python - 立即在获取和返回响应上运行脚本

Question

抱歉，但我对 Python httpserver 非常不熟悉。我也知道它的漏洞或局限性，但它非常适合我的需求。

我有大约 10 行代码在对服务器的每个获取请求上运行（基于请求标头中的内容）。这些可能需要长达 8 小时才能完全运行，但我想立即返回对请求的响应。这样做的最佳方法是什么？

from http.server import BaseHTTPRequestHandler, HTTPServer
import logging


class S(BaseHTTPRequestHandler):
    def _set_response(self):
        self.send_response(200)
        self.send_header('Content-type', 'text/html')
        self.end_headers()

    def do_GET(self):
        THIS IS WHERE MY CODE IS RUNNING
        logging.info("GET request,nPath: %snHeaders:n%sn", str(self.path), str(self.headers))
        self._set_response()
        self.wfile.write("HEY A GET request for {}".format(self.path).encode('utf-8'))


def run(server_class=HTTPServer, handler_class=S, port=8080):
    logging.basicConfig(level=logging.INFO)
    server_address = ('', port)
    httpd = server_class(server_address, handler_class)
    logging.info('Starting httpd...n')
    try:
        httpd.serve_forever()
    except KeyboardInterrupt:
        pass
    httpd.server_close()
    logging.info('Stopping httpd...n')


if __name__ == '__main__':
    from sys import argv

    if len(argv) == 2:
        run(port=int(argv[1]))
    else:
        run()

谢谢

标签： pythonbasehttpserver

Answer 1

这正是多线程的用途，我建议您阅读本教程以了解更多信息。

这是一个如何从 do_GET 函数内部创建新线程的示例：

from http.server import BaseHTTPRequestHandler, HTTPServer
import logging
import thread


class S(BaseHTTPRequestHandler):
    def _set_response(self):
        self.send_response(200)
        self.send_header('Content-type', 'text/html')
        self.end_headers()

def do_process(self):
# Your code here

def do_GET(self):
    thread.start_new_thread(self.do_process)
    logging.info("GET request,nPath: %snHeaders:n%sn", str(self.path), str(self.headers))
    self._set_response()
    self.wfile.write("HEY A GET request for {}".format(self.path).encode('utf-8'))


def run(server_class=HTTPServer, handler_class=S, port=8080):
    logging.basicConfig(level=logging.INFO)
    server_address = ('', port)
    httpd = server_class(server_address, handler_class)
    logging.info('Starting httpd...n')
    try:
        httpd.serve_forever()
    except KeyboardInterrupt:
        pass
    httpd.server_close()
    logging.info('Stopping httpd...n')


if __name__ == '__main__':
    from sys import argv

    if len(argv) == 2:
        run(port=int(argv[1]))
    else:
        run()