下面的代码是将Twitter公共时间线流式传输给变量,该变量将任何推文输出到控制台.我想将相同的变量(status.text,status.author.screen_name,status.created_at,status.source)保存到sqlite数据库中.当我的脚本看到推文并且没有任何内容写入sqlite数据库时,我收到语法错误.
错误:
$python stream-v5.py @lunchBoxhq
Filtering the public timeline for "@lunchBoxhq"RT @LunchBoxHQ: test 2 LunchBoxHQ 2012-02-29 18:03:42 Echofon
Encountered Exception: near "?": Syntax error
编码:
import sys
import tweepy
import webbrowser
import sqlite3 as lite
# Query terms
Q = sys.argv[1:]
sqlite3file='/var/www/twitter.lBox.com/html/stream5_log.sqlite'
CONSUMER_KEY = ''
CONSUMER_SECRET = ''
ACCESS_TOKEN = ''
ACCESS_TOKEN_SECRET = ''
auth = tweepy.OAuthHandler(CONSUMER_KEY, CONSUMER_SECRET)
auth.set_access_token(ACCESS_TOKEN, ACCESS_TOKEN_SECRET)
con = lite.connect(sqlite3file)
cur = con.cursor()
cur.execute("CREATE TABLE TWEETS(txt text, author text, created int, source text)")
class CustomStreamListener(tweepy.StreamListener):
def on_status(self, status):
try:
print "%s\t%s\t%s\t%s" % (status.text,
status.author.screen_name,
status.created_at,
status.source,)
cur.executemany("INSERT INTO TWEETS(?, ?, ?)", (status.text,
status.author.screen_name,
status.created_at,
status.source))
except Exception, e:
print >> sys.stderr, 'Encountered Exception:', e
pass
def on_error(self, status_code):
print >> sys.stderr, 'Encountered error with status code:', status_code
return True # Don't kill the stream
def on_timeout(self):
print >> sys.stderr, 'Timeout...'
return True # Don't kill the stream
streaming_api = tweepy.streaming.Stream(auth, CustomStreamListener(), timeout=60)
print >> sys.stderr, 'Filtering the public timeline for "%s"' % (' '.join(sys.argv[1:]),)
streaming_api.filter(follow=None, track=Q)
解决方法:
您在以下代码的最后一行缺少右括号(与您发布的内容相同的第34-37行):
cur.executemany("INSERT INTO TWEETS(?, ?, ?)", (status.text,
status.author.screen_name,
status.created_at,
status.source)
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