无法捕获 TweepError 异常

Question

我试图在 while - try - except 循环中捕获 TweepError 异常但未成功。以下代码在 TweepError/RateLimitError 发生时保持停止 运行。

    import tweepy
    import time
    name_set = ('name1','name2','name3')
    result = []
    for screen_name in name_set:
        while True:
           profile = api.get_user(screen_name = screen_name)
           try:
               print('collecting user %s'%screen_name)
               result.append(profile)
               break
           except tweepy.RateLimitError:
               print('sleep 15 minutes')
               sleep(900)
               continue
           except tweepy.TweepError as e:
               print(e)
               print('Account %s'%screen_name)
               break
           else:
               print('Account %s'%screen_name)
               break

TweepError

TweepError: [{'message': 'User not found.', 'code': 50}]

Answer 1

您应该在 try 块中放置一个 API 调用语句以捕获异常：

try:
    profile = api.get_user(screen_name = screen_name)
    print('collecting user %s'%screen_name)
    ...