烧瓶，按1个处理请求 [英] Flask, processing requests 1 by 1

问题描述

我有一个flask应用程序，它监听一些工作.这个过程很长(让我们说1分钟)，我不允许同时处理两个请求.

I have a flask application which listens for some job to do. The process is quite long (let us say 1 minute) and I would like not allow to process two requests at the same time.

如果收到请求，我可以关闭正在监听的端口烧瓶，并在完成后再次打开，我将非常高兴.另外，我可以设置一个信号灯，但不确定烧瓶如何同时运行.

I will be great if once I receive a request, I could close the port flask is listening to and open again when finish. Alternatively I could setup a semaphore but I am not sure about how flask is running concurrently.

有什么建议吗?

from flask import Flask, request
app = Flask(__name__)

@app.route("/",methods=['GET'])
def say_hi():
    return "get not allowed"

@app.route("/",methods=['POST'])
def main_process():
    # heavy process here to run alone
    return "Done"

if __name__ == "__main__":
    app.run(debug=True,host='0.0.0.0')

推荐答案

您可以为此使用信号量:

You could use a semaphore for this:

import threading
import time
sem = threading.Semaphore()

@app.route("/",methods=['POST'])
def main_process():
    sem.acquire()
    # heavy process here to run alone
    sem.release()
    return "Done"

信号灯的用法是控制对公共资源的访问.

The semaphore usage is to control the access to a common resource.

您可以在此处

这个SO问题也可以在此处

This SO question can help you as well here

正如GeorgSchölly在评论中写道:在多种服务的情况下，上述解决方案是有问题的.

As Georg Schölly wrote in comment, The above mentioned solution is problematic in a situation of multiple services.

尽管如此，您可以使用wsgi来实现您的目标.

Although, you can use the wsgi in order to accomplish your goal.

@app.route("/",methods=['POST'])
def main_process():
    uwsgi.lock()
    # Critical section
    # heavy process here to run alone
    uwsgi.unlock()
    return "Done"

uWSGI支持可配置数量的锁，可用于同步工作进程

uWSGI supports a configurable number of locks you can use to synchronize worker processes

有关更多信息，请在此处

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