在Android Webview中从浏览器打开链接

我想在带有Webview的Android应用程序中打开主机为example.com或example.de的浏览器中的链接.
我创建了这个意图:

I would like to open links in browsers with the host example.com or example.de in my Android application with webview.
I created this intent:

<intent-filter>
    <data android:scheme="https" android:host="example.com" />
    <data android:scheme="https" android:host="example.de" />
    <data android:scheme="http" android:host="example.com" />
    <data android:scheme="http" android:host="example.de" />

    <action android:name="android.intent.action.VIEW" />
    <category android:name="android.intent.category.BROWSABLE" />
    <category android:name="android.intent.category.DEFAULT" />
</intent-filter>

默认情况下，WebView加载URL example.com，这是我的onCreate:

By default, the WebView loads the URL example.com, here is my onCreate:

@Override
protected void onCreate(Bundle savedInstanceState) {

    super.onCreate(savedInstanceState);

    this.requestWindowFeature(Window.FEATURE_NO_TITLE);

    setContentView(R.layout.activity_main);

    final ProgressDialog pd = ProgressDialog.show(this, "", "Loading...", true);

    mWebView = (WebView) findViewById(R.id.activity_main_webview);

    // Enable Javascript
    WebSettings webSettings = mWebView.getSettings();
    webSettings.setJavaScriptEnabled(true);

    // Stop local links and redirects from opening in browser instead of WebView
    mWebView.setWebViewClient(new MyAppWebViewClient());

    mWebView.setWebViewClient(new WebViewClient() {

        @Override
        public boolean shouldOverrideUrlLoading(WebView view, String url) {
            pd.show();
            view.loadUrl(url);
            return true;
        }

        @Override
        public void onPageFinished(WebView view, String url) {
            if (pd.isShowing() && pd != null) {
                pd.dismiss();
            }
        }
    });

    mWebView.loadUrl(url);
}

这是我的shouldOverrideUrlLoading():

This is my shouldOverrideUrlLoading():

    @Override
    public boolean shouldOverrideUrlLoading(WebView view, String url) {

        if(Uri.parse(url).getHost().endsWith("example.com") || Uri.parse(url).getHost().endsWith("example.de") ) {
            return false;
        }

        Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
        view.getContext().startActivity(intent);
        return true;
    }

现在，我被卡住了.如果您在网络浏览器中单击URL example.com/otherurl.php，则该应用程序将打开，但会加载默认的URL example.com.如何打开应用程序并加载example.com/otherurl.php而不是example.com?

我已经阅读了此处，我需要这段代码来获取网址:

Now I am stuck. If you click on the URL example.com/otherurl.php in a web browser, the app opens, but loads the default url example.com. How can I open the app and load example.com/otherurl.php instead of example.com?

I already read here, that I need this piece of code to get the url:

Uri data = getIntent().getData();
String extUrl = data.toString();

但是我应该在哪里实现此代码? 谢谢

But where should I implement this code? Thanks