使用 tweepy 获取用户的最新消息
getting latest message of user with tweepy
所以我有这个代码 ->
import tweepy
ckey = ''
csecret = ''
atoken = ''
asecret = ''
auth = tweepy.OAuthHandler(ckey, csecret)
auth.set_access_token(atoken, asecret)
api = tweepy.API(auth)
recent_post = api.user_timeline(screen_name = 'DropSentry', count = 1, include_rts = True)
print(recent_post)
打印用户最近的 post。但是有没有办法 运行 这个代码 24/7?例如,我希望我的代码在用户 post 有新东西时再次打印。
方法user_timeline中的参数'since_id'可以帮你做这件事
您需要获取用户post最后状态的id,并在参数'since_id'
中给出
recent_id = 1388810249122062337 #hardcode the last recent post id from the user
while True:
recent_post = api.user_timeline(screen_name = 'DropSentry', count = 1, since_id=recent_id, include_rts = True)
if recent_post:
print(recent_post)
recent_id = recent_post[0].id
time.sleep(10) # To avoid spamming the API, you can put the number of seconds you want
但是如果用户 post 在 10 秒的间隔内有多个消息,这段代码将丢失消息。因此,您还可以同时获取用户的所有消息并将其全部打印出来。
recent_id = 1388810249122062337 #hardcode the last recent post id from the user
while True:
recent_posts = api.user_timeline(screen_name = 'DropSentry', since_id=recent_id, include_rts = True)
if recent_posts:
for recent_post in recent_posts:
print(recent_post)
recent_id = recent_post.id
time.sleep(10) # To avoid spamming the API, you can put the number of seconds you want
所以我有这个代码 ->
import tweepy
ckey = ''
csecret = ''
atoken = ''
asecret = ''
auth = tweepy.OAuthHandler(ckey, csecret)
auth.set_access_token(atoken, asecret)
api = tweepy.API(auth)
recent_post = api.user_timeline(screen_name = 'DropSentry', count = 1, include_rts = True)
print(recent_post)
打印用户最近的 post。但是有没有办法 运行 这个代码 24/7?例如,我希望我的代码在用户 post 有新东西时再次打印。
方法user_timeline中的参数'since_id'可以帮你做这件事
您需要获取用户post最后状态的id,并在参数'since_id'
中给出recent_id = 1388810249122062337 #hardcode the last recent post id from the user
while True:
recent_post = api.user_timeline(screen_name = 'DropSentry', count = 1, since_id=recent_id, include_rts = True)
if recent_post:
print(recent_post)
recent_id = recent_post[0].id
time.sleep(10) # To avoid spamming the API, you can put the number of seconds you want
但是如果用户 post 在 10 秒的间隔内有多个消息,这段代码将丢失消息。因此,您还可以同时获取用户的所有消息并将其全部打印出来。
recent_id = 1388810249122062337 #hardcode the last recent post id from the user
while True:
recent_posts = api.user_timeline(screen_name = 'DropSentry', since_id=recent_id, include_rts = True)
if recent_posts:
for recent_post in recent_posts:
print(recent_post)
recent_id = recent_post.id
time.sleep(10) # To avoid spamming the API, you can put the number of seconds you want