Iphone 推送通知警报查看操作?
我有一个iphone应用程序,其中我正在以alertview的形式执行推送通知。当我的应用程序处于后台状态时,推送通知就会出现,当我点击它或解锁手机时,它将直接进入应用程序,我将其保持在forground状态。我使用点击查看按钮在警报中添加操作,它将进入另一个视图控制器。我不想在点击通知时进入应用程序。我需要显示alertview和点击查看按钮,我需要执行操作。有人能帮我实现吗?这是我的代码片段`-Iphone 推送通知警报查看操作?,iphone,ios,ipad,apple-push-notifications,Iphone,Ios,Ipad,Apple Push Notifications,我有一个iphone应用程序,其中我正在以alertview的形式执行推送通知。当我的应用程序处于后台状态时,推送通知就会出现,当我点击它或解锁手机时,它将直接进入应用程序,我将其保持在forground状态。我使用点击查看按钮在警报中添加操作,它将进入另一个视图控制器。我不想在点击通知时进入应用程序。我需要显示alertview和点击查看按钮,我需要执行操作。有人能帮我实现吗?这是我的代码片段`- - (void)application:(UIApplication *)application
- (void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo
{
//check application in forground or background
if(application.applicationState == UIApplicationStateActive)
{
//NSLog(@"FOreGround");
//NSLog(@"and Showing %@",userInfo)
}
else
{
NSDictionary *curDict= [userInfo objectForKey:@"aps"];
UIAlertView *connectionAlert = [[UIAlertView alloc] initWithTitle:@"app" message:[NSString stringWithFormat:@"%@",[curDict objectForKey:@"alert"]] delegate:self cancelButtonTitle:@"View" otherButtonTitles:@"Cancel",nil];
[connectionAlert show];
[connectionAlert release];
[UIApplication sharedApplication].applicationIconBadgeNumber =[[curDict objectForKey:@"badge"] intValue];
}
}
- (void)applicationWillEnterForeground:(UIApplication *)application
{
NSLog(@"applicationWillEnterForeground");
[UIApplication sharedApplication].applicationIconBadgeNumber = 0;
}
- (void)alertView:(UIAlertView *)alertView clickedButtonAtIndex:(NSInteger)buttonIndex
{
NSString *title = [alertView buttonTitleAtIndex:buttonIndex];
if ([title isEqualToString:@"View"])
{
NSArray *mycontrollers = self.tabBarController.viewControllers;
NSLog(@"%@",mycontrollers);
[[mycontrollers objectAtIndex:0] popToRootViewControllerAnimated:NO];
mycontrollers = nil;
tabBarController.selectedIndex = 0;
}
}
您向用户显示UIAlertView,当收到通知时,将调用DidReceiveMemoteNotification:函数
- (void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo
{
NSLog(@"userInfo :%@",userInfo);
NSString* msg = [userInfo valueForKey:@"aps"];
if (self._VCObj.isViewLoaded && self._VCObj.view.window) {
// viewController is visible don't show.
}
else { // viewController is not visible
[[[UIAlertView alloc]initWithTitle:@"Title" message:msg delegate:self cancelButtonTitle:@"ok" otherButtonTitles: nil] show];
}
}
}
对不起,你能说得更准确些吗?您不想在触摸通知时进入应用程序?那根本不可能。。。