iOS Format String into minutes and seconds

微笑、不失礼 提交于 2019-11-27 03:54:05

问题


I am receiving a string from the YouTube JSONC api, but the duration is coming as a full number i.e 2321 instead of 23:21 or 2 instead of 0:02. How would I go about fixing this?

JSON C

EDIT:

int duration = [videos valueForKey:@"duration"];
int minutes = duration / 60;
int seconds = duration % 60;

NSString *time = [NSString stringWithFormat:@"%d:%02d", minutes, seconds];

回答1:


Assuming the duration value is really the duration in seconds, then you can calculate the number of minutes and seconds and then format those into a string.

int duration = ... // some duration from the JSON
int minutes = duration / 60;
int seconds = duration % 60;

NSString *time = [NSString stringWithFormat:@"%d:%02d", minutes, seconds];



回答2:


You should use DateComponentsFormatter if the duration is intended to be user-facing:

let formatter = DateComponentsFormatter()
formatter.allowedUnits = [ .minute, .second ]
formatter.zeroFormattingBehavior = [ .pad ]
let formattedDuration = formatter.string(from: duration)!



回答3:


Try this very optimized

+ (NSString *)timeFormatConvertToSeconds:(NSString *)timeSecs
{
    int totalSeconds=[timeSecs intValue];

    int seconds = totalSeconds % 60;
    int minutes = (totalSeconds / 60) % 60;
    int hours = totalSeconds / 3600;

    return [NSString stringWithFormat:@"%02d:%02d:%02d",hours, minutes, seconds];
}



回答4:


int sec = diff;//INFO: time in seconds

int a_sec = 1;
int a_min = a_sec * 60;
int an_hour = a_min * 60;
int a_day = an_hour * 24;
int a_month = a_day * 30;
int a_year = a_day * 365;

NSString *text = @"";
if (sec >= a_year)
{
    int years = floor(sec / a_year);
    text = [NSString stringWithFormat:@"%d year%@ ", years, years > 0 ? @"s" : @""];
    sec = sec - (years * a_year);
}

if (sec >= a_month)
{
    int months = floor(sec / a_month);
text = [NSString stringWithFormat:@"%@%d month%@ ", text, months, months > 0 ? @"s" : @""];
    sec = sec - (months * a_month);

}

if (sec >= a_day)
{
    int days = floor(sec / a_day);
text = [NSString stringWithFormat:@"%@%d day%@ ", text, days, days > 0 ? @"s" : @""];

    sec = sec - (days * a_day);
}

if (sec >= an_hour)
{
    int hours = floor(sec / an_hour);
text = [NSString stringWithFormat:@"%@%d hour%@ ", text, hours, hours > 0 ? @"s" : @""];

    sec = sec - (hours * an_hour);
}

if (sec >= a_min)
{
    int minutes = floor(sec / a_min);
text = [NSString stringWithFormat:@"%@%d minute%@ ", text, minutes, minutes > 0 ? @"s" : @""];

    sec = sec - (minutes * a_min);
}

if (sec >= a_sec)
{
    int seconds = floor(sec / a_sec);
text = [NSString stringWithFormat:@"%@%d second%@", text, seconds, seconds > 0 ? @"s" : @""];
}
   NSLog(@"<%@>", text);



回答5:


Here is the great code I finds for this

int duration = 1221;
int minutes = floor(duration/60)
int seconds = round(duration - (minutes * 60))
NSString * timeStr = [NSString stringWithFormat:@"%i:%i",minutes,seconds];
NSLog(@"Dilip timeStr : %@",timeStr);

And the output will belike this

Dilip timeStr : 20:21



回答6:


You can subString the 2321 and get the first string as 23 and the second as 21 and convert them to int. Also check for the length of the text:

if (text.length < 4)
   //add zeros on the left of String until be of length 4



回答7:


Objective C:

NSDateComponentsFormatter * formatter = [[NSDateComponentsFormatter alloc]init];
[formatter setUnitsStyle:NSDateComponentsFormatterUnitsStyleShort];
[formatter setAllowedUnits:NSCalendarUnitSecond | NSCalendarUnitMinute];
[formatter setZeroFormattingBehavior:NSDateComponentsFormatterZeroFormattingBehaviorPad];
return [formatter stringFromTimeInterval:duration];


来源:https://stackoverflow.com/questions/15122883/ios-format-string-into-minutes-and-seconds

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