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我试图从 Holly Schinsky http://devgirl.org/2012/10/19/tutorial-apple-push-notifications-with-phonegap-part-1/上做这个教程

插件已安装，我正在尝试第一次推送

在我的 index.html 中，我从我的 src index.js 调用函数 app.initialize()

所以这是在我的 index.js 中：

var app = {

initialize: function() {
this.bindEvents();
},

bindEvents: function() {
document.addEventListener('deviceready', this.onDeviceReady, false);
},

onDeviceReady: function() {
app.receivedEvent('deviceready');
},
tokenHandler:function(msg) {
console.log("Token Handler " + msg);
},
errorHandler:function(error) {
console.log("Error Handler  " + error);
alert(error);
},

successHandler: function(result) {
alert('Success! Result = '+result)
},

receivedEvent: function(id) {
var pushNotification = window.plugins.pushNotification;
// TODO: Enter your own GCM Sender ID in the register call for Android
if (device.platform == 'android' || device.platform == 'Android') {
    pushNotification.register(this.successHandler, this.errorHandler,    {"senderID":"554205989074","ecb":"app.onNotificationGCM"});
}
else {
    pushNotification.register(this.tokenHandler,this.errorHandler,{"badge":"true","sound":"true","alert":"true","ecb":"app.onNotificationAPN"});
}
var parentElement = document.getElementById(id);
var listeningElement = parentElement.querySelector('.listening');
var receivedElement = parentElement.querySelector('.received');

listeningElement.setAttribute('style', 'display:none;');
receivedElement.setAttribute('style', 'display:block;');

console.log('Received Event: ' + id);
},
// iOS
onNotificationAPN: function(event) {
var pushNotification = window.plugins.pushNotification;
console.log("Received a notification! " + event.alert);
console.log("event sound " + event.sound);
console.log("event badge " + event.badge);
console.log("event " + event);
if (event.alert) {
    navigator.notification.alert(event.alert);
}
if (event.badge) {
    console.log("Set badge on  " + pushNotification);
    pushNotification.setApplicationIconBadgeNumber(this.successHandler, event.badge);
}
if (event.sound) {
    var snd = new Media(event.sound);
    snd.play();
}
},
// Android
onNotificationGCM: function(e) {
switch( e.event )
{
    case 'registered':
        if ( e.regid.length > 0 )
        {
            // 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.
            alert('registration id = '+e.regid);
        }
        break;

    case 'message':
        // this is the actual push notification. its format depends on the data model
        // of the intermediary push server which must also be reflected in GCMIntentService.java
        alert('message = '+e.message+' msgcnt = '+e.msgcnt);
        break;

    case 'error':
        alert('GCM error = '+e.msg);
        break;

    default:
        alert('An unknown GCM event has occurred');
        break;
    }
}

};

对不起，如果它的代码太多，但我不知道从哪里开始。我对此很陌生，但我会尝试描述发生了什么：

  • 我调用函数 app.initialize
  • 我在 XCode 中的输出给了我这个：[LOG] Received Event: deviceready
  • 所以因为我的事件是“设备就绪”，所以 app.recievedEvent 应该开始，对吧？
  • 并且因为我的设备是 IOs-设备 pushNotification.register 应该启动，对吗？

--> 但什么也没发生（除了我在 Xcode 中得到输出：[LOG] Received Event: deviceready）

我没有在我的设备上收到推送通知

这里是我的 php 文件中的代码：

$streamContext = stream_context_create();
stream_context_set_option($streamContext, 'ssl', 'local_cert', 'cert.pem');


$socketClient = stream_socket_client('ssl://gateway.sandbox.push.apple.com:2195', $error, $errorString, 60, STREAM_CLIENT_CONNECT, $streamContext);


$payload['aps'] = array('alert' => 'test message', 'sound' => 'default', 'badge' => '2');
$payload['id'] = 666;

$payload = json_encode($payload);


$deviceToken = str_replace(' ', '', $data['deviceToken']);


$message = pack('CnH*', 0, 32, $deviceToken);


$message = $message . pack('n', strlen($payload));

$message = $message . $payload;


fwrite($socketClient, $message);


fclose($socketClient);
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1 回答 1

1

要接收推送通知，您需要发送通知。您是否已经设置了服务器来执行此操作。在 DevGirl 的示例中，她使用 node.js 服务器发送推送通知。如果您关注她的帖子，您的手机应该设置正确以接收通知。

如果您需要推送服务，我有一个 Csharp 代码可以做到这一点

于 2013-05-02T10:17:51.323 回答