我正在寻找一个标准的Python 2.7软件包,它提供了一个HTTP服务器,可以在同一个端口号上同时进行流式连接。
嘿嘿那些主持人,请停止将我的问题标记为想要以非流式方式提供的问题的副本,例如: Multithreaded web server in python。不,我不想要像ThreadingMixIn
这样的黑客只收集响应并将其作为一个整体返回。
换句话说,我正在寻找执行以下示例程序所做的标准方法 - 但是没有自己编写整个HTTP服务器。
import time, socket, threading
sock = socket.socket (socket.AF_INET, socket.SOCK_STREAM)
host = socket.gethostname()
port = 8000
sock.bind((host, port))
sock.listen(1)
# my OWN HTTP server... Oh man, this is bad style.
HTTP = "HTTP/1.1 200 OK\nContent-Type: text/html; charset=UTF-8\n\n"
class Listener(threading.Thread):
def __init__(self):
threading.Thread.__init__(self)
self.daemon = True # stop Python from biting ctrl-C
self.start()
def run(self):
conn, addr = sock.accept()
conn.send(HTTP)
# serve up an infinite stream
i = 0
while True:
conn.send("%i " % i)
time.sleep(0.1)
i += 1
[Listener() for i in range(100)]
time.sleep(9e9)
首先我尝试了:
# run with this command:
# gunicorn -k gevent myapp:app
import time
def app(environ, start_response):
data = b"Hello, World!\n"
start_response("200 OK", [
("Content-Type", "text/plain"),
("Content-Length", str(len(data)))
])
for i in range(5):
time.sleep(1)
yield "Hello %i\n" % i
# https://stackoverflow.com/questions/22739394/streaming-with-gunicorn
但遗憾的是,即使使用-k gevent
,它也不会流式传输。
更新:似乎gunicorn正在尝试使用keepalive,这需要使用last-chunk
位进行Chunked Transfer Coding。对消息来源的快速了解表明,它没有实现这一点。因此,我可能需要一个更加高级的HTTP服务器,或者更简单的HTTP服务器(就像我上面的第一个例子,基于socket
),它不会对keepalive感到烦恼(无论如何,这对于大流来说都是非常愚蠢的)。 / p>
然后我尝试了:
import time
import threading
import BaseHTTPServer
class Handler(BaseHTTPServer.BaseHTTPRequestHandler):
def do_GET(self):
if self.path != '/':
self.send_error(404, "Object not found")
return
self.send_response(200)
self.send_header('Content-type', 'text/html; charset=utf-8')
self.end_headers()
# serve up an infinite stream
i = 0
while True:
self.wfile.write("%i " % i)
time.sleep(0.1)
i += 1
class Listener(threading.Thread):
def __init__(self, i):
threading.Thread.__init__(self)
self.i = i
self.daemon = True
self.start()
def run(self):
server_address = ('', 8000+self.i) # How to attach all of them to 8000?
httpd = BaseHTTPServer.HTTPServer(server_address, Handler)
httpd.serve_forever()
[Listener(i) for i in range(100)]
time.sleep(9e9)
这是非常好的,但我必须分配100个端口号有点烦人。这将需要一个讨厌的客户端重定向来使浏览器进入下一个可用的端口(好吧,好吧,我可以用JavaScript隐藏它,但它不是那么优雅。我宁愿编写我自己的HTTP服务器比那样)。
必须有一种干净的方法来在一个端口上获取所有BaseHTTPServer
侦听器,因为它是设置Web服务器的标准方法。或者也许可以使gunicorn
或某些包裹可靠地流式传输?
答案 0 :(得分:6)
默认的BaseHTTPServer
设置会在每个侦听器上重新绑定一个新套接字,如果所有侦听器都在同一个端口上,则无法在Linux中工作。在BaseHTTPServer.HTTPServer()
来电和serve_forever()
来电之间更改这些设置。
以下示例在同一端口上启动100个处理程序线程,每个处理程序都通过BaseHTTPServer
启动。
import time, threading, socket, SocketServer, BaseHTTPServer
class Handler(BaseHTTPServer.BaseHTTPRequestHandler):
def do_GET(self):
if self.path != '/':
self.send_error(404, "Object not found")
return
self.send_response(200)
self.send_header('Content-type', 'text/html; charset=utf-8')
self.end_headers()
# serve up an infinite stream
i = 0
while True:
self.wfile.write("%i " % i)
time.sleep(0.1)
i += 1
# Create ONE socket.
addr = ('', 8000)
sock = socket.socket (socket.AF_INET, socket.SOCK_STREAM)
sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
sock.bind(addr)
sock.listen(5)
# Launch 100 listener threads.
class Thread(threading.Thread):
def __init__(self, i):
threading.Thread.__init__(self)
self.i = i
self.daemon = True
self.start()
def run(self):
httpd = BaseHTTPServer.HTTPServer(addr, Handler, False)
# Prevent the HTTP server from re-binding every handler.
# https://stackoverflow.com/questions/46210672/
httpd.socket = sock
httpd.server_bind = self.server_close = lambda self: None
httpd.serve_forever()
[Thread(i) for i in range(100)]
time.sleep(9e9)