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Home/ Questions/Q 700983
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T03:33:32+00:00 2026-05-14T03:33:32+00:00

I want to build a REST web service on app engine. Currently i have

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I want to build a REST web service on app engine. Currently i have this:

from google.appengine.ext import webapp
from google.appengine.ext.webapp import util

class UsersHandler(webapp.RequestHandler):  

def get(self, name):
    self.response.out.write('Hello '+ name+'!') 

def main():
util.run_wsgi_app(application)

#Map url like /rest/users/johnsmith
application = webapp.WSGIApplication([(r'/rest/users/(.*)',UsersHandler)]                                      
                                   debug=True)
if __name__ == '__main__':
    main()

And i would like to retreive for example all my users when the path /rest/users is accessed. I Imagine I can do this by building another handler, but I want to know if is possible to do it inside of this handler.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T03:33:32+00:00Added an answer on May 14, 2026 at 3:33 am

    Sure, you can — change your handler’s get method to

    def get(self, name=None):
        if name is None:
            """deal with the /rest/users case"""
        else:
            # deal with the /rest/users/(.*) case
            self.response.out.write('Hello '+ name+'!') 
    

    and your application to

    application = webapp.WSGIApplication([(r'/rest/users/(.*)', UsersHandler),
                                          (r'/rest/users', UsersHandler)]                                      
                                         debug=True)
    

    In other words, map your handler to all the URL patterns you want it to handle, and make sure the handler’s get method can distinguish among them easily (typically via its arguments).

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