iOS拿到推送从而跳转到指定页面

最新推荐文章于 2025-02-18 16:45:38 发布

转载最新推荐文章于 2025-02-18 16:45:38 发布 · 1.5k 阅读

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本文介绍了一个iOS应用如何处理从后台或杀死状态返回时接收到的远程推送通知，并展示了具体的实现代码，包括清除应用角标数字及如何根据推送信息展示特定视图控制器。

//这是程序杀死后再通过点击通知进入时调用的方法

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

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<br>   
if

(launchOptions)

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 {

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<br>       
 [[UIApplication sharedApplication] setApplicationIconBadgeNumber:0];

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<br>       
 NSDictionary *userInfo = [launchOptions objectForKey: UIApplicationLaunchOptionsRemoteNotificationKey];

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 [self presentViewControllerWithUserInfo:userInfo];

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 }

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<br>}

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<br>//App状态为正在前台或者后台运行，那么此函数将被调用

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<br>-
 (void)application:(UIApplication
 *)application didReceiveRemoteNotification:(NSDictionary *)userInfo

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<br>{

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<br>   

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<br>   
if

(application.applicationState != UIApplicationStateActive)

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<br>   
 {

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 [self presentViewControllerWithUserInfo:userInfo];

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 }

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<br>}

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<br>#pragma
 mark - 消息跳转页面

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<br>-
 (void)presentViewControllerWithUserInfo:(NSDictionary
 *)userInfo

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<br>{

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<br>      
//拿出你的消息内容，跳转即可~~

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<br>[self.window.rootViewController
 presentViewController:fyfNavCtrl animated:YES completion:^{

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<br>                   
//

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 }];

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<br>}