tweepy错误python 2.7

Question

我一直收到这个错误:

tweepy.error.TweepError: [{u'message': u'Status is a duplicate.', u'code': 187

我不知道为什么我得到这个错误我已经尝试了一切!

我的主要代码是:

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!"

我出于安全原因删除了API!我还删除了指向我服务器的链接.任何帮助将不胜感激!

谢谢

Answer 1

问题是tweepy不允许你发推特两次相同的推文,所以要修复它我添加了这些代码行:

for status in tweepy.Cursor(api.user_timeline).items():
    try:
        api.destroy_status(status.id)
    except:
        pass

上面的代码删除了以前的推文,以便我的下一条推文不会失败.

我希望这有助于其他人!