Python BaseHTTPServer.HTTPServer - 启动和停止事件的回调

时间:2013-05-24 08:35:50

标签: python http callback basehttpserver

参考:http://docs.python.org/2/library/basehttpserver.html

我有以下代码片段,它使用Python BaseHTTPServer来运行基本的HTTP服务器。

from BaseHTTPServer import HTTPServer
from BaseHTTPServer import BaseHTTPRequestHandler

# http request handler
class HttpHandler(BaseHTTPRequestHandler):
    def do_POST(self):
        print "I have just received a HTTP request through POST"

try:
    server = HTTPServer((<ip>, <port>), HttpHandler)

    # wait forever for incoming http requests!
    server.serve_forever()

except KeyboardInterrupt:
    server.socket.close()

我正在寻找的是一种在使用server.serve_forever() / server.socket.close()方法启动/停止http服务器时获取回调的方法。

假设我们有以下功能:

def http_server_start_callback():
    print "http server has just been started" 

def http_server_stop_callback():
    print "http server has just been stopped" 

我想在我启动服务器之后立即调用http_server_start_callback函数即server.serve_forever()并且我希望在我停止服务器之后立即调用http_server_stop_callback函数即server.socket.close()

使用以下回调配置http服务器非常好:

  • 启动服务器之前
  • 启动服务器后
  • 停止服务器之前
  • 停止服务器后

有没有办法在Python BaseHTTPServer.HTTPServer中设置这些回调?!

2 个答案:

答案 0 :(得分:4)

  

使用以下内容配置http服务器会很棒   回调:

     
      
  • 启动服务器之前
  •   
  • 启动服务器后
  •   
  • 停止服务器之前
  •   停止服务器后
  •   

请记住,操作系统将在套接字开始侦听时开始接受和排队TCP连接,这在BaseHTTPServer的构造函数中完成,因此如果您想在启动服务器之前执行冗长的任务,那么在OS开始接受连接之前,最好先做它们。

server_activate()方法调用socket.listen(),因此最好覆盖它。

类似地,操作系统将继续接受连接,直到调用socket.close()为止,因此如果您希望能够定义一个能够防止自身被关闭的“预停止”处理程序,那么最好使用server_close()方法,而不是直接调用socket.close()

我已经汇总了一个简单的例子,在请求处理程序上使用类方法来处理这四个新事件,尽管你可以将它们移动到其他地方......

from BaseHTTPServer import HTTPServer
from BaseHTTPServer import BaseHTTPRequestHandler

# Subclass HTTPServer with some additional callbacks
class CallbackHTTPServer(HTTPServer):

    def server_activate(self):
        self.RequestHandlerClass.pre_start()
        HTTPServer.server_activate(self)
        self.RequestHandlerClass.post_start()

    def server_close(self):
        self.RequestHandlerClass.pre_stop()
        HTTPServer.server_close(self)
        self.RequestHandlerClass.post_stop()


# HTTP request handler
class HttpHandler(BaseHTTPRequestHandler):

    @classmethod
    def pre_start(cls):
        print 'Before calling socket.listen()'

    @classmethod
    def post_start(cls):
        print 'After calling socket.listen()'

    @classmethod
    def pre_stop(cls):
        print 'Before calling socket.close()'

    @classmethod
    def post_stop(cls):
        print 'After calling socket.close()'

    def do_POST(self):
        print "I have just received an HTTP POST request"


def main():

    # Create server
    try:
        print "Creating server"
        server = CallbackHTTPServer(('', 8000), HttpHandler)
    except KeyboardInterrupt:
        print "Server creation aborted"
        return

    # Start serving
    try:
        print "Calling serve_forever()"
        server.serve_forever()
    except KeyboardInterrupt:
        print "Calling server.server_close()"
        server.server_close()


if __name__ == '__main__':
    main()

请注意,我还将对构造函数的调用移动到其自己的try...except块中,因为如果在构造期间按CTRL-C,server变量将不存在。

答案 1 :(得分:3)

您必须继承HTTPServer并使用您的班级HTTPServer

from __future__ import print_function
from BaseHTTPServer import HTTPServer
from BaseHTTPServer import BaseHTTPRequestHandler


class MyHTTPServer(HTTPServer):


    def __init__(self, *args, **kwargs):

        self.on_before_serve = kwargs.pop('on_before_serve', None)
        HTTPServer.__init__(self, *args, **kwargs)

    def serve_forever(self, poll_interval=0.5):

        if self.on_before_serve:
            self.on_before_serve(self)
        HTTPServer.serve_forever(self, poll_interval)


# http request handler
class HttpHandler(BaseHTTPRequestHandler):
    def do_POST(self):
        print("I have just received a HTTP request through POST")

try:
    server = MyHTTPServer(('0.0.0.0', 8080), HttpHandler,
                          on_before_serve = lambda server: print('Server will start to serve in the moment...'))

    # wait forever for incoming http requests!
    server.serve_forever()

except KeyboardInterrupt:
    server.socket.close()