android webview 打開谷歌地圖應用程序

Question

我有一個 android 網絡視圖，它可以正確顯示網站，但是當我按下帶有地理數據的鏈接時，我收到以下錯誤：“網站不可用，網站地理：48.9596430,9.125670?q=48.959630,9.125670 無法加載”

net::ERR_UNKNOWN_URL_SCHEME

當我在 chrome 瀏覽器上打開同一個網站時，它會用所需的坐標打開我的谷歌地圖。

這是網絡視圖代碼：

public class MainActivity extends Activity {

    /**
     * WebViewClient subclass loads all hyperlinks in the existing WebView
     */
    public class GeoWebViewClient extends WebViewClient {
        @Override
        public boolean shouldOverrideUrlLoading(WebView view, String url) {
            // When user clicks a hyperlink, load in the existing WebView
            view.loadUrl(url);
            return true;
        }
    }

    /**
     * WebChromeClient subclass handles UI-related calls
     * Note: think chrome as in decoration, not the Chrome browser
     */
    public class GeoWebChromeClient extends WebChromeClient {
        @Override
        public void onGeolocationPermissionsShowPrompt(String origin,
                                                       GeolocationPermissions.Callback callback) {
            // Always grant permission since the app itself requires location
            // permission and the user has therefore already granted it
            callback.invoke(origin, true, false);
        }
    }

    WebView mWebView;

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        mWebView = (WebView) findViewById(R.id.webView);
        // Brower niceties -- pinch / zoom, follow links in place
        mWebView.getSettings().setJavaScriptCanOpenWindowsAutomatically(true);
        mWebView.getSettings().setBuiltInZoomControls(true);
        mWebView.setWebViewClient(new GeoWebViewClient());
        // Below required for geolocation
        mWebView.getSettings().setJavaScriptEnabled(true);
        mWebView.getSettings().setGeolocationEnabled(true);
        mWebView.setWebChromeClient(new GeoWebChromeClient());

        mWebView.loadUrl("http://www.bda-bawue.de");
    }



    @Override
    public void onBackPressed() {
        // Pop the browser back stack or exit the activity
        if (mWebView.canGoBack()) {
            mWebView.goBack();
        }
        else {
            super.onBackPressed();
        }
    }

應該打開這個鏈接：

<a href="geo:48.959630, 9.125670?q=48.959630, 9.125670">

權限：

<uses-permission android:name="android.permission.INTERNET"/>
<uses-permission android:name="android.permission.ACCESS_GPS" />
<uses-permission android:name="android.permission.ACCESS_ASSISTED_GPS" />
<uses-permission android:name="android.permission.ACCESS_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_COARSE_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />

我是否必須更改 android 網絡視圖的鏈接或有什么意義？

Answer 1

可能是這個幫助。

@Override
        public boolean shouldOverrideUrlLoading(WebView view, String url) {
            // When user clicks a hyperlink, load in the existing WebView
            if(url.contains("geo:")) {
                Uri gmmIntentUri = Uri.parse(url);
                Intent mapIntent = new Intent(Intent.ACTION_VIEW, gmmIntentUri);
                mapIntent.setPackage("com.google.android.apps.maps");
                if (mapIntent.resolveActivity(getPackageManager()) != null) {
                      startActivity(mapIntent);
                }
            return true;
        }  
            view.loadUrl(url);
            return true;
        }

Answer 2

if(url.contains("https://www.google.com/maps/"))
{
  Uri IntentUri = Uri.parse(url);
  Intent mapIntent = new Intent(Intent.ACTION_VIEW, IntentUri);
  mapIntent.setPackage("com.google.android.apps.maps");

  if (mapIntent.resolveActivity(getPackageManager()) != null) {
     startActivity(mapIntent);
  }
  return true;
 }

Answer 3

如果您不希望用戶只使用 google 地圖，您可以嘗試下面的代碼。

public boolean shouldOverrideUrlLoading(WebView view, String url) {
    if(url.contains("geo:")){
        view.getContext().startActivity(new Intent(Intent.ACTION_VIEW,Uri.parse(url)));
        return true;
            else
                {
                view.loadUrl(url);
                return true;
                }

    }
}