颤抖的webview恢复状态

Question

Hi I have one button in my home screen when clicking that button..it will open a new page with webview widget.(ex.webview widget initial URL is google,so it will open google.com).After that I am doing some interaction with that webpage,then Clicking back button will take me to home screen.If I click the button now,I want to open webview with resume interaction.(ex. I click on that button and it takes me google then I search for 'computer' in Google and it shows me the results.now clicking back button will take me home screen.again now I click on home screen button it will have to open webview with computer searched results).any idea?.......oh. And I found out this one method OnUrlChanged.listen()...but I don't know where to put it exactly??

Answer 1

Declaring variables

final flutterWebviewPlugin = FlutterWebviewPlugin();
String oldUrl;

...

@override
Widget build(BuildContext context) {

flutterWebviewPlugin.onUrlChanged.listen((String url) {
oldUrl = url;} //You might have to use async here
.....

...

In Your WebView Class

return SafeArea(
          child: WebviewScaffold(
            url: oldURl==null? "www.google.com" : oldUrl ))

I assume you'll have a MaterialPageRoute to navigate to Webview from your button. You can create the class with a parameter for url, and pass oldUrl or "google.com" as a parameter in it accordingly.