Question

在此示例代码中，应用程序的URL似乎由应用程序中的以下行确定：

application = webapp.WSGIApplication([('/mailjob', MailJob)], debug=True)

也可以通过app.yaml的app处理程序中的这一行：

- url: /.*
  script: main.py

但是，cron任务的URL由以下行设置：

url: /tasks/summary

因此，cron实用程序似乎会调用“/tasks/summary”，并且由于应用程序处理程序，这将导致main.py被调用。这是否意味着，就cron而言，应用程序中设置URL的行是无关紧要的：

application = webapp.WSGIApplication([('/mailjob', MailJob)], debug=True)

。。。因为cron任务所需的唯一URL是app.yaml中定义的URL。

app.yaml
application: yourappname
version: 1
runtime: python
api_version: 1

handlers:

- url: /.*
  script: main.py

cron.yaml
cron:
    - description: daily mailing job
    url: /tasks/summary
    schedule: every 24 hours

main.py
#!/usr/bin/env python  

import cgi
from google.appengine.ext import webapp
from google.appengine.api import mail
from google.appengine.api import urlfetch 

class MailJob(webapp.RequestHandler):
    def get(self):

        # Call your website using URL Fetch service ...
        url = "http://www.yoursite.com/page_or_service"
        result = urlfetch.fetch(url)

        if result.status_code == 200:
            doSomethingWithResult(result.content)

        # Send emails using Mail service ...
        mail.send_mail(sender="admin@gmail.com",
                to="someone@gmail.com",
                subject="Your account on YourSite.com has expired",
                body="Bla bla bla ...")
        return

application = webapp.WSGIApplication([
        ('/mailjob', MailJob)], debug=True)

def main():
    wsgiref.handlers.CGIHandler().run(application)

if __name__ == '__main__':
    main()

Answer 1

你可以这样做：

app.yaml
application: yourappname
version: 1
runtime: python
api_version: 1

handlers:

- url: /tasks/.*
  script: main.py

cron.yaml
cron:
    - description: daily mailing job
    url: /tasks/summary
    schedule: every 24 hours

main.py
#!/usr/bin/env python  

import cgi
from google.appengine.ext import webapp
from google.appengine.api import mail
from google.appengine.api import urlfetch 

class MailJob(webapp.RequestHandler):
    def get(self):

        # Call your website using URL Fetch service ...
        url = "http://www.yoursite.com/page_or_service"
        result = urlfetch.fetch(url)

        if result.status_code == 200:
                doSomethingWithResult(result.content)

        # Send emails using Mail service ...
        mail.send_mail(sender="admin@gmail.com",
                        to="someone@gmail.com",
                        subject="Your account on YourSite.com has expired",
                        body="Bla bla bla ...")
        return

application = webapp.WSGIApplication([
        ('/tasks/summary', MailJob)], debug=True)

def main():
    wsgiref.handlers.CGIHandler().run(application)

if __name__ == '__main__':
    main()

Answer 2

看起来你正在阅读this page（即使你没有给我们提供网址）。所提供的配置和代码将无法成功运行：cron任务将尝试访问URL路径/任务/摘要，app.yaml将使其执行main.py，但后者仅为/ mailjob设置处理程序，因此cron任务的尝试将失败，并带有404状态代码。

main.py或app.yaml是否确定此示例中App Engine cron任务使用的URL？

2 个答案: