将当前时间与两个时间字符串进行比较[关闭]

我怎样才能得到像“11:30”这样的时间，以便我想将它与以下内容进行比较：

strOpenTime = @"10:00";
strCloseTime = @"2:00";

那么我怎样才能像上面的打开/关闭时间格式一样获取当前时间，并且我想要当前时间是否在间隔打开/关闭时间之内？

提前致谢..！！

首先，您必须将字符串“10:00”、“2:00”转换为日期当前日期。 这可以例如完成使用以下方法（为简洁起见，省略错误检查）：

- (NSDate *)todaysDateFromString:(NSString *)time
{
    // Split hour/minute into separate strings:
    NSArray *array = [time componentsSeparatedByString:@":"];

    // Get year/month/day from today:
    NSCalendar *cal = [NSCalendar currentCalendar];
NSDateComponents *comp = [cal components:NSYearCalendarUnit|NSMonthCalendarUnit|NSDayCalendarUnit fromDate:[NSDate date]];

    // Set hour/minute from the given input:
    [comp setHour:[array[0] integerValue]];
    [comp setMinute:[array[1] integerValue]];

    return [cal dateFromComponents:comp];
}

然后转换您的开门和关门时间：

NSString *strOpenTime = @"10:00";
NSString *strCloseTime = @"2:00";

NSDate *openTime = [self todaysDateFromString:strOpenTime];
NSDate *closeTime = [self todaysDateFromString:strCloseTime];

现在你必须考虑到关闭时间可能是第二天：

if ([closeTime compare:openTime] != NSOrderedDescending) {
    // closeTime is less than or equal to openTime, so add one day:
    NSCalendar *cal = [NSCalendar currentCalendar];
    NSDateComponents *comp = [[NSDateComponents alloc] init];
    [comp setDay:1];
    closeTime = [cal dateByAddingComponents:comp toDate:closeTime options:0];
}

然后你就可以按照@visualication 在他的回答中所说的那样继续：

NSDate *now = [NSDate date];

if ([now compare:openTime] != NSOrderedAscending &&
    [now compare:closeTime] != NSOrderedDescending) {
    // now should be inside = Open
} else {
    // now is outside = Close
}