How to open application and go to given url in WebView?

Question

JavaScript

I can't figure out how to use the link to launch the application on my phone and go to the desired URL.
I downloaded the plugin: https://www.npmjs.com/package/cordova-plugin-custo...

The app opens from the link but doesn't go to the WebView. That is, it just opened and that's it. And you need to go to the specified link.

Now I use links of this kind: MYAPP.https://site.ru/
first I put the application identifier MYAPP, and then I write the link through a dot. The identifier turns out to work and opens the application, but does not follow the link. How to write correctly to follow the link?

Answer 1

Understood. It was necessary to process links, cutting off unnecessary from them.
This code helped:

handleOpenURL = OpenPush;
          function OpenPush (url) {
          var strValue = url;
          strValue = strValue.replace('auspapp://','');
          window.location = strValue;
          }

auspapp://This is what should have been cut off.