下面的代码正在为一个变量流式传输 Twitter 公共(public)时间线,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。当我的脚本看到一条推文但没有任何内容写入 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
代码:
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)
最佳答案
您在以下代码的最后一行(您发布的第 34-37 行)缺少右括号:
cur.executemany("INSERT INTO TWEETS(?, ?, ?)", (status.text,
status.author.screen_name,
status.created_at,
status.source)
只需在您的元组参数后添加一个括号即可立即关闭方法调用。
关于python - tweepy 流到 sqlite 数据库 - 语法无效,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9434205/