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我为我的 iPhone 应用程序制作了自己的推送通知服务器。我正在向客户端设备发送不同的推送通知。现在有了一个特殊的通知，我想在 appDelegate 或任何地方调用一个特定的函数。

我该如何实施？

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2 回答 2

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您不能指定要直接调用的函数

当应用程序启动时，-(BOOL)application:(UIApplication *)app didFinishLaunchingWithOptions:(NSDictionary *)launchOptions被调用并通过选项下的注释UIApplicationLaunchOptionsRemoteNotificationKey

when the app is running, you have application:didReceiveRemoteNotification:

you could pass the NAME of a method to call with the notification! so:

...
NSString *methodName = [notificationUserInfo objectForKey:@"methodName"];
[self performSelector:NSSelectorFromString(methodName)];
...

the server side JSON would contain the methodName key: as seen here we can Include all we like APNS JSON PAYLOAD - more arguments

{"aps":{"alert":"APP_NAME': BLA_BLA_BLA","sound":"default"}, "methodName":"xy"}
于 2013-01-29T17:16:24.597 回答
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当用户通过通知启动应用程序时，它可能有不同的场景：

它没有启动，然后应用程序以默认方式启动，您可以通过以下方式处理通知：

-(BOOL)application:(UIApplication *)app didFinishLaunchingWithOptions:(NSDictionary *)launchOptions {
    UILocalNotification *remoteNotif =
        [launchOptions objectForKey:UIApplicationLaunchOptionsRemoteNotificationKey];
    if (remoteNotif) {
        //handle remote notification
    }
    ....
}

如果应用程序在后台或前台，则调用委托方法

- application:didReceiveRemoteNotification: {
    if (application.applicationState == UIApplicationStateActive){
        //application was in foreground
    } else if (application.applicationState == UIApplicationStateInactive){
        //application was in background
    }
}

此外，如果应用程序在前台 - 系统不显示警报，不要更改徽章图标或播放声音 - 您应该完全自己处理通知

于 2013-01-29T17:01:43.060 回答