Chrome发出阻止请求

编程问答 2022-06-11

问题描述

我做了一个最小的例子，完全重复了我的代码的行为。我从firefox和chrome发送请求。我注意到，在chrome发出请求后，firefox不再收到答案。经过一些研究，我意识到服务器响应localhost：8000 / favicon.ico路径请求时出现错误。一次收到错误404后，chrome在来自服务器的每个响应之后都会创建到该服务器的另一个连接，但不发送数据，这会导致recv函数锁定。

  File "/usr/lib/python3.7/socket.py",line 589,in readinto
    return self._sock.recv_into(b)

我发现可以为处理程序类设置连接超时，它在StreamRequestHandler：r中已考虑在内：

if self.timeout is not None:
    self.connection.settimeout(self.timeout)

但是我很尴尬的是在文档中没有关于此的信息 https://docs.python.org/3/library/socketserver.html#socketserver.BaseRequestHandler.handle 

import logging
import json
import http.server
from http import HTTPStatus
from typing import Optional
from urllib.parse import urlparse,parse_qs
import socketserver
from threading import Thread
import traceback
from functools import wraps
import sys,os

project_dir = os.path.abspath(os.curdir)
sys.path.append(project_dir)


logging.getLogger().setLevel("DEBUG")


class RESTHandler(http.server.BaseHTTPRequestHandler):
    """
    Rest router for api methods
    """
    def __init__(self,*args,**kwargs):
        logging.info(f"Creating RESTHandler obj. Args: {args},kwargs: {kwargs}")
        super().__init__(*args,**kwargs)
    
    def end_headers(self) -> None:
        self.send_header('Access-Control-Allow-Origin','*')
        http.server.BaseHTTPRequestHandler.end_headers(self)

    # noinspection PyPep8Naming
    def do_GET(self):
        logging.info(self.path)
        url = urlparse(self.path)

        if "favicon.ico" in url.path:
            self.send_error(HTTPStatus.NOT_FOUND,message='UnkNown api path.')
            return
        self.send_response(HTTPStatus.OK)
        self.send_header('Content-Type','application/json')
        self.end_headers()
        self.wfile.write(json.dumps({"resp":"I am OK","int": 5},ensure_ascii=False).encode('utf-8'))

class ApiService():

    DEFAULT_API_PORT = 8000
    DEFAULT_API_HOST = ''

    def __init__(self,ui_service = None,host: Optional[str] = None,port: Optional[int] = None):
        self.ui_service = ui_service
        self.host = host or self.DEFAULT_API_HOST
        self.port = port or self.DEFAULT_API_PORT

    def _run(self):
        while True:
            try:
                with socketserver.Tcpserver((self.host,self.port),RESTHandler,bind_and_activate=False) as httpd:
                    logging.info("Starting server....")
                    httpd.allow_reuse_address = True
                    httpd.server_bind()
                    httpd.server_activate()
                    logging.info(f"Serving API at {self.host}:{self.port}")
                    httpd.serve_forever()
                    break
            except Exception as e:
                tb_list = traceback.format_exception( type(e),e,tb=e.__traceback__)
                tb_list = [ s.replace("\n","") for s in tb_list ]
                tb_str = "; ".join(tb_list)
                logging.error(f"Unexpected exception while http-server was working: {tb_str}")

    def run(self,in_thread=True):
        if in_thread:
            t = Thread(target=self._run)
            t.start()
        else:
            self._run()

if __name__ == '__main__':
    ApiService().run(in_thread=False)

解决方法

我猜想，Chrome浏览器会使用预打开套接字的网络浏览器，在我看来，TCPServer会无限期等待。但是我仍然很有趣，为什么只有在404之后，使用请求处理程序超时的合法性又如何呢？

httphttppython-3.xtcpserver