Question

我正在处理一个通知项目，我的要求是在通知上传递一个url链接，当用户点击通知时，它会导航到一些活动X并在活动X中的Web视图中打开该URL。 / p>

这就是我正在做的......

Context context = this.getApplicationContext();
notificationManager = (NotificationManager) context.getSystemService(Context.NOTIFICATION_SERVICE);
//  Intent mIntent = new Intent(this, MainActivity.class);
Intent notificationIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("http://www.google.com"));
Bundle bundle = new Bundle();
bundle.putString("test", "test");
notificationIntent.putExtras(bundle);
pendingIntent = PendingIntent.getActivity(context, 0, notificationIntent, PendingIntent.FLAG_UPDATE_CURRENT);

打开网址，但不在我的网页浏览中，而是要求选择手机中的浏览器..但我只想在我的网页浏览中打开该链接！

由于

Answer 1

通知服务

@Override
public void onMessageReceived(RemoteMessage remoteMessage) {
    String title = remoteMessage.getNotification().getTitle();
    String body = remoteMessage.getNotification().getBody();
    String intentUri = remoteMessage.getData().get("link");

    sendNotification(intentUri , title , body);
}


//This method is only generating push notification
//It is same as we did in earlier posts
private void sendNotification(String url, String title, String body) {
    Intent intent = new Intent(this, MainActivity.class);
    intent.addFlags(Intent.FLAG_ACTIVITY_SINGLE_TOP | Intent.FLAG_ACTIVITY_CLEAR_TOP);

    if (url != null) {
        intent.putExtra("URL", url);
    }
    intent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK | Intent.FLAG_ACTIVITY_CLEAR_TASK);

    PendingIntent pendingIntent = PendingIntent.getActivity(this, 0, intent,
            PendingIntent.FLAG_UPDATE_CURRENT);

    Uri defaultSoundUri = RingtoneManager.getDefaultUri(RingtoneManager.TYPE_NOTIFICATION);
    String channelId = "Default";
    NotificationCompat.Builder notificationBuilder = new NotificationCompat.Builder(this, channelId);
    notificationBuilder.setSmallIcon(R.mipmap.ic_launcher);
    notificationBuilder.setContentTitle(title);
    notificationBuilder.setContentText(body);
    notificationBuilder.setAutoCancel(true);
    notificationBuilder.setSound(defaultSoundUri);
    notificationBuilder.setContentIntent(pendingIntent);

    NotificationManager notificationManager =
            (NotificationManager) getSystemService(Context.NOTIFICATION_SERVICE);

    notificationManager.notify(0, notificationBuilder.build());
}

WEBVIEWACTIVITY

  Intent intent = this.getIntent();

    Bundle extras = intent.getExtras();
    if(extras != null){
        if(extras.containsKey("URL"))
        {
            Url = extras.getString("URL");
        }
    }
    if (Url != null) {
        webView.loadUrl(Url);
    } else {
        webView.loadUrl("https://www.example.com");
    }

Answer 2

无需在通知中传递网址。在通知中将其作为String传递，在接收端将该字符串转换为url。

例如：

Bundle bundle = new Bundle();
bundle.putString("url", "www.google.com");
notificationIntent.putExtras(bundle);

在receiverEnd上将其转换为

String url = getArguments.getString("url");
Url myUrl = new Url(url);

在webView中打开myUrl。

Answer 3

您必须使用webview创建新活动。

在您的网页浏览活动中使用此代码

public class WebViewActivity extends Activity {

    private WebView webView;
Bundle extras;
String loadurl;
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.webview);
extras=getIntent().getExtras();
if(extras!=null){
loadurl=extras.getString("url","your default web page");
}

        webView = (WebView) findViewById(R.id.webView1);
        webView.getSettings().setJavaScriptEnabled(true);
        webView.loadUrl(loadurl);
    }
}

如何在通知中传递URL链接，并在点击它时应该在webview中打开？

3 个答案: