Question

I have been trying to get the request url as follows

import webapp2

class MainPage(webapp2.RequestHandler):
    def get(self):
        print self.request.get('url')

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

when the request is

http://localhost:8080/index.html

it gives me something like

Status: 200 Content-Type: text/html; charset=utf-8 Cache-Control: no-cache Content-Length: 70

what I need to get is something like

index.html

edit: so that I can check the string and display the correct html/template file accordingly.

I have already checked Request documentation and tried many alternatives yet I can’t seem to find a solution. I’m quite new to web development. What am I missing?

Answer 1

you should start here: https://developers.google.com/appengine/docs/python/gettingstartedpython27/helloworld

you are not using a template or a templating module/engine so its irrelevant what path you are accessing because you are mapping everything with /.*

use self.response.write() not print.
again its better for you if you start reading the documentation from the beginning before checking out the request class.

EDIT:

if you want to get the urlpath in the requesthandler and serve different templates based on the urlpath then do something like this:

def get(self):
    if self.request.path == '/foo':
        # here i write something out
        # but you would serve a template
        self.response.write('urlpath is /foo')
    elif self.request.path == '/bar':
        self.response.write('urlpath is /bar')
    else:
        self.response.write('urlpath is %s' %self.request.path)

a better approach would be to have multiple ReqeustHandlers and map them in the WSGIApplication:

app = webapp2.WSGIApplication([('/', MainPage),
                               ('/foo', FooHandler),
                               ('/bar', BarHandler),
                               ('/.*', CatchEverythingElseHandler)], debug=True)