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Python 异步线程超时_Python_Timeout_Asyncore - Fatal编程技术网

Python 异步线程超时

Python 异步线程超时,python,timeout,asyncore,Python,Timeout,Asyncore,简单异步http客户端,在站点不可用的情况下长时间挂起 例如,在网站www.evtur.ru上,它会等待很长时间,十分钟或更长时间 我找不到如何最小化超时的方法,例如,是否可以在5秒内完成超时 # coding=utf-8 import asyncore import string, socket import StringIO import mimetools, urlparse class AsyncHTTP(asyncore.dispatcher): # HTTP request

简单异步http客户端,在站点不可用的情况下长时间挂起

例如,在网站www.evtur.ru上,它会等待很长时间,十分钟或更长时间

我找不到如何最小化超时的方法,例如,是否可以在5秒内完成超时

# 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)

您必须在dispatcher套接字上设置超时。 打完电话

self.create_socket(socket.AF_INET, socket.SOCK_STREAM)
你可以打电话

self.socket.settimeout(nn)
在套接字上设置超时。您还可以像设置任何其他套接字一样设置任何其他套接字选项

self.socket.settimeout(nn)
这似乎对我不起作用

但是,asyncore.loop()的参数名为count。下面是asyncore.loop()的伪代码:

因此,如果您需要5秒钟,您需要执行以下操作:

asyncore.loop(timeout=1, count=5)
但不要建议以这种确切的方式工作。请注意,如果存在“事件”,则可能有5个以上的“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

我认为你的代码需要一些格式化。你是在混合制表符和空格吗?如果是这样,这是个坏主意。如本主题的其他问题中所述,问题不在于代码,而在于系统设置。可能与否重复,此问题不同(仅代码相同)。这里的问题不在于dns呼叫阻塞。站点访问中的问题(长时间没有服务器响应)和我通过asyncore使用非阻塞套接字搜索最小化超时,是否可能?
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