GAE: Error 404 when creating a new script

综合编程 2018-03-21 阅读原文

I'm using Google App Engine with Python environment.

I have my main code in the file. I want to create a new .py file for a different page. I created the .py file, added the path to the yaml file. But I still get a '404 Error, resource not found'.

Here is my yaml file

application: myapp
version: 1
runtime: python27
api_version: 1
threadsafe: yes

- url: .*

- url: /hello

- name: webapp2
  version: "2.5.2"

When the user goes to I want the file to be executed.

Here's the current content of

import webapp2

class HeyPage(webapp2.RequestHandler):
  def get(self):
      self.response.headers['Content-Type'] = 'text/html'
      self.response.out.write('Hello, All!')

app = webapp2.WSGIApplication([('/hello', HeyPage)],

Here is the log:

INFO     2014-01-10 06:15:31,150] default: "GET /hello HTTP/1.1" 404 154

You should list your handlers from most specific to least specific. Your handler:

- url: .*

basically says that
should handle every url. Since it is the first in the list,
will try to handle every request regardless of the handlers that follow it in app.yaml
. Change it to:

- url: /hello

- url: .*

And all should work.

Hello, buddy!

责编内容by:Hello, buddy!阅读原文】。感谢您的支持!


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