Question

我的Json是

{
 "MinPerAppointment": "30",
 "OpeningTime": "12:05"    
}

现在，我想添加30到12:05，应该是12:35，我无法获得如何格式化，所以，我尝试使用NSdate

NSString *str_MinPerAppointment = timeDetailData.minPerAppointment;
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:@"HH:mm"];
NSDate *myDate = [dateFormatter dateFromString:str_MinPerAppointment];

NSString *dateInString = [dateFormatter stringFromDate:myDate];    
NSLog(@"New AppointMentValue :%@",dateInString);

我将新的AppointMentValue改为 nil 。

Answer 1

我认为您不需要使用NSDate格式。只需将HH:MM转换为分钟（或秒，如果你处理它们），做数学然后转换回来。

类似的东西：

NSString *time = @"12:30";
int addMinutes = 30;
int hh, mm;
if (sscanf([time UTF8String], "%d:%d", &hh, &mm) == 2) {
    int minutes = (hh * 60) + mm;
    minutes += addMinutes;
    hh = minutes / 60;
    mm = minutes - (hh * 60);
    hh %= 24;    // day roll-over
    NSString *newTime = [NSString stringWithFormat:@"%02:%02d", hh, mm];
} else {
    NSLog(@"Invalid time value: %@", time);
}

Answer 2

如果你想使用NSDateFormatter/NSDate试试这个 -

NSString *str_MinPerAppointment = @"30";
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:@"HH:mm"];
NSDate *myDate = [dateFormatter dateFromString:@"12:05"] ;

NSString *dateInString = [dateFormatter stringFromDate:[myDate dateByAddingTimeInterval:60*[str_MinPerAppointment integerValue]]];

NSLog(@"New AppointMentValue :%@",dateInString);

将字符串转换为时间格式将返回nil数据

2 个答案: