Google AppEngine Python Cron作业urllib [英] Google AppEngine Python Cron job urllib
本文介绍了Google AppEngine Python Cron作业urllib的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我需要使用Google AppEngine设置cron作业,这将使用urllib2执行托管在我的另一个服务器上的网页。
我知道
#!/ usr / bin / env python
来自google.appengine.ext import webapp
来自google.appengine.ext.webapp import util
import logging
import urllib
import urllib2
class MainHandler(webapp.RequestHandler):
def get(self):
logging.info('Starting backups。')
def main():
application = webapp.WSGIApplication([('/',MainHandler)],
debug = True)
util.run_wsgi_app b $ b urlStr =http://example.com/cron.php
request = urllib2.Request(urlStr)
try:
response = urllib2.urlopen请求)
logging.info(备份完成!)
除了URLError,e:
logging.error('出现错误%s',e.reason)
if __name__ =='__main__':
main()
<
>应该在
util.run_wsgi_app(application)
后结束。其余的属于你的处理程序类。 I'm in need of setting up a cron job using Google AppEngine which will use urllib2 to execute a web page hosted on another server of mine.
I know that the script is executing (checked the logs) but my logging inside my python script never seems to be outputted.
#!/usr/bin/env python
from google.appengine.ext import webapp
from google.appengine.ext.webapp import util
import logging
import urllib
import urllib2
class MainHandler(webapp.RequestHandler):
def get(self):
logging.info('Starting backups.')
def main():
application = webapp.WSGIApplication([('/', MainHandler)],
debug=True)
util.run_wsgi_app(application)
urlStr = "http://example.com/cron.php"
request = urllib2.Request(urlStr)
try:
response = urllib2.urlopen(request)
logging.info("Backup complete!")
except URLError, e:
logging.error('There was an error %s', e.reason)
if __name__ == '__main__':
main()
Can anyone see anything wrong with this?
解决方案
main()
should end after util.run_wsgi_app(application)
. The rest of that belongs in your handler class.
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