flutter - How do I get the the query parameters of the url of the web-app?

Question

I am planning to create an online game that uses some kind of lobby system. I want the host to be able to create an invite link http://randomgame.io/join?code=1234abcd.

How do I get the code parameter of the URL so that I can show content based on this code?

Any help is much appreciated :D

Edit:

I can get the URL by using Uri.base but the problem now is that I can't enter any query parameters into the url because they get instantly removed.

question from:https://stackoverflow.com/questions/65869086/how-do-i-get-the-the-query-parameters-of-the-url-of-the-web-app

Answer 1

With this package you can get current link from users browser, using function getHref(). So the final code could look like this:

import 'package:window_location_href/window_location_href.dart';

final String url = getHref();
Uri uri = new Uri.dataFromString(url);
String codeParam = uri.queryParameters['code'];