1

在下面的代码中，如果我将其中一个 url 更改为无效的，整个过程将停止，并且我无法使用 ctrl+c 退出表单终端。所以我的问题是我应该如何在我的主线程运行方法中处理异常，如果发生错误触发它并转到下一个列表元素而不会失败整个过程：

#!/usr/bin/env python
import Queue
import threading
import urllib2
import time

hosts = ["http://yahoo.com", "http://google.com", "http://amazon.com","http://apple.com"]

queue = Queue.Queue()

class ThreadUrl(threading.Thread):
    """Threaded Url Grab"""
    def __init__(self, queue):
        threading.Thread.__init__(self)
        self.queue = queue

    def run(self):
        while True:
            #grabs host from queue
            host = self.queue.get()

            #grabs urls of hosts and prints first 1024 bytes of page
            url = urllib2.urlopen(host)
            print "connected"

            #signals to queue job is done
            self.queue.task_done()


start = time.time()
def main():

    #spawn a pool of threads, and pass them queue instance 
    for i in range(5):
        t = ThreadUrl(queue)
        t.setDaemon(True)
        t.start()


    #populate queue with data
    for host in hosts:
        queue.put(host)

    #wait on the queue until everything has been processed
    queue.join()
main()
print "Elapsed Time: %s" % (time.time() - start)
4

1 回答 1

3

使用finally块来确保即使出现错误，线程也始终发出信号。

def run(self):
    while True:
        #grabs host from queue
        host = self.queue.get()

        #grabs urls of hosts and prints first 1024 bytes of page
        try:
            url = urllib2.urlopen(host)
            print "connected"
        except urllib2.URLError:
            print "couldn't connect to %s" % host

        finally:
            #signals to queue job is done
            self.queue.task_done()
于 2013-05-27T11:36:58.043 回答