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python线程异常导致进程停止

转载 作者:行者123 更新时间:2023-11-30 23:36:08 25 4
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在下面的代码中,如果我将其中一个网址更改为无效的内容,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处理主线程运行方法中的异常,如果发生错误则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)

最佳答案

使用 finally block 来确保线程即使在出现错误时也始终发出信号。

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()

关于python线程异常导致进程停止,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16771822/

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