python - 如何查看 python 方法的参数(和类型)？

Question

$ py twitterDump2.py 
Traceback (most recent call last):
  File "twitterDump2.py", line 30, in <module>
    stream=tweepy.Stream(username,password,listener)
TypeError: __init__() takes exactly 3 arguments (4 given)

我的代码:

username="abc"
password="abc"
listener = StreamWatcherListener()
stream=tweepy.Stream(username,password,listener)

Answer 1

__init__ 的第一个参数通常是 self，因此它希望您只传递两个参数。

令人惊讶的是 tweepy.streaming.py 代码建议:

class Stream(object):

    host = 'stream.twitter.com'

    def __init__(self, auth, listener, **options):
        self.auth = auth
        self.listener = listener

授权是这样创建的:

auth = tweepy.BasicAuthHandler(username, password)

你的代码应该是这样的

username="abc"
password="abc"
listener = StreamWatcherListener()
auth = tweepy.BasicAuthHandler(username, password)
stream=tweepy.Stream(auth,listener)

查看代码:http://github.com/joshthecoder/tweepy/blob/master/tweepy/streaming.py