Question

我正在为我的项目使用带有RPC，Spring和Postgresql的GWT（2.5）。我的问题是关于HttpSession处理。所有对服务器的查询都由Spring（DispatchServlet）调度到我的GwtController（扩展RemoteServiceServlet）。特定的RemoteService注入GwtController。在GwtContorller中很容易得到HttpSession。例如，getThreadLocalRequest（）。getSession（）或仅来自request.getSession（）。我的问题是如何在RemoteService中获取HttpSession对象？

public class GwtRpcController extends RemoteServiceServlet {
……………
private RemoteService remoteService;
private Class remoteServiceClass;
………………

public ModelAndView handleRequest(HttpServletRequest arg0, HttpServletResponse arg1) throws Exception {
    …………
} 

public String processCall(String payload) throws SerializationException {
    …………
}

public void setRemoteService(RemoteService remoteService) {
    …………….
}

}

我的界面 - 实现RemoteService的DataService

public class DataServiceImpl implements DataService {

public Data getData(){

    !!!!! Here I want to get HttpSession !!!!!
    …………………………

}

}

Answer 1

您可以在ThreadLocal中维护Servlet并将当前Request存储在那里，然后使用静态方法展示您的Request。

public class GwtRpcController extends RemoteServiceServlet {

  static ThreadLocal<HttpServletRequest> perThreadRequest = 
         new ThreadLocal<HttpServletRequest>();

  @Override
  public String processCall(String payload) throws SerializationException {
    try {
      perThreadRequest.set(getThreadLocalRequest());
      return super.processCall(payload);
    } finally {
      perThreadRequest.set(null);
    }
  }

  public static HttpServletRequest getRequest() {
    return perThreadRequest.get();
  }
}


 public class DataServiceImpl implements DataService {
    public Data getData(){
       HttpServletRequest request = GwtRpcController.getRequest();
       HttpSession session = request.getSession();
    }
 }

GWT - RemoteService接口和Spring - 如何获取HttpSession？

1 个答案: