问题
I keep getting this error:
tweepy.error.TweepError: [{u'message': u'Status is a duplicate.', u'code': 187
I have no clue why I am getting this error I have tried everything!
My main code is:
import socket
from urllib2 import urlopen, URLError, HTTPError
socket.setdefaulttimeout( 23 ) # timeout in seconds
url = 'http://google.co.uk'
try :
response = urlopen( url )
except HTTPError, e:
tweet_text = "Raspberry Pi Server is DOWN!"
textfile = open('/root/Documents/server_check.txt','w')
textfile.write("down")
textfile.close()
except URLError, e:
tweet_text = "Raspberry Pi Server is DOWN!"
textfile = open('/root/Documents/server_check.txt','w')
textfile.write("down")
textfile.close()
else :
textfile = open('/root/Documents/server_check.txt','r')
if 'down' in open('/root/Documents/server_check.txt').read():
tweet_text = "Raspberry Pi Server is UP!"
textfile = open('/root/Documents/server_check.txt','w')
textfile.write("up")
textfile.close()
elif 'up' in open('/root/Documents/server_check.txt').read():
tweet_text = ""
if len(tweet_text) <= 140:
if tweet_text == "Raspberry Pi Server is DOWN!" or tweet_text == "Raspberry Pi Server is UP!":
api.update_status(status=tweet_text)
else:
pass
else:
print "Your message is too long!"
I have removed the API's for security reasons! I have also removed the link to my server. Any help will be appreciated!
Thanks
回答1:
The problem was that tweepy doesn't let you tweet the same tweet twice so to fix it i added these lines of code:
for status in tweepy.Cursor(api.user_timeline).items():
try:
api.destroy_status(status.id)
except:
pass
the code above deletes previous tweets so that my next tweet doesn't fail.
I hope this helps someone else!
来源:https://stackoverflow.com/questions/29635085/tweepy-error-python-2-7