Java 带JPA的OSGi：虽然代码似乎正常，但服务未启动

是的，我知道我应该使用

声明性服务

或

蓝图

，但是我正试图找出一个简单的例子，使用较低级别的API来感受OSGi

我只想知道代码有什么问题，为什么我根本无法启动服务

我在这里使用了两种方法：一种是使用

ServiceTracker

，另一种是使用

ServiceReference

，我知道这两种方法都不可靠，但有人能帮我让这个示例代码正常工作吗。我会非常感激的

这是我的密码：

我有一个简单的账户实体类：

package model.account;

import javax.persistence.*;

@Entity
public class Account {

@Id @GeneratedValue
int id;
double balance;

public int getId() {
    return id;
}

public void setId(int id) {
    this.id = id;
}

public double getBalance() {
    return balance;
}

public void setBalance(double balance) {
    this.balance = balance;
}

@Override
public String toString() {
    return "Account{" + "id=" + id + ", balance=" + balance + '}';
}

}

帐户客户端为：

package client;

public class AccountClient {

public void run(EntityManagerFactory emf) {
    EntityManager em = emf.createEntityManager();
    em.getTransaction().begin();

    Account a = new Account();
    a.setBalance(100.0);
    em.persist(a);

    em.getTransaction().commit();

    TypedQuery<Account> q = em.createQuery("SELECT a FROM Account a", Account.class);
    List<Account> results = q.getResultList();
    System.out.println("\n*** Account Report ***");
    for (Account acct : results) {
        System.out.println("Account: " + acct);
    }
    em.close();
}
}

ServiceTracker

：

package client;
public class Activator implements BundleActivator, ServiceTrackerCustomizer {

BundleContext ctx;
ServiceTracker emfTracker;

public void start(BundleContext context) throws Exception {
    ctx = context;
    System.out.println("Gemini JPA Basic Sample started");

    /* We are in the same bundle as the persistence unit so the services should be 
     * available when we start up (if nothing bad happened) and the tracker is really 
     * just saving us the lookup, but this is the idea of how you would listen for a 
     * persistence unit coming from another bundle.
     */
    emfTracker = new ServiceTracker(ctx, EntityManagerFactory.class.getName(), this);
    emfTracker.open();
    System.out.println("Started finally!!");
}

public void stop(BundleContext context) throws Exception {
    emfTracker.close();
    System.out.println("Gemini JPA Basic Sample stopped");
}

/*========================*/
/* ServiceTracker methods */
/*========================*/

public Object addingService(ServiceReference ref) {
    System.out.println("reached in add");
    Bundle b = ref.getBundle();
    System.out.println("Got ref");
    Object service = b.getBundleContext().getService(ref);
    System.out.println("service");
    String unitName = (String)ref.getProperty(EntityManagerFactoryBuilder.JPA_UNIT_NAME);
    System.out.println("search");

    if (unitName.equals("Accounts")) {
        new AccountClient().run((EntityManagerFactory)service);
        System.out.println("Found and started");
    }
    return service;
}
public void modifiedService(ServiceReference ref, Object service) {}
public void removedService(ServiceReference ref, Object service) {}    
}

---

您的清单是否包含元数据持久性标头和中描述的所有依赖项？ 所有包裹都准备好了吗

---

也许这会有帮助

是的，我也是这样做的。我尝试了另一种方法，它将实例化EntityManagerFactory的对象并调用Persistence.createEntityManagerFactory（“Accounts”）；其中给出了“未找到命名帐户的持久性单元”的错误。您对此有任何输入吗？Persistence.createEntityManagerFactory（“Accounts”）进行类路径扫描，在OSGi环境中不起作用。Gemini JPA使用Extender模式定位Persistence.xml文件。因此，JPA扩展器将等待清单中包含元持久性头的包。如果找到一个，它将使用这个bundles上下文来查找这个bundle中的persistence.xml文件。然后Gemini JPA读取persistence.xml文件，并使用persistence bunlde上下文加载实体。然后将配置的EntityManagerFactory作为服务添加到上下文中。如果这不起作用，那么可能是您的持久性捆绑包或JPA extender没有启动，元持久性丢失了……我检查了MANIFEST.MF是否仅在构建项目时创建。但在执行maven clean时，它并不存在。是pom.xml正在生成清单吗？如果是，我如何更新它以使其包含元持久性？我通过确保所有依赖项都已启动但未创建EMF来尝试代码。服务不返回。现在我如何更新构建，使清单包含元持久性。请帮忙。非常感谢。

package client;

public class Activator implements BundleActivator {

BundleContext ctx;
ServiceReference[] serviceReferences;
EntityManagerFactory emf;

public void start(BundleContext context) throws Exception {
    ctx = context;
    System.out.println("Gemini JPA Basic Sample started");

    try{
        serviceReferences = context.getServiceReferences(
                      EntityManagerFactory.class.getName(),
                      "(osgi.unit.name=Accounts)");
    }catch(Exception e){
        e.printStackTrace();
    }
    if(serviceReferences != null){
        emf = (EntityManagerFactory)context.getService(serviceReferences[0]);
    }
    if(emf != null){
        new AccountClient().run(emf);
    }
}

public void stop(BundleContext context) throws Exception {
    if(serviceReferences != null){
        context.ungetService(serviceReferences[0]);
    }
    System.out.println("Gemini JPA Basic Sample stopped");
}
}

package client;
public class Activator implements BundleActivator, ServiceTrackerCustomizer {

BundleContext ctx;
ServiceTracker emfTracker;

public void start(BundleContext context) throws Exception {
    ctx = context;
    System.out.println("Gemini JPA Basic Sample started");

    /* We are in the same bundle as the persistence unit so the services should be 
     * available when we start up (if nothing bad happened) and the tracker is really 
     * just saving us the lookup, but this is the idea of how you would listen for a 
     * persistence unit coming from another bundle.
     */
    emfTracker = new ServiceTracker(ctx, EntityManagerFactory.class.getName(), this);
    emfTracker.open();
    System.out.println("Started finally!!");
}

public void stop(BundleContext context) throws Exception {
    emfTracker.close();
    System.out.println("Gemini JPA Basic Sample stopped");
}

/*========================*/
/* ServiceTracker methods */
/*========================*/

public Object addingService(ServiceReference ref) {
    System.out.println("reached in add");
    Bundle b = ref.getBundle();
    System.out.println("Got ref");
    Object service = b.getBundleContext().getService(ref);
    System.out.println("service");
    String unitName = (String)ref.getProperty(EntityManagerFactoryBuilder.JPA_UNIT_NAME);
    System.out.println("search");

    if (unitName.equals("Accounts")) {
        new AccountClient().run((EntityManagerFactory)service);
        System.out.println("Found and started");
    }
    return service;
}
public void modifiedService(ServiceReference ref, Object service) {}
public void removedService(ServiceReference ref, Object service) {}    
}