Question

使用Spring 4和Hibernate 4，我可以使用Reflection从当前环境中获取Hibernate Configuration对象，使用以下代码：

@Autowired LocalContainerEntityManagerFactoryBean lcemfb;

EntityManagerFactoryImpl emf = (EntityManagerFactoryImpl) lcemfb.getNativeEntityManagerFactory();
SessionFactoryImpl sf = emf.getSessionFactory();
SessionFactoryServiceRegistryImpl serviceRegistry = (SessionFactoryServiceRegistryImpl) sf.getServiceRegistry();
Configuration cfg = null;

try {
    Field field = SessionFactoryServiceRegistryImpl.class.getDeclaredField("configuration");
    field.setAccessible(true);
    cfg = (Configuration) field.get(serviceRegistry);
} catch (NoSuchFieldException | SecurityException | IllegalArgumentException | IllegalAccessException e) {
    e.printStackTrace();
}

SchemaUpdate update = new SchemaUpdate(serviceRegistry, cfg);

对于Hibernate 5，我必须使用一些MetadataImplementor，这些似乎不能从任何这些对象中获得。我还尝试将MetadataSources与serviceRegistry一起使用。但确实说它是错误的ServiceRegistry。

还有其他方法让这个工作吗？

Answer 1

这个问题的基本思路是：

实施org.hibernate.integrator.spi.Integrator，将所需数据存储到某个持有者。将实施注册为服务，并在您需要的地方使用它。

您可以在此处找到工作示例https://github.com/valery-barysok/spring4-hibernate5-stackoverflow-34612019

创建org.hibernate.integrator.api.integrator.Integrator类

import hello.HibernateInfoHolder;
import org.hibernate.boot.Metadata;
import org.hibernate.engine.spi.SessionFactoryImplementor;
import org.hibernate.service.spi.SessionFactoryServiceRegistry;

public class Integrator implements org.hibernate.integrator.spi.Integrator {

    @Override
    public void integrate(Metadata metadata, SessionFactoryImplementor sessionFactory, SessionFactoryServiceRegistry serviceRegistry) {
        HibernateInfoHolder.setMetadata(metadata);
        HibernateInfoHolder.setSessionFactory(sessionFactory);
        HibernateInfoHolder.setServiceRegistry(serviceRegistry);
    }

    @Override
    public void disintegrate(SessionFactoryImplementor sessionFactory, SessionFactoryServiceRegistry serviceRegistry) {
    }
}

创建META-INF/services/org.hibernate.integrator.spi.Integrator文件

org.hibernate.integrator.api.integrator.Integrator

import org.hibernate.boot.spi.MetadataImplementor;
import org.hibernate.tool.hbm2ddl.SchemaExport;
import org.hibernate.tool.hbm2ddl.SchemaUpdate;
import org.springframework.boot.CommandLineRunner;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;

@SpringBootApplication
public class Application implements CommandLineRunner {

    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }

    @Override
    public void run(String... args) throws Exception {
        new SchemaExport((MetadataImplementor) HibernateInfoHolder.getMetadata()).create(true, true);
        new SchemaUpdate(HibernateInfoHolder.getServiceRegistry(), (MetadataImplementor) HibernateInfoHolder.getMetadata()).execute(true, true);
    }
}

Answer 2

我想加上Aviad的答案，按照OP的要求完成它。

内幕：

为了获取MetadataImplementor的实例，解决方法是通过Java的SessionFactoryBuilderFactory工具注册ServiceLoader的实例。当hibernate被引导时，getSessionFactoryBuilder方法会调用此注册服务的MetadataImplementor方法及其自身实例。代码参考如下：

Service Loading
Invocation of getSessionFactoryBuilder

因此，最终要获得MetadataImplementor的实例，您必须实现SessionFactoryBuilderFactory并注册，以便ServiceLoader可以识别此服务：

SessionFactoryBuilderFactory的实现：

public class MetadataProvider implements SessionFactoryBuilderFactory {

    private static MetadataImplementor metadata;

    @Override
    public SessionFactoryBuilder getSessionFactoryBuilder(MetadataImplementor metadata, SessionFactoryBuilderImplementor defaultBuilder) {
        this.metadata = metadata;
        return defaultBuilder; //Just return the one provided in the argument itself. All we care about is the metadata :)
    }

    public static MetadataImplementor getMetadata() {
        return metadata;
    }
}

为了注册上述内容，请在以下路径中创建简单的文本文件（假设它是一个maven项目，最终我们需要'META-INF'文件夹在类路径中可用）：

src/main/resources/META-INF/services/org.hibernate.boot.spi.SessionFactoryBuilderFactory

文本文件的内容应该是一行（如果需要注册多个实例，甚至可以是多行），说明SessionFactoryBuilderFactory实现的完全限定类路径。例如，对于上面的类，如果您的包名是'com.yourcompany.prj'，则以下内容应该是该文件的内容。

com.yourcompany.prj.MetadataProvider

就是这样，如果您运行应用程序，弹簧应用程序或独立休眠，一旦hibernate被引导，您将通过静态方法获得MetadataImplementor实例。

更新1：

无法通过Spring注入它。我深入研究了Hibernate的源代码，并且元数据对象没有存储在SessionFactory中的任何地方（这是我们从Spring获得的）。所以，它不可能注入它。但如果你想以Spring的方式使用它，有两种选择：

扩展现有类并自定义

LocalSessionFactoryBean - ＆gt;元数据源 - ＆gt; MetadataBuilder

LocalSessionFactoryBean是您在Spring中配置的，它具有MetadataSources的对象。 MetadataSources创建MetadataBuilder，后者又创建MetadataImplementor。以上所有操作都不存储任何内容，它们只是动态创建对象并返回。如果您想拥有MetaData实例，则应扩展并修改上述类，以便它们在返回之前存储相应对象的本地副本。这样您就可以引用MetadataImplementor。但我不会真的推荐这个，除非它真的需要，因为API可能会随着时间而改变。

另一方面，如果您不介意从SessionFactory构建MetaDataImplemetor，以下代码将帮助您：

EntityManagerFactoryImpl emf=(EntityManagerFactoryImpl)lcemfb.getNativeEntityManagerFactory();
SessionFactoryImpl sf=emf.getSessionFactory();
StandardServiceRegistry serviceRegistry = sf.getSessionFactoryOptions().getServiceRegistry();
MetadataSources metadataSources = new MetadataSources(new BootstrapServiceRegistryBuilder().build());
Metadata metadata = metadataSources.buildMetadata(serviceRegistry);
SchemaUpdate update=new SchemaUpdate(serviceRegistry,metadata); //To create SchemaUpdate

// You can either create SchemaExport from the above details, or you can get the existing one as follows:
try {
    Field field = SessionFactoryImpl.class.getDeclaredField("schemaExport");
    field.setAccessible(true);
    SchemaExport schemaExport = (SchemaExport) field.get(serviceRegistry);
} catch (NoSuchFieldException | SecurityException | IllegalArgumentException | IllegalAccessException e) {
    e.printStackTrace();
}

Answer 3

看一下这个：

public class EntityMetaData implements SessionFactoryBuilderFactory {

    private static final ThreadLocal<MetadataImplementor> meta = new ThreadLocal<>();

    @Override
    public SessionFactoryBuilder getSessionFactoryBuilder(MetadataImplementor metadata, SessionFactoryBuilderImplementor defaultBuilder) {
        meta.set(metadata);
        return defaultBuilder;
    }

    public static MetadataImplementor getMeta() {
        return meta.get();
    }
}

看看似乎满足您需求的This Thread

Answer 4

好吧，我接受这个：

public class SchemaTranslator {
    public static void main(String[] args) throws Exception {
        new SchemaTranslator().run();
    }
    private void run() throws Exception {    
        String packageName[] = { "model"};    
        generate(packageName);
    }   
    private List<Class<?>> getClasses(String packageName) throws Exception {
        File directory = null;
        try {
            ClassLoader cld = getClassLoader();
            URL resource = getResource(packageName, cld);
            directory = new File(resource.getFile());
        } catch (NullPointerException ex) {
            throw new ClassNotFoundException(packageName + " (" + directory + ") does not appear to be a valid package");
        }
        return collectClasses(packageName, directory);
    }
    private ClassLoader getClassLoader() throws ClassNotFoundException {
        ClassLoader cld = Thread.currentThread().getContextClassLoader();
        if (cld == null) {
            throw new ClassNotFoundException("Can't get class loader.");
        }
        return cld;
    }
    private URL getResource(String packageName, ClassLoader cld) throws ClassNotFoundException {
        String path = packageName.replace('.', '/');
        URL resource = cld.getResource(path);
        if (resource == null) {
            throw new ClassNotFoundException("No resource for " + path);
        }
        return resource;
    }
    private List<Class<?>> collectClasses(String packageName, File directory) throws ClassNotFoundException {
        List<Class<?>> classes = new ArrayList<>();
        if (directory.exists()) {
            String[] files = directory.list();
            for (String file : files) {
                if (file.endsWith(".class")) {
                    // removes the .class extension
                    classes.add(Class.forName(packageName + '.' + file.substring(0, file.length() - 6)));
                }
            }
        } else {
            throw new ClassNotFoundException(packageName + " is not a valid package");
        }
        return classes;
    }
    private void generate(String[] packagesName) throws Exception {
        Map<String, String> settings = new HashMap<String, String>();
        settings.put("hibernate.hbm2ddl.auto", "drop-create");
        settings.put("hibernate.dialect", "org.hibernate.dialect.PostgreSQL94Dialect");
        MetadataSources metadata = new MetadataSources(
                new StandardServiceRegistryBuilder()
                        .applySettings(settings)
                        .build());    
        for (String packageName : packagesName) {
            System.out.println("packageName: " + packageName);
            for (Class<?> clazz : getClasses(packageName)) {
                System.out.println("Class: " + clazz);
                metadata.addAnnotatedClass(clazz);
            }
        }
        SchemaExport export = new SchemaExport(
                (MetadataImplementor) metadata.buildMetadata()
        );
        export.setDelimiter(";");
        export.setOutputFile("db-schema.sql");
        export.setFormat(true);
        export.execute(true, false, false, false);
    }
}

使用Hibernate 5和Spring 4的编程SchemaExport / SchemaUpdate

4 个答案: