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Home/ Questions/Q 642467
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T21:10:29+00:00 2026-05-13T21:10:29+00:00

Here is a portion of my app.yaml file: handlers: – url: /remote_api script: $PYTHON_LIB/google/appengine/ext/remote_api/handler.py

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Here is a portion of my app.yaml file:

handlers:
- url: /remote_api
  script: $PYTHON_LIB/google/appengine/ext/remote_api/handler.py
  login: admin
- url: /detail/(\d)+
  script: Detail.py
- url: /.*
  script: Index.py

I want that capture group (the one signified by (\d)) to be available to the script Detail.py. How can I do this?

Do I need to figure out a way to access GET data from Detail.py?

Also, when I navigate to a URL like /fff, which should match the Index.py handler, I just get a blank response.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T21:10:30+00:00Added an answer on May 13, 2026 at 9:10 pm

    I see two questions, how to pass elements of the url path as variables in the handler, and how to get the catch-all to render properly.

    Both of these have more to do with the main() method in the handler than the app.yaml

    1) to pass the id in the /detail/(\d) url, you want something like this:

    class DetailHandler(webapp.RequestHandler):
        def get(self, detail_id):
          # put your code here, detail_id contains the passed variable
    
    def main():
      # Note the wildcard placeholder in the url matcher
      application = webapp.WSGIApplication([('/details/(.*)', DetailHandler)]
      wsgiref.handlers.CGIHandler().run(application)
    

    2) to ensure your Index.py catches everything, you want something like this:

    class IndexHandler(webapp.RequestHandler):
        def get(self):
          # put your handler code here
    
    def main():
      # Note the wildcard without parens
      application = webapp.WSGIApplication([('/.*', IndexHandler)]
      wsgiref.handlers.CGIHandler().run(application)
    

    Hope that helps.

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