Getting URL as a parameter in Flask - but Flask stripping out the params from the URL

Question

I am passing in a URL as a parameter

GET http://127.0.0.1:5000/?url=https://url.com/?param1=123&param2=456

When I grab the URL parameter with request.args.get('url') the value returned is https://url.com/?param1=123 so its stripping off the second parameter in the URL

How can I get all parameters?

If it's possible, I think you'd want to URIencode that url param. Is that an option? — RichS
– RichS, Commented Apr 30, 2019 at 3:16

QIFENG LI · Accepted Answer · 2019-04-30 03:33:03Z

1

You should encode your url, like this

http://127.0.0.1:5000/?url=https%3A%2F%2Furl.com%2F%3Fparam1%3D123%26param2%3D456

use python3

from urllib.parse import urlencode

get_url = '?'.join(['http://127.0.0.1:5000/', urlencode({'url': 'https://url.com/?param1=123&param2=456'})])

edited Apr 30, 2019 at 3:33

answered Apr 30, 2019 at 3:27

QIFENG LI

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