asyncore timeout

Question

简单的异步http客户端，由于没有可用的站点而长时间挂起。

例如在网站www.evtur.ru等待很长时间，十分钟或更长时间。

我无法找到如何最小化超时的方法，例如有可能在5秒内完成超时吗？

# coding=utf-8 
import asyncore 
import string, socket 
import StringIO 
import mimetools, urlparse 

class AsyncHTTP(asyncore.dispatcher): 
    # HTTP requestor 

    def __init__(self, uri): 
     asyncore.dispatcher.__init__(self) 

     self.uri = uri 


     # turn the uri into a valid request 
     scheme, host, path, params, query, fragment = urlparse.urlparse(uri) 
     assert scheme == "http", "only supports HTTP requests" 
     try: 
      host, port = string.split(host, ":", 1) 
      port = int(port) 
     except (TypeError, ValueError): 
      port = 80 # default port 
     if not path: 
      path = "/" 
     if params: 
      path = path + ";" + params 
     if query: 
      path = path + "?" + query 

     self.request = "GET %s HTTP/1.0\r\nHost: %s\r\n\r\n" % (path, host) 

     self.host = host 
     self.port = port 

     self.status = None 
     self.header = None 
     self.http_code = None 
     self.data = "" 

     # get things going! 
     self.create_socket(socket.AF_INET, socket.SOCK_STREAM) 
     #self.connect((host, port)) 
     #return 
     try: 
      self.connect((host, port)) 
     except Exception,e: 
      self.close() 
      self.handle_connect_expt(e) 

    def handle_connect(self): 
     self.send(self.request) 

    def handle_expt(self): 
     print "handle_expt error!" 
     self.close() 

    def handle_error(self): 
     print "handle_error error!" 
     self.close() 

    def handle_connect_expt(self,expt): 
     print "connection error:",expt 

    def handle_code(self):   
     print self.host," : ","recv http code: ",self.http_code 


    def handle_read(self): 
     data = self.recv(2048) 
     #print data 
     if not self.header: 
      self.data = self.data + data 
      try: 
       i = string.index(self.data, "\r\n\r\n") 
      except ValueError: 
       return # continue 
      else: 
       # parse header 
       fp = StringIO.StringIO(self.data[:i+4]) 
       # status line is "HTTP/version status message" 
       status = fp.readline() 
       self.status = string.split(status, " ", 2) 
       self.http_code = self.status[1] 
       self.handle_code()  

       # followed by a rfc822-style message header 
       self.header = mimetools.Message(fp) 
       # followed by a newline, and the payload (if any) 
       data = self.data[i+4:] 
       self.data = "" 
       #header recived 
       #self.close() 


    def handle_close(self): 
     self.close() 



c = AsyncHTTP('http://www.python.org') 
c = AsyncHTTP('http://www.evtur.ru') 
asyncore.loop(timeout=0.05) 

Answer 1

您必须在调度程序套接字上设置超时。 调用

self.create_socket(socket.AF_INET, socket.SOCK_STREAM) 

后，您可以拨打

self.socket.settimeout(nn) 

设置超时的插座上。您也可以像设置其他套接字一样设置其他套接字选项。

Answer 2

self.socket.settimeout(nn) 

这似乎不适合我。

但是，asyncore.loop（）具有称为count的参数。这里是伪代码什么asyncore.loop（）这样做：

for i in range(count): 
    ... 
    select(... , timeout) 
    ... 

所以，如果你想5秒，你需要做的事：

asyncore.loop(timeout=1, count=5) 

但不建议在这个确切的方式工作。请注意，如果有“事件”，则可能有超过5个“计数”。我使用以下代码：

start = int(time.time()) 

while True: 
    asyncore.loop(timeout=1, count=5) 
    print "LOOP : %d enqueued, waiting to finish" % len(asyncore.socket_map) 

    if len(asyncore.socket_map) == 0 : 
     break 

    if int(time.time()) - start > timeout : 
     print "LOOP : Timeout" 
     break