如果我安排两个UILocalNotification
s 并将它们都设置为在完全相同的 fireDate 触发。然后在设备上(这不是模拟器错误)上的 fireDateapplication:didReceiveLocalNotification:
将触发 4 次(每个通知 2 次)。这是一个已知的错误?因为我一直无法找到有关它的任何信息。
问问题
3660 次
2 回答
1
请向http://bugreport.apple.com报告错误。
话虽如此,之前已经注意到虽然模拟器中存在错误,但设备上似乎也存在错误。
请参阅有关此 SO 问题的评论和答案:本地通知“didReceiveLocalNotification”调用两次
于 2012-07-11T11:09:29.017 回答
0
试试这个它在我的应用程序中工作:
-(IBAction)setRemind:(id)sender{
NSCalendar *calendar = [NSCalendar autoupdatingCurrentCalendar];
NSDateFormatter *dateFormatter2 = [[NSDateFormatter alloc] init];
[dateFormatter2 setDateFormat:@"YYYY-MM-dd HH:mm:ss"];
//Gets our picker
NSDate *selectedTime = [datePicker date];
strDate2 = [dateFormatter2 stringFromDate:selectedTime];
NSDate *Date=[dateFormatter2 dateFromString:strDate2];
NSDateComponents *dateComponents = [calendar components:(NSYearCalendarUnit | NSMonthCalendarUnit | NSDayCalendarUnit ) fromDate:Date];
// Set up the fire time
NSDateComponents *dateComp = [[NSDateComponents alloc] init];
[dateComp setDay:[dateComponents day]];
[dateComp setMonth:[dateComponents month]];
[dateComp setYear:[dateComponents year]];
[dateComp setHour:9];
[dateComp setMinute:00];
[dateComp setSecond:00];
[dateComp release];
NSDate *date = [calendar dateFromComponents:dateComp];
[self scheduleAlarmForDate:date message:txtDescri.text];
}
-(IBAction)scheduleAlarmForDate:(NSDate*)date message:(NSString*)msg
{
//====== TO SEE OLD NOTIFI=======
UIApplication *Ap = [UIApplication sharedApplication];
NSArray *arr = [Ap scheduledLocalNotifications];
NSLog(@"Old Notifications :>> %@",arr);
UIApplication* app = [UIApplication sharedApplication];
UILocalNotification *alarm = [[UILocalNotification alloc] init];
// Create a new notification
alarm.fireDate = date;
NSLog(@"fireDate IS >> %@", alarm.fireDate);
alarm.timeZone = [NSTimeZone localTimeZone];
alarm.alertBody = msg;
NSLog(@"msg IS >> %@",msg);
alarm.alertAction = @"Show";
alarm.repeatInterval = 0;
alarm.soundName = UILocalNotificationDefaultSoundName;
alarm.applicationIconBadgeNumber = 1;
[app scheduleLocalNotification:alarm];
[alarm release];
}
我希望它对你有帮助。
于 2012-07-11T11:13:17.013 回答