YAML file url and script in GAE python

Question

I\'m using Python 2.7 in Google App Engine and can\'t seem to get my app.yaml file set up right.

My goal is so that if I go to http://localhost/carlos/ I ge

Answer 1

I was able to determine the solution and figured I'd post it for anyone out there.

In my carlos.py file I needed to replace:

app = webapp2.WSGIApplication([('/', MainHandler)],
                              debug=True)

with

app = webapp2.WSGIApplication([('/carlos/', MainHandler)],
                              debug=True)

It appears that the first argument for the WSGIApplication is referring to the TOTAL path from your root web address as opposed to the INCREMENTAL path from which it was originally directed.

I'm selecting this answer over what was provided by Littm because I'd like to keep using WSGI

Answer 2

It worked with the following modifications:

1 - Replacing "carlos.app" by "carlos.py" and "main.app" by "main.py" in the yaml file.

2 - Add a slash ("/") after " /carlos " in the file "carlos.py".

3 - Add the following part of code in the end of each python files (carlos.py and main.py)

def main():
    app.run()

---

Here is an example of modified files:

app.yaml :

application: myapp
version: 1
runtime: python27
api_version: 1
threadsafe: no

handlers:
- url: /carlos/.*
  script: carlos.py

- url: .*
  script: main.py

carlos.py : import webapp2

class MainHandler(webapp2.RequestHandler):
  def get(self):
    self.response.out.write("Hello, Carlos!")

app = webapp2.WSGIApplication([('/carlos/', MainHandler)],
                          debug=True)

def main():
    app.run()

main.py :

import webapp2

class MainHandler(webapp2.RequestHandler):
  def get(self):
    self.response.out.write("Hello, MAIN!")

app = webapp2.WSGIApplication([('/', MainHandler)],
                          debug=True)

def main():
    app.run()

You can try navigating to:

localhost:8080/carlos/ and localhost:8080/ to see the results

Hope it helps ;)