Question

目前,我正通过我所有排定的当地通知,根据用户Info报标的值,找到一个“事项”。当我提出30+地方通知时,这似乎非常缓慢。是否有办法在不穿透阵列的情况下查阅个人的地方通知?

在这方面,我已经:

NSArray *notificationArray = [[UIApplication sharedApplication]     scheduledLocalNotifications];
UILocalNotification *row = nil;
for (row in notificationArray) {
            NSDictionary *userInfo = row.userInfo;
            NSString *identifier = [userInfo valueForKey:@"movieTitle"];
            NSDate *currentAlarmDateTime = row.fireDate;
if([identifier isEqualToString:myLookUpName]) {
 NSLog(@"Found a match!");
}
}

这里我想的是:

NSArray *notificationArray = [[UIApplication sharedApplication]     scheduledLocalNotifications];
UILocalNotification *row = " The row in notificationArray where [userInfo valueForKey:@"movieTitle"]=myLookUpName" ;

Answer 1

你们也许能够为此使用前提,但我没有测试:

NSPredicate *predicate = [NSPredicate predicateWithFormat:@"userInfo.movieTitle = %@", myLookUpName];

And then use that predicate to filter the array and grab the first element:

UILocalNotification *row = [[notificationArray filteredArrayUsingPredicate:predicate]objectAtIndex:0];

同样,这是未经测试的,可能不可行。

http://www.ohchr.org。

如果这种工作不成功,你可以使用试验组:

UILocalNotification *row = [[notificationArray objectsAtIndexes:[notificationArray indexesOfObjectsPassingTest:^(id obj, NSUInteger idx, BOOL *stop){
    return [[[obj userInfo]valueForKey:@"movieTitle"] isEqualToString:myLookUpName];
}]]objectAtIndex:0];

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