python: 遇到异常重试X次，否则退出

Question

现在连接成功一次，如果遇到异常，没有像我希望的那样重试，只是抛出：

Will retry: [Errno 111] Connection refused

如果所有尝试都不成功，它应该 return 为假，如果至少有一个 return 回答

，它应该为真

似乎需要一些复杂的 'while'，例如

for attempt in range(attempts) and while True

这是我的代码：

attempts = 10
for attempt in range(attempts):
   try:
       conn = httplib.HTTPConnection("server:80", timeout=5)
       conn.request("GET","/url")
       r = conn.getresponse()
   except socket.error, serr:
       print("Will retry: %s" % serr)
       conn.close()
   else:
       print("OK")
   finally:
       return False

我也试过：

for attempt in range(attempts):
   while True:
       try:

同样的结果...

Answer 1

尝试在 while 循环中使用计数器和标志。

def funct():
    flag = False
    counter = 0
    while True:
        counter += 1
        try:
           conn = httplib.HTTPConnection("server:80", timeout=5)
           conn.request("GET","/url")
           r = conn.getresponse()
           flag = True
           break
        except socket.error, serr:
           print("Will retry: %s" % serr)
           conn.close() 
        if counter>9:
           break
    return flag

python: 遇到异常重试X次，否则退出

python: retry X times if exception met, exit if not

python

exception-handling

exception

httplib

python-2.7