如何比较两个给定的时间与当前时间？

我想比较当前时间和不同date的两个不同的时间。我的代码是

NSString *time1 = record.trigger_from_time; NSString *time2 = record.trigger_to_time; NSDateFormatter *formatter = [[NSDateFormatter alloc] init]; [formatter setDateFormat:@"HH:mm:ss"]; NSDate *date1= [formatter dateFromString:time1]; NSDate *date2 = [formatter dateFromString:time2]; NSLog(@"%@,%@",date1,date2); 

我得到日志为2000-01-01 09:12:00 + 0000,2000-01-01 12:30:00 +0000

但我目前的时间是2014-04-01 17:06:18。

我正在从当前date

 NSDate *now = [NSDate date]; NSDateFormatter *timeFormatter = [[NSDateFormatter alloc] init]; timeFormatter.dateFormat = @"HH:mm"; [timeFormatter setTimeZone:[NSTimeZone systemTimeZone]]; NSLog(@"The Current Time is %@",[timeFormatter stringFromDate:now]); 

我想比较当前时间与record.trigger_from_time和record.trigger_to_time

Answer 1

比较2个NSDates。

  ... NSDate *date1= [formatter dateFromString:time1]; NSDate *date2 = [formatter dateFromString:time2]; if ([date2 compare:date1] == NSOrderedDescending) { NSLog(@"date1 is later than date2"); } else if ([date2 compare:date1] == NSOrderedAscending) { NSLog(@"date1 is earlier than date2"); } else { NSLog(@"dates are the same"); } 

UPD：通过开关

  ... NSDate *date1= [formatter dateFromString:time1]; NSDate *date2 = [formatter dateFromString:time2]; NSComparisonResult result = [date2 compare:date1]; switch(result){ case NSOrderedDescending: NSLog(@"date1 is later than date2"); break; case NSOrderedAscending: NSLog(@"date1 is earlier than date2"); break; default: NSLog(@"dates are the same"); break; } 

Answer 2

 NSString *time1 = @"09:10:02"; NSString *time2 = @"17:10:16"; NSDateFormatter *formatter = [[NSDateFormatter alloc] init]; [formatter setDateFormat:@"HH:mm:"]; NSDate *date1= [formatter dateFromString:time1]; NSDate *date2 = [formatter dateFromString:time2]; NSComparisonResult result = [date1 compare:date2]; if(result == NSOrderedDescending) { NSLog(@"time1 is later than time2"); } else if(result == NSOrderedAscending) { NSLog(@"time2 is later than time1"); } else { NSLog(@"time1 is equal to time2"); }