由于没有互联网连接而无法打开网址时的android事件

Question

I am properly checking if device has internet connection or not but for any other reasons is not suitable for me. I need to capture event when a webview tries to open a webpage but this one could not be opened due to internet conection. Now, webview displays standard browser page "www.mypage.com could not be opened...". Which event can be handled for this? Thank you

Answer 1

尝试使用附加到WebView的WebViewClient拦截事件。

Answer 2

I generally check what response the req sent to internet connection gives and if it fails I throw a customised Exception .And in my catch block I handle it the way I want like give customised message, alert or Toast .

The conditions I check for which url can't be opened:

1. Server is not reachable(Eg: response code other than 200)
2. No internet connection
3. Connection Timed out
4. Malformed URL
5. Unsupported Encoding or content type
6. Any other exception

Each of them I handle differently using my own exception classes.

Answer 3

You should be getting connection timeout response of sort. Normally, webview would display default result but you can create your own handler to manage it.