ios：通过点击通知中心的消息启动检测应用程序

Question

有没有办法通过点击通知中心的消息来了解应用是否已启动？

我想通过点击通知中心的消息启动应用程序时，只对服务器进行一些调用。

Answer 1

在应用程序委托的- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions方法中，您将收到launchOptions字典中的通知信息。这样你就可以知道应用程序是从通知托盘启动的。

Answer 2

是的，您可以在

中找到应用程序启动原因

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{

// keys can be UIApplicationLaunchOptionsLocalNotificationKey

NSDictionary *notificationPayload = launchOptions[UIApplicationLaunchOptionsRemoteNotificationKey];
if(notificationPayload)
{
    // application launch because of notification
    // do some stuff here
}

return YES;

}

Answer 3

您可以处理推送通知，例如

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
    NSDictionary *pushNotification = [launchOptions objectForKey:UIApplicationLaunchOptionsRemoteNotificationKey];
    if (pushNotification) {
        //Application did started by clicking push notification. Do whatever you want to do
    }

  ....//Your rest code
  ....
}

有些时候应用程序处于活动状态，我们仍然希望处理推送通知，而不是以下方法将被称为

- (void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo
{
    //Application did receive push notification. Do whatever you want to do
}