如何在python 2中启动super

Question

我需要使这个python 3代码python 2兼容：

class AlarmThread(threading.Thread):
    def __init__(self, file_name="beep.wav"):
        super().__init__()
        self.file_name = file_name
        self.ongoing = None

当我尝试在python 2中使用该类时出现此错误：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "plugin.py", line 144, in run
    alert_after_timeout(seconds, message)
  File "plugin.py", line 96, in alert_after_timeout
    thread = AlarmThread()
  File "plugin.py", line 78, in __init__
    super().__init__()
TypeError: super() takes at least 1 argument (0 given)

Answer 1

您需要传入当前班级和self：

super(AlarmThread, self).__init__()

另见Why is Python 3.x's super() magic?为什么以及Python 3如何放弃指定这些内容的需要。