新海报在这里，如果这已经被覆盖，请原谅我。我正在尝试在 python 中编写一个小脚本，这将使我能够自动化我的 Twitter 帐户并直接向新关注者发送消息。大部分代码都有效，但是在尝试实现直接消息功能时我收到了一些错误代码。
这是我的代码 -
#!C:\Python27\python.exe
import tweepy
import webbrowser
# Make constants to hold values of our keys
CONSUMER_KEY = "xxxxxx"
CONSUMER_SECRET = "xxxxxxx"
ACCESS_TOKEN = "xxxxxxxxxx"
ACCESS_SECRET = "xxxxxxxxxx"
# set auth variables
auth = tweepy.OAuthHandler(CONSUMER_KEY, CONSUMER_SECRET)
auth.set_access_token(ACCESS_TOKEN, ACCESS_SECRET)
# create a new api
api = tweepy.API(auth)
# create an instance of the twitter api class
auth = tweepy.OAuthHandler(CONSUMER_KEY, CONSUMER_SECRET)
auth_url = auth.get_authorization_url()
# open the window for authorization, twitter will generate the pin
webbrowser.open(auth_url)
print "Copy PIN from the window that opens"
# get the pin number from the user
verifier = raw_input('PIN: ').strip()
auth.get_access_token(verifier)
# get the access key and secret returned from twitter
access_key = auth.access_token.key
access_secret = auth.access_token.secret
# set authorization token
auth.set_access_token(access_key, access_secret)
api = tweepy.API(auth)
# make a tweet
def send_tweet():
to_tweet = True
tweet_text = raw_input("Enter your tweet content below... Only the first 140 characters will be used.\n>>> ")
api.update_status(tweet_text[0:140])
print "You tweeted \n'" + tweet_text[0:140] + "'"
restart = raw_input("Do you want to tweet again? (Y/N)\n>>> ")
if restart.lower() == "y":
send_tweet()
else:
print "Returning to the Main Menu...\n"
# search twitter
def keyword_follow():
search_phrase = raw_input("What do you want to search for?\n>>> ").strip()
search_number = raw_input("How many results do you want to return?\n>>> ")
search_result = api.search(search_phrase, rpp=search_number)
for i in search_result:
print i.from_user + " said " + i.text + "\n"
to_follow = raw_input("Do you want to follow " + i.from_user + "? (Y/N)\n>>> ")
if to_follow.lower() == "n":
print i.from_user + " was not followed!"
else:
api.create_friendship(i.from_user)
print "You followed " + i.from_user + "!\n"
# check if the user wants to search again
restart = raw_input("Do you want to search again? (Y/N)\n>>> ")
if restart.lower() == "n":
print "Returning to the Main Menu...\n"
else:
return keyword_follow()
def keyword_retweet():
search_phrase = raw_input("What do you want to search for?\n>>> ").strip()
search_number = raw_input("How many results do you want to return?\n>>> ")
search_result = api.search(search_phrase, rpp=search_number)
for i in search_result:
print i.from_user + " said " + i.text + "\n"
to_retweet = raw_input("Do you want to retweet" + i.from_user + "? (Y/N)\n>>> ")
if to_retweet.lower() == "n":
print i.from_user + " was not retweeted!"
else:
api.retweet(i.id)
print "Retweeted!\n"
again = raw_input("See more? (Y/N)\n>>> ")
if again.lower() == "n":
break
# check if the user wants to search again
restart = raw_input("Do you want to search again? (Y/N)\n>>> ")
if restart.lower() == "n":
print "Returning to the Main Menu...\n"
else:
return keyword_retweet()
def mass_unfollow():
hits_left = api.rate_limit_status()['remaining_hits']
print "You can unfollow " + str(hits_left) + " people this hour...\n"
print "Checking who doesn't follow you back. This will take a minute.\n"
# first, create some lists to hold the followers
followers = []
friends = []
# we have to use a Cursor for pagination purposes
for follower in tweepy.Cursor(api.followers).items():
followers.append(follower)
for friend in tweepy.Cursor(api.friends).items():
friends.append(friend)
# create a non_reciprocals list, these are non-followers (set - set)
non_reciprocal = list(set(friends) - set(followers))
print str(len(non_reciprocal)) + " non-reciprocal followers.\n"
# first, double check that we want to unfollow
double_check = raw_input("Unfollow them? (Y/N) ***WARNING, THIS ACTION CANNOT BE UNDONE***\n>>> " )
if double_check.lower() == "y":
# count the number of people we unfollow, just for fun
counter = 0
for i in non_reciprocal:
if hits_left > 0:
api.destroy_friendship(i.screen_name)
print "Successfully unfollowed " + i.screen_name
hits_left -= 1
else:
print "You ran out of hits! Try again in an hour!"
counter += 1
print "You unfollowed " + str(counter) + " people!\n"
print "Now returning to the Main Menu."
else:
print "Returning to the Main Menu...\n"
#todo - automatically DM new followers
def direct_messages():
new_followers = API.followers(user)
for i in new_followers:
newDM = raw_input (i.from_user + "send follower DM?" + "Y/N" )
if newDM.lower() == "n":
print i.from_user + " was not messaged"
print "Now returning to the Main Menu."
else:
api.send_direct_message(user_id = i.from_user, text = "message text here"
print "You messaged " + i.from_user
# create the menu
keep_running = True
while keep_running:
print "Main Menu"
print "---------\n"
selection = raw_input("(1)Tweet | (2)Keyword Follow | (3)Keyword Retweet | (4)Mass Unfollow | (5)End | (6)Direct Message\n\n>>> ")
if selection == "1":
print "New Tweet"
print "---------\n"
send_tweet()
elif selection == "2":
print "Keyword Follow"
print "--------------\n"
keyword_follow()
elif selection == "3":
print "Keyword Retweet"
print "---------------\n"
keyword_retweet()
elif selection == "4":
print "Mass Unfollow"
print "-------------\n"
print "WARNING: MASS UNFOLLOW IS AGAINST THE TOS OF TWITTER. YOU'VE BEEN WARNED\n"
mass_unfollow()
elif selection == "5":
print "Direct Message"
print "------------\n"
direct_messages()
else:
print "BYE!\n\n"
keep_running = False
我在第 159 行遇到的第一个错误说明打印的语法错误，当删除此打印功能时，我在第 163 行收到错误，说明 keep_running 函数语法无效。
我对此很陌生，所以这可能是我忽略的非常简单的事情。
任何帮助，将不胜感激