Python cannot get url params

Question

I want to get the param of a url that is sent back to the server when the user submits the form. However I get this error AttributeError: 'HTTPServerRequest' object has no attribute 'get'

My url is http://127.0.0.1:8000/reset?key=0OeKkQcSRXiy6yAvtgd9GGv4DhO1t0EYuybjHG5Buzo=

The url string is created as:

keyVal=base64.b64encode(uuid.uuid4().bytes + uuid.uuid4().bytes)
emailLink = 'http://127.0.0.1:8000/reset?key=%s'%keyVal

The emailLink is a used as a link.

I tried to retrieving the value of the param key in the above url as:

class ResetPwdHandler(tornado.web.RequestHandler):       
    def get(self):
        self.render("reset.html")

    def post(self):
        value = self.request.get('key')
        print value

I receive the other form data fine, using self.get_argument(), if I try to use it here then it says 'key' is missing argument.

I see examples that use urlparse but how could you use this to get the param of the url when the form is submitted.

Maybe the context is obvious to other users, but can you show the full class definition? The error is that the 'HTTPServerRequest' object has no attribute 'get', but where is `request` set and where is its class defined? — Nicholas Flees, Nov 06 '14 at 19:54
Where is `self`'s `request` assigned? Where is `get` defined? — Nicholas Flees, Nov 06 '14 at 20:15

score 2 · Accepted Answer · answered Nov 06 '14 at 20:30

2

I've never used Tornado or its web framework, but the documentation is quite clear that the request class doesn't define a get method. You are perhaps confusing it with Flask or webapp2. It seems like you can either do self.request.arguments['key'] or self.get_arguments('key').

answered Nov 06 '14 at 20:30

Daniel Roseman

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Thanks; Yes I had to use `self.get_arguments('key')` as well, I had it incorrectly in `def post` not `def post`. – user94628 Nov 06 '14 at 21:10

Python cannot get url params

1 Answers1