从数组中删除过去的时间

Question

我有这样的一系列时间:

(
    "2000-01-01 23:48:00 +0000",
    "2000-01-01 02:15:00 +0000",
    "2000-01-01 04:39:00 +0000",
    "2000-01-01 17:23:00 +0000",
    "2000-01-01 13:02:00 +0000",
    "2000-01-01 21:25:00 +0000"
)

这是从这段代码生成的:

//loop through array and convert the string times to NSDates
                NSDateFormatter *timeFormatter = [[NSDateFormatter alloc] init];
                [timeFormatter setDateFormat:@"hh:mm a"];
                NSMutableArray *arrayOfDatesAsDates = [NSMutableArray array];
                for (NSObject* o in arrayTimes)
                {
                    NSLog(@"%@",o);
                    NSDate *nsDateTime = [timeFormatter dateFromString:o];
                    [arrayOfDatesAsDates addObject:nsDateTime];
                }
                NSLog(@"times array: %@", arrayOfDatesAsDates);//

我这样做是因为我试图在接下来的数组中获得时间.我的计划是删除过去的时间,订购它们,然后在下次使用第一个.

我怎样才能删除过去的？

谢谢

Answer 1

像这样的东西会......

NSArray *dateArray = ...

NSArray *filteredArray = [dateArray filteredArrayUsingPredicate:[NSPredicate predicateWithBlock:^BOOL(NSDate *date, NSDictionary *bind){
    NSComparisonResult result = [[NSDate date] compare:date];
    return result == NSOrderedAscending; 
}]];

filteredArray将是dateArray中现在或将来的所有日期.