Capture URL with query string as parameter in Flask route

Question

Is there a way for Flask to accept a full URL as a URL parameter?

I am aware that <path:something> accepts paths with slashes. However I need to accept everything including a query string after ?, and path doesn't capture that.

http://example.com/someurl.com?andother?yetanother

I want to capture someurl.com?andother?yetanother. I don't know ahead of time what query args, if any, will be supplied. I'd like to avoid having to rebuild the query string from request.args.

Answer 1

the path pattern will let you capture more complicated route patterns like URLs:

@app.route('/catch/<path:foo>')
def catch(foo):
    print(foo)
    return foo

The data past the ? indicate it's a query parameter, so they won't be included in that patter. You can either access that part form request.query_string or build it back up from the request.args as mentioned in the comments.

Answer 2

Due to the way routing works, you will not be able to capture the query string as part of the path. Use <path:path> in the rule to capture arbitrary paths. Then access request.url to get the full URL that was accessed, including the query string. request.url always includes a ? even if there was no query string. It's valid, but you can strip that off if you don't want it.

@app.route("/<path:path>")
def index(path=None):
    return request.url.rstrip("?")

For example, accessing http://127.0.0.1:5000/hello?world would return http://127.0.0.1:5000/hello?world.