3

我有一个简单 webview 的代码,并且在运行代码时有效,但在 logcat 中显示错误“捕获安全异常”。这是什么意思?

WebActivity.java

public class WebActivity extends Activity {
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_web);
        WebView mainWebView = (WebView) findViewById(R.id.mainWebView);
        WebSettings webSettings = mainWebView.getSettings();
        webSettings.setJavaScriptEnabled(true);
        mainWebView.setWebViewClient(new MyCustomWebViewClient());
        mainWebView.loadUrl("http://google.com");
        mainWebView.setScrollBarStyle(View.SCROLLBARS_INSIDE_OVERLAY);
    }
    private class MyCustomWebViewClient extends WebViewClient {
        @Override
        public boolean shouldOverrideUrlLoading(WebView view, String url) {
            view.loadUrl(url);
            return true;
            }
        }
    }

日志猫

08-22 11:37:01.569: E/geolocationService(537): Caught security exception registering for location updates from system. This should only happen in DumpRenderTree.
4

2 回答 2

3

您需要在清单中添加它。

<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_COARSE_LOCATION" />
于 2013-08-22T06:35:54.707 回答
0

此链接具有密钥 http://cordova.apache.org/docs/en/2.5.0/cordova_device_device.md.html

我正在使用 Android 开发 - Phonegap

我添加了

--> app/res/xml/config.xml

<plugin name="Device" value="org.apache.cordova.Device" />

--> 应用程序/AndroidManifest.xml

< 使用权限 android:name="android.permission.READ_PHONE_STATE" />

并且工作正常

PS:也许这与Android 2.0+上的WebView中使用navigator.geolocation.getCurrentPosition的答案相同 (PhoneGap相关)

于 2013-12-18T18:38:23.670 回答