我有一个脚本在线程中设置BasicHTTPServer
,以便主脚本可以自动打开指向服务器url
的Web浏览器以下载文件。下载文件后,我想关闭该服务器,但我不知道如何去做。这是我目前所做的一个例子:
def server():
HandlerClass = SimpleHTTPRequestHandler
ServerClass = BaseHTTPServer.HTTPServer
Protocol = 'HTTP/1.0'
server_address = ('127.0.0.1', 8000)
HandlerClass.protocol_version = Protocol
httpd = ServerClass(server_address, HandlerClass)
httpd.serve_forever()
def download():
t = threading.Thread(name='server', target=server)
t.start()
webbrowser.open('safari-http://127.0.0.1:8000/')
我想在webbrowser.open()之后关闭服务器。
谢谢
答案 0 :(得分:1)
我尝试了here给出的示例。你能检查一下它是否适合你。
runFlag = True
def server(server_class=BaseHTTPServer.HTTPServer,
handler_class=BaseHTTPServer.BaseHTTPRequestHandler):
global runFlag
server_address = ('127.0.0.1', 8000)
HandlerClass.protocol_version = Protocol
httpd = ServerClass(server_address, HandlerClass)
while runFlag:
httpd.handle_request()
httpd.shutdown()
def download():
t = threading.Thread(name='server', target=server)
t.start()
webbrowser.open('https:\\www.google.com')
global runFlag
runFlag = False
答案 1 :(得分:0)
以下是cryptoassets.core project, status server的示例:
class StatusHTTPServer(threading.Thread):
def __init__(self, ip, port):
threading.Thread.__init__(self)
self.running = False
self.ready = False
self.ip = ip
self.port = port
def run(self):
self.running = True
self.ready = True
self.httpd.serve_forever()
self.running = False
def start(self):
server_address = (self.ip, self.port)
try:
self.httpd = HTTPServer(server_address, StatusGetHandler)
except OSError as e:
raise RuntimeError("Could not start cryptoassets helper service status server at {}:{}".format(self.ip, self.port)) from e
threading.Thread.start(self)
def stop(self):
if self.httpd and self.running:
self.httpd.shutdown()
self.httpd = None
答案 2 :(得分:0)
因此,在深入研究了许多文章之后,我设法找到了一种对我来说完全关闭服务器的凌乱解决方案。为此,我整合了以下来源的代码:
UI是iOS Python IDE Python独有的模块,它基本上只创建按钮" start"," stop"和"访问"它们绑定到各自的_t函数。 ui.in_background装饰器只是让ui在后台发生事件时保持响应。
self.httpd.socket.close()
真正关闭了服务器,但是它很乱并且向stdout / err输出一个丑陋的错误所以我别无选择,只能通过将stdout / err重定向到死类来抑制它,这样就会丢失错误。标准stdout / err行为在之后立即恢复。感谢大家花时间和精力帮助我,我非常感激。
import console
import BaseHTTPServer
import SocketServer
from SimpleHTTPServer import SimpleHTTPRequestHandler
import sys
import threading
import webbrowser
from time import sleep
import ui
original_stdout = sys.stdout
original_stderr = sys.stderr
class BasicServer(SocketServer.TCPServer):
allow_reuse_address = True
class NullifyOutput():
def write(self, s):
pass
class ServerThread(threading.Thread):
def __init__(self, ip, port):
super(ServerThread, self).__init__()
self.ip = ip
self.port = port
self.HandlerClass = SimpleHTTPRequestHandler
self.Protocol = 'HTTP/1.0'
self.server_address = (self.ip, self.port)
self.HandlerClass.protocol_version = self.Protocol
try:
self.httpd = BasicServer(self.server_address, self.HandlerClass)
except:
self.port += 1
self.server_address = (self.ip, self.port)
self.httpd = BasicServer(self.server_address, self.HandlerClass)
self.stoptheserver = threading.Event()
def run(self):
while not self.stoptheserver.isSet():
self.httpd.handle_request()
def join(self, timeout=None):
self.stoptheserver.set()
self.httpd.socket.close()
super(ServerThread, self).join(timeout)
server = ServerThread('127.0.0.1', 8000)
def start_t(sender):
print server.isAlive()
if not server.isAlive():
server.start()
def visit_t(sender):
webbrowser.open('http://127.0.0.1:' + str(server.port))
#webbrowser.open('safari-http://127.0.0.1' + str(server.port))
# Use the safari- prefix to open in safari. You may need to switch to
# pythonista then back to safari to get the page to load.
@ui.in_background
def stop_t(sender):
sys.stdout, sys.stderr = NullifyOutput(), NullifyOutput()
server.join(3)
sys.stdout, sys.stderr = original_stdout, original_stderr
ui.load_view('SimpleServer').present('sheet')