如何在python中異常後重試？

Question

我有一個從for i in range(0, 100)開始的循環。通常它運行正常，但有時由於網絡條件而失敗。目前，我已經設置了它，以便在失敗時，它將在除外條款中爲continue（繼續到i的下一個數字）。

我可以重新分配相同的數字到i並再次運行循環的失敗迭代嗎？

Answer 1

在您的for循環中執行while True，將您的try代碼放在裏面，並且只有在代碼成功時才從while循環中斷開。

for i in range(0,100): 
    while True: 
     try: 
      # do stuff 
     except SomeSpecificException: 
      continue 
     break 

Answer 2

最明確的方法是明確設置i。例如：

i = 0 
while i < 100: 
    i += 1 
    try: 
     # do stuff 

    except MyException: 
     continue 

Answer 3

增量只有當try子句成功

Answer 4

更「實用」法不使用那些醜陋的循環變量，同時循環：

def tryAgain(retries=0): 
    if retries > 10: return 
    try: 
     # Do stuff 
    except: 
     retries+=1 
     tryAgain(retries) 

tryAgain() 

Answer 5

有一個在類似的東西Python Decorator Library。

請注意，它不會測試例外情況，而是測試返回值。它重試，直到裝飾函數返回True。

稍作修改後的版本應該可以做到。

Answer 6

我傾向於限制重試次數，所以，如果有與特定項目的問題，你最終會繼續到下一個，即：

for i in range(100): 
    for attempt in range(10): 
    try: 
     # do thing 
    except: 
     # perhaps reconnect, etc. 
    else: 
     break 
    else: 
    # we failed all the attempts - deal with the consequences. 

Answer 7

的retrying package是重試的好方法代碼塊失敗。

例如：

@retry(wait_random_min=1000, wait_random_max=2000) 
def wait_random_1_to_2_s(): 
    print "Randomly wait 1 to 2 seconds between retries" 

Answer 8

一個超時的通用解決方案：

import time 

def onerror_retry(exception, callback, timeout=2, timedelta=.1): 
    end_time = time.time() + timeout 
    while True: 
     try: 
      yield callback 
      break 
     except exception: 
      if time.time() > end_time: 
       raise 
      elif timedelta > 0: 
       time.sleep(timedelta) 

用法：

for retry in onerror_retry(SomeSpecificException, do_stuff): 
    retry() 

Answer 9

下面是我對如何解決這一問題的想法：

j = 19 
def calc(y): 
    global j 
    try: 
     j = j + 8 - y 
     x = int(y/j) # this will eventually raise DIV/0 when j=0 
     print("i = ", str(y), " j = ", str(j), " x = ", str(x)) 
    except: 
     j = j + 1 # when the exception happens, increment "j" and retry 
     calc(y) 
for i in range(50): 
    calc(i) 

Answer 10

這是一個類似於其他的解決方案，但是如果它在規定的編號或重試中沒有成功，它將引發異常。

tries = 3 
for i in range(tries): 
    try: 
     do_the_thing() 
    except KeyError as e: 
     if i < tries - 1: # i is zero indexed 
      continue 
     else: 
      raise 
    break 

Answer 11

使用一段時間，計數器：

count = 1 
while count <= 3: # try 3 times 
    try: 
     # do_the_logic() 
     break 
    except SomeSpecificException as e: 
     # If trying 3rd time and still error?? 
     # Just throw the error- we don't have anything to hide :) 
     if count == 3: 
      raise 
     count += 1 

Answer 12

使用遞歸

for i in range(100): 
    def do(): 
     try: 
      ## Network related scripts 
     except SpecificException as ex: 
      do() 
    do() ## invoke do() whenever required inside this loop 

Answer 13

您可以使用Python重試包。 Retrying

它是用Python編寫的，以簡化向任何事物添加重試行爲的任務。