Question

我无法接收任何类型的用于phonegap构建的push notifications plugin的回调，我已将该插件包含在config.xml中。

我已注册GCM并获取了pushNotification.register（）所需的项目编号。

我也可以访问window.plugins.pushNotification对象，所以我知道它包含在插件中。

PhoneGap Build版本： 3.1
保湿：已停用
调试：已启用
设备：Samsung Tab 2

我的 index.html js文件包括：

<script type="text/javascript" src="phonegap.js"></script>
<script type="text/javascript" src="PushNotification.js"></script>
<script type="text/javascript" src="js/lib/jquery.js" ></script>
<script type="text/javascript" src="js/lib/handlebars.js"></script>
<script type="text/javascript" src="js/handlebars/helpers.js"></script>
<script type="text/javascript" src="js/plugins/fastclick.js"></script>
<script type="text/javascript" src="js/app.js"></script>

我的 config.xml 插件包括：

// plugins
<gap:plugin name="org.apache.cordova.console" />
<gap:plugin name="org.apache.cordova.device" />
<gap:plugin name="org.apache.cordova.geolocation" />
<gap:plugin name="org.apache.cordova.dialogs" />
<gap:plugin name="org.apache.cordova.inappbrowser" />
<gap:plugin name="org.apache.cordova.splashscreen" />
<gap:plugin name="com.phonegap.plugins.pushplugin" />
// access to external domains
<access origin="*"/>

我的app.js调用了pushNotification.register（）

var app = {
init: function() {
    document.addEventListener("deviceready", this.onDeviceReady, false);
    },

    onDeviceReady: function(){
       // DO STUFF
       // ....

       // ENABLE PUSH
       this.push_init();
    },

    push_init: function(){
        app.SENDER_ID = 123456789; // replaced by my actual GCM project no

        var pushNotification = window.plugins.pushNotification;

        pushNotification.register( 
            function(){alert('Push: win');}, // never called
            function(){alert('Push: Error');},  // never called
            { senderID: app.SENDER_ID, ecb: "app.push_android" }
        );
    },
   // never called
   push_android: function(e){
       alert('connection established...');
   console.log( 'successfully started android' );
   console.log( e );
   }

};

// start the app
app.init();

在调用之后没有执行任何操作，app.push_android（）是app对象的一个功能。

如果我没有输入senderID，我会收到错误，说明没有发件人ID，所以我知道某些内容正在运行。任何想法都令人沮丧吗？

PS - 我也注意到了一些奇怪的东西，当我在console.log中时，window.plugins.pushNotification它返回一个空对象，但是我仍然可以调用window.plugins.pushNotification.register（），但我想我将在console.log中显示。

Answer 1

我想我找到了解决方案。

我在对象

中传递属性senderID的整数而不是字符串

无效

pushNotification.register( 
    function(){alert('Push: win');}, // NOT called
    function(){alert('Push: Error');},  // NOT called
    { senderID: 123456789, ecb: "app.push_android" }
);

工作

pushNotification.register( 
    function(){alert('Push: win');}, // called
    function(){alert('Push: Error');},  // called
    { senderID: "123456789", ecb: "app.push_android" }
);

Answer 2

尝试此推送通知代码 -

var pushNotification;

document.addEventListener('deviceready', onDeviceReady, true);

function onDeviceReady() {

   try {
       pushNotification = window.plugins.pushNotification;
       if (device.platform == 'android' || device.platform == 'Android') {
           pushNotification.register(successHandler, errorHandler, { "senderID": "123456789", "ecb": "onNotificationGCM" });    // required!
          }
      else {
          pushNotification.register(tokenHandler, errorHandler, { "badge": "true", "sound": "true", "alert": "true", "ecb": "onNotificationAPN" }); // required!
          }
      }  
    catch (err) {
        txt = "There was an error on this page.\n\n";
        txt += "Error description: " + err.message + "\n\n";
        alert(txt);
     }
   };

// handle GCM notifications for Android
function onNotificationGCM(e) {
    switch (e.event) {
        case 'registered':
            if (e.regid.length > 0) {
                alert(e.regid);
                storeToken(e.regid);
            }
            break;

        case 'message':
            if (e.foreground) {
                var my_media = new Media("beep.wav");
                my_media.play();
            }
            else {  
              // otherwise we were launched because the user touched a notification in the notification tray.
            }

            break;

       case 'error':
           break;
       default:
          break;
    }
}

参考Link

Refer Devgirl's Weblog

PhoneGap构建推送通知（Android）

2 个答案: