Android：让Link始终保持在程序的WebView中跳转

1 WebView webView = (WebView) findViewById(R.id.webView1);
2 webView.setWebViewClient(new WebViewClient());

 1 private class MyWebViewClient extends WebViewClient {
 2     @Override
 3     public boolean shouldOverrideUrlLoading(WebView view, String url) {
 4         if (Uri.parse(url).getHost().equals("192.168.3.95")) {
 5             // This is my web site, so do not override; let my WebView load the page
 6             return false;
 7         }
 8         // Otherwise, the link is not for a page on my site, so launch another Activity that handles URLs
 9         Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
10         startActivity(intent);
11         return true;
12     }
13 }