python:如果遇到异常则重试X次，如果没有则退出

Question

现在，如果连接成功，它会工作一次，但如果遇到异常，它会不会按照我的意愿重试，只是抛出:

Will retry: [Errno 111] Connection refused

如果所有尝试均未成功，则应返回 False；如果至少有一次尝试返回答案，则应返回 True

似乎需要“while”有些复杂，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

我也尝试过:

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

相同的结果...

Answer 1

尝试在 while 循环内使用计数器和标志。

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