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java - OSGi 与 JPA : Service not starting although code seems ok

转载 作者:行者123 更新时间:2023-12-01 13:42:56 25 4
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是的,我知道我应该使用 Declarative ServicesBlueprint,但是我试图找出一个使用较低级别 API 的简单示例来感受 OSGi。

我只是想知道代码出了什么问题以及为什么我根本无法启动服务!!??

我在这里使用了两种方法:一种使用 ServiceTracker,另一种使用 ServiceReference,我知道这两种方法都不可靠> 但有人可以帮我让这个示例代码正常工作吗?对此将非常感激!!

这是我的代码:

我有一个简单的帐户实体类:

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 + '}';
}

}

AccountClient 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();
}
}

persistence.xml

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="Accounts" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<class>model.account.Account</class>
<exclude-unlisted-classes>true</exclude-unlisted-classes>
<properties>
<property name="eclipselink.target-database" value="Derby"/>
<property name="javax.persistence.jdbc.driver" value="org.apache.derby.jdbc.ClientDriver"/>
<property name="javax.persistence.jdbc.url" value="jdbc:derby://localhost:1527/accountDB;create=true"/>
<property name="javax.persistence.jdbc.user" value="app"/>
<property name="javax.persistence.jdbc.password" value="app"/>

<property name="eclipselink.logging.level" value="FINE"/>
<property name="eclipselink.logging.timestamp" value="false"/>
<property name="eclipselink.logging.thread" value="false"/>
<property name="eclipselink.logging.exceptions" value="true"/>
<property name="eclipselink.orm.throw.exceptions" value="true"/>
<property name="eclipselink.jdbc.read-connections.min" value="1"/>
<property name="eclipselink.jdbc.write-connections.min" value="1"/>
<property name="eclipselink.ddl-generation" value="drop-and-create-tables"/>
<property name="eclipselink.weaving" value="true"/>

</properties>

最后是Activator,其ServiceReference为:

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");
}
}

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) {}
}

最佳答案

您的 list 是否包含元持久性 header 以及 JPA Documentation 中所述的所有依赖项?所有 bundle 都启动了吗?

也许这有帮助Installing and Starting Gemini JPA Applications

关于java - OSGi 与 JPA : Service not starting although code seems ok,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20560589/

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