如何获得2个stringforms的string值之间的时间

我有店铺工作小时JSON

{ "MinPerAppointment": "30", "OpeningTime": "10:30", "BreakStartTime": "12:00", "BreakEndTime": "13:00", "ClosingTime": "19:30" } 

我想通过忽略BreakStartTime和BreakEndTime来获得时间间隔为30（MinPerAppointment） OpeningTime的所有数据为ClosingTime 。

我想以这种格式获取数据

10:30

11:00

11:30

BreakStartTime到BreakEndTime ..之间的数据，是12点到13点

13:00

13:30

…..

…..

19:00

Answer 1

尝试这个-

 NSString *dateInString=@"11:15";// your OpeningTime NSMutableArray *array=[[NSMutableArray alloc]init]; NSString *str_MinPerAppointment = @"30"; // your MinPerAppointment NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init]; [dateFormatter setDateFormat:@"HH:mm"]; NSDate *Breaktime = [dateFormatter dateFromString:@"14:00"] ; // your breakStartTime NSDate *BreaktimeEnd = [dateFormatter dateFromString:@"15:00"] ; // your breakEndTime NSDate *closeTime = [dateFormatter dateFromString:@"20:30"] ; // your closeTime while ([[dateFormatter dateFromString:dateInString] compare:closeTime] == NSOrderedAscending) { NSDate *myDate = [dateFormatter dateFromString:dateInString] ; if([myDate compare:Breaktime] == NSOrderedDescending && [myDate compare:BreaktimeEnd] == NSOrderedAscending){ }else{ if ([myDate compare:Breaktime]==NSOrderedSame) { }else{ [array addObject:dateInString]; } } dateInString = [dateFormatter stringFromDate:[myDate dateByAddingTimeInterval:60*[str_MinPerAppointment integerValue]]]; } NSLog(@"your data :%@",array); 

OP： –

 your data :( "10:30", "11:00", "11:30", "13:00", "13:30", "14:00", "14:30", "15:00", "15:30", "16:00", "16:30", "17:00", "17:30", "18:00", "18:30", "19:00" ) 

Answer 2

根据我对前一个问题的回答 ，我会采取以下的方法。 你不想比较string的时间数据：

1. 将所有stringhh：mm转换为整数分钟。
2. 从开始时间到结束时间（减去预约时间间隔）以预约时间段的步骤进行迭代。
3. 如果时间在rest开始和结束之间，则忽略。

4. 将每个值转换为hh：mmstring并将其添加到数组中。

    + (int)minutesFromHhmm:(NSString *)hhmm { int minutes = 0, hh, mm; if (sscanf([hhmm UTF8String], "%d:%d", &hh, &mm) == 2) minutes = (hh * 60) + mm; return minutes; } + (NSString *)hhmmFromMinutes:(int)minutes { int hh = minutes / 60; int mm = minutes - (hh * 60); hh %= 24; // day roll-over return [NSString stringWithFormat:@"%02:%02d", hh, mm]; } - (NSMutableArray *)yourMethod:(NSDictionary *)jsonData { // 1. int minAppointmentTime = [self minutesFromHhmm:jsonData[@"MinPerAppointment"]]; int openingTime = [self minutesFromHhmm:jsonData[@"OpeningTime"]]; int breakStartTime = [self minutesFromHhmm:jsonData[@"BreakStartTime"]]; int breakEndTime = [self minutesFromHhmm:jsonData[@"BreakEndTime"]]; int closingTime = [self minutesFromHhmm:jsonData[@"ClosingTime"]; NSMutableArray *appointmentSlots = [NSMutableArray new]; // 2. for (int time = openingTime; time <= closingTime - minAppointmentTime; time += minAppointmentTime) { if (time >= breakStartTime && time <= breakEndTime) continue; // 3. [appointmentSlots addObject:[self hhmmFromMinutes:time]]; // 4. } return appointmentSlots; } 

Answer 3

  NSDateFormatter * df = [[NSDateFormatter alloc] init]; [df setDateFormat: @"yyyy-MM-dd"]; NSDate * dt1 ; NSDate * dt2 ; dt1 = [df dateFromString: txt1.text]; dt2 = [df dateFromString: txt2.text]; NSComparisonResult result = [dt1 compare: dt2]; switch (result) { case NSOrderedAscending: NSLog(@"%@ is greater than %@", dt2, dt1); break; case NSOrderedDescending: NSLog(@"%@ is less %@", dt2, dt1); break; case NSOrderedSame: NSLog(@"%@ is equal to %@", dt2, dt1); break; default: NSLog(@"erorr dates %@, %@", dt2, dt1); break; }