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我有一個iphone應用程序,我在做推送通知作爲alertview.When我的應用程序處於後臺狀態時,推送通知即將到來,當我點擊它或解鎖手機時,它直接進入應用程序,我已經離開它在forground狀態。我在警報中添加一個動作,點擊查看按鈕,它將去往另一個視圖控制器。我不想進入應用程序,當我點擊通知。我需要顯示alertview和視圖按鈕點擊,當我需要做我的action.Can人幫助我實現this.This是我的代碼片段` -推送通知警報視圖操作?
- (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;
}
}
很抱歉,但你能更準確?觸摸通知時,您不想進入該應用程序?這根本不可能...... – Yorxxx 2012-07-19 08:49:52