Question

使用Phonegap Build为Android设备使用推送通知请求，在注册请求时，看看发生了什么！

    var pushNotification=window.plugins.pushNotification;

    var return = pushNotification.register(successHandler , errorHandler,{"senderID":GCMProject,"ecb":"onNotification"});


function successHandler (result) {
    alert('result = ' + result); // returned 'result = ok' //
}

function errorHandler (error) {
    alert('error = ' + error);
}


function onNotification(e) {
    $("#app-status-ul").append('<li>EVENT -> RECEIVED:' + e.event + '</li>');

    switch( e.event )
    {
    case 'registered':
        if ( e.regid.length > 0 )
        {
            $("#app-status-ul").append('<li>REGISTERED -> REGID:' + e.regid + "</li>");
            // Your GCM push server needs to know the regID before it can push to this device
            // here is where you might want to send it the regID for later use.
            console.log("regID = " + e.regid);
        }
    break;

    case 'message':
        // if this flag is set, this notification happened while we were in the foreground.
        // you might want to play a sound to get the user's attention, throw up a dialog, etc.
        if ( e.foreground )
        {
            $("#app-status-ul").append('<li>--INLINE NOTIFICATION--' + '</li>');

            // on Android soundname is outside the payload. 
            // On Amazon FireOS all custom attributes are contained within payload
            var soundfile = e.soundname || e.payload.sound;
            // if the notification contains a soundname, play it.
            var my_media = new Media("/android_asset/www/"+ soundfile);
            my_media.play();
        }
        else
        {  // otherwise we were launched because the user touched a notification in the notification tray.
            if ( e.coldstart )
            {
                $("#app-status-ul").append('<li>--COLDSTART NOTIFICATION--' + '</li>');
            }
            else
            {
                $("#app-status-ul").append('<li>--BACKGROUND NOTIFICATION--' + '</li>');
            }
        }

       $("#app-status-ul").append('<li>MESSAGE -> MSG: ' + e.payload.message + '</li>');
           //Only works for GCM
       $("#app-status-ul").append('<li>MESSAGE -> MSGCNT: ' + e.payload.msgcnt + '</li>');

    break;

    case 'error':
        $("#app-status-ul").append('<li>ERROR -> MSG:' + e.msg + '</li>');
    break;

    default:
        alert('EVENT -> Unknown, an event was received and we do not know what it is');
    break;
  }
}

使用（'ok'）消息结果触发successHandler()！

为什么the onNotification()方法不会在任何时候被触发???

该应用程序尚未部署在Play商店中。它在本地使用，但连接到互联网。

感谢您的建议。

Answer 1

问题在于JS库，当我们将代码直接放入index.js时，它可以正常工作。因此问题包含在PushNotification.js脚本中的对象定义中。

GCM推送通知注册调用返回正常但设备未注册

1 个答案: