完成HttpServletResponse但继续处理 [英] Finishing a HttpServletResponse but continue processing

问题描述

我的情况似乎符合Async Servlet 3.0 / Comet情况但我需要做的就是在接受传入参数后返回200响应代码（或其他）。

I have a situation that seems to fit the Async Servlet 3.0 / Comet situation but all I need to do is return a 200 response code (or other) after accepting the incoming parameters.

HttpServlet有没有办法完成http请求/响应握手并继续处理？

Is there a way for a HttpServlet to complete the http request/response handshake and yet continue processing?

类似于......

doPost( req, response ) {
   // verify input params...
   response.setStatus( SC_OK );
   response.close();
   // execute long query
}     

编辑：查看javax。 servlet包 - 我的问题的正确措辞是

Looking at the javax.servlet package - the proper phrasing to my question is

如何提交响应？

与Servlet一样.isCommitted（）

as in Servlet.isCommitted()

推荐答案

以下是我处理这种情况的方法：

Here's how I've handled this situation:

1. 当应用启动时，创建 ExecutorService ，带 Executors.newFixedThreadPool（numThreads）（还有其他类型）执行者，但我建议从这个开始）


2. 在 doPost（）中，创建一个 Runnable 将执行所需的处理 - 您的任务 - 并将其提交到 ExecutorService ，如下所示： executor.execute（任务）


3. 最后，您应该返回HTTP状态 202接受 ，如果可能的话，位置标题表示客户可以检查处理状态的位置。

1. When the app starts up, create an ExecutorService with Executors.newFixedThreadPool(numThreads) (there are other types of executors, but I suggest starting with this one)
2. In doPost(), create an instance of Runnable which will perform the desired processing - your task - and submit it to the ExecutorService like so: executor.execute(task)
3. Finally, you should return the HTTP Status 202 Accepted, and, if possible, a Location header indicating where a client will be able to check up on the status of the processing.

我高度建议您阅读 Java Concurrency in Practice ，这是一本非常棒且非常实用的书。

I highly recommend you read Java Concurrency in Practice, it's a fantastic and very practical book.

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