory).getConfiguration();
    MetaDataRepository repositoryInstance = configuration.getMetaDataRepositoryInstance();

    repositoryInstance.addPersistenceAware(Category.class);
    repositoryInstance.addPersistenceAware(Order.class);
    repositoryInstance.addPersistenceAware(Product.class);
    repositoryInstance.addPersistenceAware(User.class);

    manager = factory.createEntityManager();    
}

但是我收到了错误

javax.persistence.PersistenceException: No persistence providers available for "test" after trying the following discovered implementations: org.apache.openjpa.persistence.PersistenceProviderImpl
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:182)
    at de.jpa.demo.dto.OpenJpaUserRepositoryIT.init(OpenJpaUserRepositoryIT.java:56)
    at de.jpa.demo.dto.OpenJpaUserRepositoryIT.before(OpenJpaUserRepositoryIT.java:22)

我做错了什么？

Answer 1

你必须有一个META-INF / persistence.xml文件，其中包含一个名为＆＃39; test＆＃39;的持久性单元。在它。

另外，请取悦删除openjpa.RuntimeUnenhancedClasses属性。这是一个错误的功能，如果你花时间找出另一个增强策略，你会更开心。

OpenJPA：如何以编程方式配置EntityManager？

1 个答案: