java - 同步访问 REST Web 服务

Question

我在使用这段代码的简单 REST 服务上遇到了麻烦:

@GET
@Path("next/{uuid}")
@Produces({"application/xml", "application/json"})
public synchronized Links nextLink(@PathParam("uuid") String uuid) {
    Links link = null;
    try {
        link = super.next();
        if (link != null) {
            link.setStatusCode(5);
            link.setProcessUUID(uuid);
            getEntityManager().flush(); 
            Logger.getLogger("Glassfish Rest Service").log(Level.INFO, "Process {0} request url : {1} #id  {2} at {3} #", new Object[]{uuid, link.getLinkTxt(), link.getLinkID(), Calendar.getInstance().getTimeInMillis()});
        }
    } catch (NoResultException ex) {
    } catch (IllegalArgumentException ex) {
    }
    return link;
}

这应该提供一个链接对象，并将其标记为已使用 (setStatusCode(5)) 以防止下次访问服务时发送相同的对象。问题是，当有很多快速客户端访问 Web 服务时，这一个向不同客户端提供 2 或 3 倍的相同链接对象。我该如何解决这个问题？？

这是使用 to 的请求: @NamedQuery(name = "Links.getNext", query = "SELECT l FROM Links l WHERE l.statusCode = 2")

和 super.next() 方法:

public T next() {

    javax.persistence.Query q = getEntityManager().createNamedQuery("Links.getNext");
    q.setMaxResults(1);
    T res = (T) q.getSingleResult();
    return res;
}

谢谢

Answer 1

(根)JAX-RS 资源的生命周期是每个请求，所以(否则是正确的)synchronized nextLink 上的关键字遗憾的是，该方法无效。

您需要的是同步访问/更新的方法。这可以通过多种方式完成:

I) 您可以同步外部对象，由框架注入(inject)(例如:注入(inject)@ApplicationScoped 的 CDI)，如下所示:

@ApplicationScoped
public class SyncLink{
    private ReentrantLock lock = new ReentrantLock();
    public Lock getLock(){
       return lock;
    }
}
....
public class MyResource{
  @Inject SyncLink sync;

  @GET
  @Path("next/{uuid}")
  @Produces({"application/xml", "application/json"})
  public Links nextLink(@PathParam("uuid") String uuid) {
    sync.getLock().lock();
    try{
      Links link = null;
      try {
        link = super.next();
        if (link != null) {
            link.setStatusCode(5);
            link.setProcessUUID(uuid);
            getEntityManager().flush(); 
            Logger.getLogger("Glassfish Rest Service").log(Level.INFO, "Process {0} request url : {1} #id  {2} at {3} #", new Object[]{uuid, link.getLinkTxt(), link.getLinkID(), Calendar.getInstance().getTimeInMillis()});
        }
      } catch (NoResultException ex) {
      } catch (IllegalArgumentException ex) {
      }
      return link;
    }finally{
       sync.getLock().unlock();
    }
  }
}

II)你可以偷懒，在类里面保持同步

public class MyResource{
  @Inject SyncLink sync;

  @GET
  @Path("next/{uuid}")
  @Produces({"application/xml", "application/json"})
  public Links nextLink(@PathParam("uuid") String uuid) {
     Links link = null;
    synchronized(MyResource.class){
      try {
        link = super.next();
        if (link != null) {
            link.setStatusCode(5);
            link.setProcessUUID(uuid);
            getEntityManager().flush(); 
            Logger.getLogger("Glassfish Rest Service").log(Level.INFO, "Process {0} request url : {1} #id  {2} at {3} #", new Object[]{uuid, link.getLinkTxt(), link.getLinkID(), Calendar.getInstance().getTimeInMillis()});
        }
      } catch (NoResultException ex) {
      } catch (IllegalArgumentException ex) {
      }

    }
    return link;
  }
}

III) 您可以使用数据库进行同步。在那种情况下，您将研究 JPA2 中可用的悲观 锁定。