Question

我想在Java中使用doPost方法结束之前获取HTTP状态代码。例如在Python中有self.send_response(200)，它以异步方式发送状态？

 /**
 * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse
 * response)
 *
 */
@SuppressWarnings("unchecked")
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    //Final result object


    //I want to send a http status code 200 before the task running

    myThreadPool = Executors.newFixedThreadPool(2);
    Future taskOne = myThreadPool.submit(new Runnable() {
        @Override
        public void run() {
            try {
                try {
                    // My first task
                } catch (ServletException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }

    });

    // Task two 
    .
    .

Answer 1

我不完全确定您发布的代码的意图是什么，但我猜你想要某种异步处理，以便在请求完成之前将一些数据发送到客户端？

为此，请参阅异步servlet，例如：https://blogs.oracle.com/enterprisetechtips/entry/asynchronous_support_in_servlet_3

有一个很好的例子似乎正在服务于我想要实现的目标：

@WebServlet("/foo" asyncSupported=true)
   public class MyServlet extends HttpServlet {
        public void doGet(HttpServletRequest req, HttpServletResponse res) {
            ...
            AsyncContext aCtx = request.startAsync(req, res);
            ScheduledThreadPoolExecutor executor = new ThreadPoolExecutor(10);
            executor.execute(new AsyncWebService(aCtx));
        }
   }

   public class AsyncWebService implements Runnable {
        AsyncContext ctx;
        public AsyncWebService(AsyncContext ctx) {
            this.ctx = ctx;
        }
        public void run() {
            // Invoke web service and save result in request attribute
            // Dispatch the request to render the result to a JSP.
            ctx.dispatch("/render.jsp");
   }
}

以异步方式获取HTTP状态代码

1 个答案: