How to get just the arguments of a request in Flask?

Question

I have found this post on stackoverflow :
How do I get the different parts of a Flask request's url?
We can get the different parts of the request like this :

With this URL :

  http://www.example.com/myapplication/page.html?x=y

We can get these values :

  path             /page.html
  script_root      /myapplication
  base_url         http://www.example.com/myapplication/page.html
  url              http://www.example.com/myapplication/page.html?x=y
  url_root         http://www.example.com/myapplication/

Is there a way to get the last part (the arguments) in a string ?
--> Get this : ?x=y

Thanks !

Does this answer your question? [How can I get the named parameters from a URL using Flask?](https://stackoverflow.com/questions/24892035/how-can-i-get-the-named-parameters-from-a-url-using-flask) — LimorAbv, Aug 08 '20 at 13:43

score 3 · Accepted Answer · answered Aug 08 '20 at 13:44

3

Use request.args

@app.route(...)
def some_method():
    args = request.args
    print(args)

answered Aug 08 '20 at 13:44

LimorAbv

56
5

Thanks but I want to store all arguments in a string like this : `?x=y` – PacCol Aug 08 '20 at 13:46
@PacCol why? What are you trying to do? – roganjosh Aug 08 '20 at 13:47
I want to edit the url_root. – PacCol Aug 08 '20 at 13:48
@PacCol that still doesn't explain much to me. I don't have enough context and it seems like you have an [XY Problem](https://en.wikipedia.org/wiki/XY_problem) – roganjosh Aug 08 '20 at 13:52

How to get just the arguments of a request in Flask?

1 Answers1