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社区首页 >问答首页 >如何防止EntityManager与guice-持久化和@事务性的重用?

如何防止EntityManager与guice-持久化和@事务性的重用?
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Stack Overflow用户
提问于 2019-08-20 15:02:20
回答 1查看 680关注 0票数 0

根据这个问题,当使用guice-persist时,EntityManager是事务作用域。如果我正确理解,这意味着将为每个事务创建一个新的EntityManager。在使用guice-persist时,建议使用JpaPersistModule,它提供所有绑定,只需将Provider<EntityManager>注入某个类,如下所示:

代码语言:javascript
复制
public class ProjectDAO {

  private final Provider<EntityManager> entityManagerProvider;

  @Inject
  public ProjectDAO(Provider<EntityManager> entityManagerProvider) {

    this.entityManagerProvider = entityManagerProvider;
  }
} 

Note:在这个回答中,它说EntityManager不应该直接注入,而是使用Provider<EntityManager>,以避免这个问题,因此要注入Provider<EntityManager>。此外,通过查看JpaPersistService的代码,EntityManager实例存储在ThreadLocal中。同时,@Transactional注释及其对应的JpaLocalTxnInterceptor应该确保在每个事务之后在ThreadLocal<EntityManager>字段上调用.set().remove()

现在,我已经尝试过了,每个线程都有自己的EntityManager。但是,它似乎没有再次被删除和设置,而是用于后续事务,即Hibernate的第一级缓存未被清除。

下面是一个完整的示例,它从两个不同的线程(顺序,而不是并行)中插入和删除一些实体,从而导致一个线程具有陈旧的信息:

项目(一个简单的实体)

代码语言:javascript
复制
    @NamedQueries({
        @NamedQuery(name = "project.findAll", query = "from project"),
        @NamedQuery(name = "project.deleteByProjectName", query = "delete from project p where p.name = :project_name")
    }
    )
    @Entity(name = "project")
    public class Project {

        @Id
        @GeneratedValue
        private Long id;

        @Column(name="name")
        private String name;

        // ... getters/setters
    }

ProjectDAO

代码语言:javascript
复制
    public class ProjectDAO {

      private final Provider<EntityManager> entityManagerProvider;

      @Inject
      public ProjectDAO(Provider<EntityManager> entityManagerProvider) {
        this.entityManagerProvider = entityManagerProvider;
      }

      public void insert(Project project) {
        entityManagerProvider.get().persist(project);
      }

      public List<Project> findAll() {

        return entityManagerProvider.get()
            .createNamedQuery("project.findAll", Project.class)
            .getResultList();
      }

      public void delete(String projectName) {

        entityManagerProvider.get()
            .createNamedQuery("project.deleteByProjectName")
            .setParameter("project_name", projectName)
            .executeUpdate(); 
      }

      public Project findById(Long id) {

        return entityManagerProvider.get().find(Project.class, id);
      }
    }

ProjectService

代码语言:javascript
复制
    public class ProjectService {

      private final ProjectDAO projectDAO;

      @Inject
      public ProjectService(ProjectDAO projectDAO) {

        this.projectDAO = projectDAO;
      }

      @Transactional
      public void addProject(Project project) {
        projectDAO.insert(project);
      }

      @Transactional
      public List<Project> findAll() {
        return projectDAO.findAll();
      }

      @Transactional
      public void delete(String projectName) {
        projectDAO.delete(projectName);
      }

      @Transactional
      public Project findById(Long id) {
        return projectDAO.findById(id);
      }

      public EntityManager getEntityManager() {
        return projectDAO.getEntityManager();
      }
    }

主类

代码语言:javascript
复制
    public class Start {

      public static void main(String[] args) throws InterruptedException {

        Injector injector = Guice.createInjector(new AbstractModule() {
          @Override 
          protected void configure() {
            install(new JpaPersistModule("hibernatetesting"));
            bind(ProjectService.class).in(Scopes.SINGLETON);
          }
        });

        ProjectService projectService = injector.getInstance(ProjectService.class);
        PersistService persistService = injector.getInstance(PersistService.class);

        persistService.start();

        // For the purpose of making transactions from different threads, we
        // create two single threaded executors
        ExecutorService executorService1 = Executors.newSingleThreadExecutor();

        ExecutorService executorService2 = Executors.newSingleThreadExecutor();

        // Execute a few queries from Thread 1
        CountDownLatch countDownLatch1 = new CountDownLatch(1);

        executorService1.execute(() -> {
        System.out.println("TEST: " + Thread.currentThread().getName());
        System.out.println("TEST: " +"EntityManager: " + projectService.getEntityManager().hashCode());
          projectService.addProject(new Project("project1"));
          projectService.addProject(new Project("project2"));
          countDownLatch1.countDown();
        });

        countDownLatch1.await();


        // Execute a few queries from Thread 2
        CountDownLatch countDownLatch2 = new CountDownLatch(1);

        executorService2.execute(() -> {
          System.out.println("TEST: " + Thread.currentThread().getName());
          System.out.println("TEST: " +"EntityManager: " + projectService.getEntityManager().hashCode());
          projectService.addProject(new Project("project3"));
          projectService.addProject(new Project("project4"));

          //----
          projectService.delete("project1");
          //----

          // project3 is not shown in this list
          projectService.findAll().forEach(System.out::println);
          countDownLatch2.countDown();
        });

        countDownLatch2.await();

        // Execute a few more queries from Thread 1
        CountDownLatch countDownLatch3 = new CountDownLatch(1);

        executorService1.execute(() -> {
          System.out.println("TEST: " + Thread.currentThread().getName());
          System.out.println("TEST: " +"EntityManager: " + projectService.getEntityManager().hashCode());
          projectService.addProject(new Project("project5"));
          projectService.addProject(new Project("project6"));

          // project3, which was deleted in Thread 2 is still visible in
          // this EntityManager
          // ----
          Project project = projectService.findById(3L);
          System.out.println("Project still exists " + project);
          // ----

          projectService.findAll().forEach(System.out::println);
          countDownLatch3.countDown();
        });

        countDownLatch3.await();

      }
    }

pom.xml

代码语言:javascript
复制
    ...
    <dependencies>

      <dependency>
        <groupId>org.hibernate</groupId>
        <artifactId>hibernate-core</artifactId>
        <version>4.3.11.Final</version>
      </dependency>

      <dependency>
        <groupId>com.google.inject</groupId>
        <artifactId>guice</artifactId>
        <version>4.2.2</version>
      </dependency>

      <dependency>
        <groupId>com.google.inject.extensions</groupId>
        <artifactId>guice-persist</artifactId>
        <version>4.2.2</version>
      </dependency>

      <dependency>
        <groupId>org.hsqldb</groupId>
        <artifactId>hsqldb</artifactId>
        <version>2.5.0</version>
      </dependency>

      <dependency>
        <groupId>org.hibernate</groupId>
        <artifactId>hibernate-entitymanager</artifactId>
        <version>4.3.11.Final</version>
      </dependency>

    </dependencies>
    ...

persistence.xml

代码语言:javascript
复制
    <?xml version="1.0" encoding="UTF-8"?>
    <persistence xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd" version="1.0">

    <persistence-unit name="hibernatetesting" transaction-type="RESOURCE_LOCAL">
      <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>

      <properties>
        <property name="hibernate.connection.driver_class" value="org.hsqldb.jdbcDriver"/>
        <property name="hibernate.connection.url" value="jdbc:hsqldb:mem:testDB"/>

        <property name="hibernate.show_sql" value="true" />
        <property name="hibernate.dialect" value="org.hibernate.dialect.HSQLDialect" />
        <property name="hibernate.hbm2ddl.auto" value="create" />

      </properties>
    </persistence-unit>
    </persistence>

1)这是在guice中使用EntityManager的常用方法吗?持久化和处理不同线程可能具有不同状态的事实?

2)如果不是,如何确保在每个事务之后在ThreadLocal上重新设置ThreadLocal?

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2019-08-26 10:10:34

上述代码中存在两个问题:

1)下一行

代码语言:javascript
复制
System.out.println("TEST: " +"EntityManager: " + projectService.getEntityManager().hashCode());

为调试目的添加的。但是,方法ProjectService.getEntityManager()调用ProjectDAO.getEntityManager(),而后者又调用entityManagerProvider.get(),它没有使用@Transactional进行注释。这会导致EntityManager在每个线程上设置一次,并且永远不会取消设置,即使其他来自ProjectService的具有@Transactional注释的方法稍后会被调用。简单地添加这个注释就可以解决这个问题。

2)在一个线程中,删除了名为"project1“的实体。

代码语言:javascript
复制
   //----
   projectService.delete("project1");
   //----

但是,在另一个线程中,将验证另一个实体的存在。

代码语言:javascript
复制
   // project3, which was deleted in Thread 2 is still visible in this EntityManager
   Project project = projectService.findById(3L);
   System.out.println("Project still exists " + project);

从一开始就没有被删除过。实体一个一个地添加- project1,project2,project3.他们的身份证是1,2,3.分别使用。所以代码应该是

代码语言:javascript
复制
   // project1, which was deleted in Thread 2 is still visible in this EntityManager
   Project project = projectService.findById(1L);
   System.out.println("Project still exists " + project);
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/57576510

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