How to pass URL with parameters in GET request in Flask?

Question

In Flask, I want to send an URL in the URL as a parameter with all its parameters

Here is a working example http://graph.facebook.com/https://www.facebook.com/photo.php?fbid=717097101644637&set=a.456449604376056.98921.367116489976035&type=1&theater

I couldn't find a way, because the moment I cal something like this, the url doesn't go true.

@app.route('/<url>', methods=['GET']) def show(url): """ do something with it """

Any idea how to do it?

Thank you

try to use this : @app.route('/<path:url>', methods=['GET']) — Odai Al-Ghamdi
– Odai Al-Ghamdi, Commented Dec 6, 2013 at 7:45
That works to the point that it will remove all the parameters from the URL. — acheruns
– acheruns, Commented Dec 6, 2013 at 7:59
are you trying to print the url on the page ? or you're trying to deal with its data "json" ? — Odai Al-Ghamdi
– Odai Al-Ghamdi, Commented Dec 6, 2013 at 8:38
Trying to use the url, I don't care about the json. I want to achieve the same functionality, not get the same data. — acheruns
– acheruns, Commented Dec 6, 2013 at 8:58
The best I could come up to this moment is url = request.url.replace(request.url_root,'') for the URL and @app.route('/<path:url>', methods=['GET']) for the app.route. — acheruns
– acheruns, Commented Dec 6, 2013 at 15:40

Sean Vieira · Accepted Answer · 2013-12-06 16:29:42Z

If you look at what Facebook returns for that URL:

{ "id": "https://www.facebook.com/photo.php", "shares": 377224, "comments": 2 }

you will note that the query string params are being consumed by graph.facebook.com - they are not treated as part of the URL that is being submitted as a search param. If you want to be able to get the whole URL including query string arguments, you will need to URL-encode the query string your.url/https://www.facebook.com/photo.php%3Ffbid=717097101644637%26set=a.456449604376056.98921.367116489976035%26type=1%26theater

I gave Facebook as the closest example. I know they don't do exactly what I wanted. Unfortunately I can't URL encode the whole string, because if you will paste the link into your browsers (at least in Safari) it will do exactly what happened there.

acheruns · Accepted Answer · 2013-12-07 20:07:03Z

Unfortunately, I can't control if the URL is URL-encoded so the best thing I came along is this.

_url = request.url.replace(request.url_root,'') url = urllib.unquote(_url)

Maybe it will help somebody.

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