Convert milliseconds into day,hour,minute and second iOS? [duplicate] - ios

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How to convert milliseconds into human readable form?
(22 answers)
Closed 6 years ago.
I need to convert an milliseconds into Days, Hours, Minutes Second.
ex: 5 Days, 4 hours, 13 minutes, 1 second.
Thanks

if you don't want to do the calculation by yourself, you could go for such a solution.
I, however, know that is a kinda costly solution, so you need to be aware of potential performance issues in runtime – depending on how frequently you intend to invoke this.
NSTimeInterval _timeInSeconds = 123456789.123; // or any other interval...;
NSCalendar *_calendar = [NSCalendar calendarWithIdentifier:NSCalendarIdentifierGregorian];
NSCalendarUnit _units = NSCalendarUnitDay | NSCalendarUnitHour | NSCalendarUnitMinute | NSCalendarUnitSecond;
NSDateComponents *_components = [_calendar components:_units fromDate:[NSDate date] toDate:[NSDate dateWithTimeIntervalSinceNow:_timeInSeconds] options:kNilOptions];
NSLog(#"%ld Days, %ld Hours, %ld Minutes, %ld Seconds", _components.day, _components.hour, _components.minute, _components.second);

You can write your own function like this:
import UIKit
let miliseconds: Int = 24 * 3600 * 1000 + 3700 * 1000
// 1 day and 1 hour 1 minute 40 seconds
func convertTime(miliseconds: Int) -> String {
var seconds: Int = 0
var minutes: Int = 0
var hours: Int = 0
var days: Int = 0
var secondsTemp: Int = 0
var minutesTemp: Int = 0
var hoursTemp: Int = 0
if miliseconds < 1000 {
return ""
} else if miliseconds < 1000 * 60 {
seconds = miliseconds / 1000
return "\(seconds) seconds"
} else if miliseconds < 1000 * 60 * 60 {
secondsTemp = miliseconds / 1000
minutes = secondsTemp / 60
seconds = (miliseconds - minutes * 60 * 1000) / 1000
return "\(minutes) minutes, \(seconds) seconds"
} else if miliseconds < 1000 * 60 * 60 * 24 {
minutesTemp = miliseconds / 1000 / 60
hours = minutesTemp / 60
minutes = (miliseconds - hours * 60 * 60 * 1000) / 1000 / 60
seconds = (miliseconds - hours * 60 * 60 * 1000 - minutes * 60 * 1000) / 1000
return "\(hours) hours, \(minutes) minutes, \(seconds) seconds"
} else {
hoursTemp = miliseconds / 1000 / 60 / 60
days = hoursTemp / 24
hours = (miliseconds - days * 24 * 60 * 60 * 1000) / 1000 / 60 / 60
minutes = (miliseconds - days * 24 * 60 * 60 * 1000 - hours * 60 * 60 * 1000) / 1000 / 60
seconds = (miliseconds - days * 24 * 60 * 60 * 1000 - hours * 60 * 60 * 1000 - minutes * 60 * 1000) / 1000
return "\(days) days, \(hours) hours, \(minutes) minutes, \(seconds) seconds"
}
}
convertTime(miliseconds)
//result is "1 days, 1 hours, 1 minutes, 40 seconds"

try this
NSTimeInterval time = <timein ms>;
NSInteger days = time / (24 * 60 * 60);
NSInteger hours = (time / (60 * 60)) - (24 * days);
NSInteger minutes =(time / 60) - (24 * 60 * days) - (hours * 60);
NSInteger seconds = (lroundf(time) % 60);

I have write the easy code to do this in both Objective-c and Swift:
Swift
var milliseconds : double_t = 568569600;
milliseconds = floor(milliseconds/1000);
let seconds : double_t = fmod(milliseconds, 60);
let minutes : double_t = fmod((milliseconds / 60) , 60);
let hours : double_t = fmod((milliseconds / (60*60)), 60);
let days : double_t = fmod(milliseconds / ((60*60)*24), 24);
NSLog("seconds : %.f minutes : %.f hours : %.f days : %.f", seconds, minutes, hours, days);
Output - seconds : 9 minutes : 56 hours : 38 days : 7
Objective
double milliseconds = 568569600;
milliseconds = milliseconds/1000;
float seconds = fmod(milliseconds, 60);
float minutes = fmod((milliseconds / 60) , 60);
float hours = fmod((milliseconds / (60*60)), 60);
float days = fmod(milliseconds / ((60*60)*24), 24);
NSLog(#"seconds : %.f minutes : %.f hours : %.f days : %.f ", seconds, minutes, hours, days);
Output - seconds : 10 minutes : 56 hours : 38 days : 7

Related

Won't print calculations for hours and minutes to seconds

I'm making a program that calculates minutes and hours into seconds but it won't print, would appreciate any help
enter_hours = int(input("Please enter number of hours: "))
enter_minutes = int(input("Please enter number of minutes: "))
def CalculateSeconds():
hours = enter_hours * 3600
minutes = enter_minutes * 60
return(hours, minutes)
Figured it out
def CalculateSeconds(enter_hours, enter_minutes):
hours = enter_hours * 3600
minutes = enter_minutes * 60
return(hours, minutes)
seconds = CalculateSeconds(enter_hours=int(input("Enter first number: ")),
enter_minutes=int(input("Enter second number: ")))
print(seconds)

Running sum in background

Modified Question.
My fitness app will calculate the number of calories burned based on a calculated value for each second. I have a timer that will allow the app to pick back up should it. I can't get the running sum to continue calculating when the app goes into the background. I tried to place the running sum inside of a DispatchQueue but not getting the sum as expected. Any guidance is greatly appreciated.
Here's the code I have placed inside the function that updates the timer.
//MARK: - Update Timer Label
func updateTimerLabel() {
interval = -Int(timerStartDate.timeIntervalSinceNow)
time = interval
let hours = interval / 3600
let minutes = interval / 60 % 60
let seconds = interval % 60
print("Current interval = \(interval)")
timerLabel.text = String(format:"%02i:%02i:%02i", hours, minutes, seconds)
DispatchQueue.global(qos: .background).async {
if self.activityArray[self.currentArrayRow].2 <= 4.5 {
self.cps = self.activityArray[self.currentArrayRow].2 * Double(self.user.userWeightInKilo) / 3600
self.runningCPS = self.runningCPS + self.cps
print("MET \(self.activityArray[self.currentArrayRow].2) <= 4.5 * KG (\(Double(self.user.userWeightInKilo))) * HR (\(Double(self.user.userHeartRate))) / MaxHR (\(Double(self.user.maxHeartRate)) * interval \(Double(self.interval)) / 3600. Gives a cps 0f \(self.cps) and a runningCPS of \(self.runningCPS) ")
} else {
self.cps = self.activityArray[self.currentArrayRow].2 * Double(self.user.userWeightInKilo) * Double(self.user.userHeartRate) / Double(self.user.maxHeartRate) / 3600
self.runningCPS = self.runningCPS + self.cps
print("MET \(self.activityArray[self.currentArrayRow].2) > 4.5 * KG (\(Double(self.user.userWeightInKilo))) * HR (\(Double(self.user.userHeartRate))) / MaxHR (\(Double(self.user.maxHeartRate)) * interval \(Double(self.interval)) / 3600. Gives a cps 0f \(self.cps) and a runningCPS of \(self.runningCPS) ")
}
}
activeLabel.text = String(format: "%0.1f", runningCPS) + " Calories Burned"
}

Swift - Hours/Minutes/Seconds to Integer conversion

I have Hours / Minutes / Seconds that I would like converted into an integer
7 Hours 30 Minutes 5 Seconds
You could use split():
import Foundation
func getSecondsFromString(timeString: String) -> (Int) {
let timeParts = timeString.replacingOccurrences(of: "[^0-9]", with: " ", options: [.regularExpression])
.split(separator: " ")
.map{Int($0)!}
return timeParts[0] * 3600 + timeParts[1] * 60 + timeParts[2]
}
print(getSecondsFromString(timeString: "7 Hours 30 Minutes 5 Seconds"))
Output:
27005
If you have the time difference between two dates in seconds using timeIntervalSince then you can directly convert this into hours as a double by doing
let hours: Double = elapsedTime / 3600 // 7.501388...
In your examples this is 7.5 which you can then multiply with the hourly rate. If you for some reason only want to use full hours you can either round to the nearest full hour
let fullHours = round(hours) // 8.0
or if you want to truncate minutes and keep the hour then you can do a integer division from the start
let hours: Int = elapsedTime / 3600 // 7

How to customize my timer display to show only minutes and seconds?

My Timer is displaying Minutes and Hours, but once it gets to 60 minutes it restarts from 0 Minute.
Should I get rid of the modulo ( % 60 ) for minutes.
I would like my timer to display for ex: 80:45 ( basically not stopping at 60 min once it reaches 1 hour)
var min = 0
var sec = 0
func stringFromTimeInterval(interval: NSTimeInterval) -> String {
let interval = Int(interval)
let seconds = interval % 60
let minutes = (interval / 60) % 60
//let hours = (interval / 3600)// I don't need the hours
return String(format: "%02d:%02d",minutes, seconds)
}
% 60 means that it will spit out a minutes value that is the remainder when divided by 60(minutes). This is most probably because for time in the form hh:mm, you want it to go from 5:59 to 6:00, not 5:60. So changing the following line will give you what you seek.
let minutes = (interval / 60) % 60 -> let minutes = interval / 60

Where is the bug in these length-of-daylight/night approximations?

I am trying to make an approximation of the length of day from sunrise to sunset, and the length of night from sunset to sunrise. My present approximation is crude (it assumes yesterday and tomorrow have equivalent values to today), but for now I am not specifically concerned with pinpointing yesterday sunset, today sunrise, today sunset, and tomorrow sunrise (yet). My goal is a calculation based on twelve equal hours per night (twelve equal to each other, not equal to a standard hour or daytime hour), and twelve equal hours per day.
What I am concerned with is that in my iOS app, the calculation is way off; a minute flies by in 5-6 (standard) seconds' time. When I use unmodified time, in other code from here, the clock moves at a standard pace, but when I try to get this code to feed the clock code, something is out of bounds.
The code I've been working on, as an approximation, is:
NSDate *now = [[NSDate alloc] init];
NSDate *factory = [[NSDate alloc] init];
NSDate *summerSolstice2013 = [factory initWithTimeIntervalSinceReferenceDate:_referenceSummerSolstice];
double distanceAlong = [now timeIntervalSinceDate:summerSolstice2013];
double angleAlong = M_PI * 2 * distanceAlong / (2 * (_referenceWinterSolstice - _referenceSummerSolstice));
double currentHeight = cos(angleAlong) * _latitudeAngle + _tiltAngle;
...
if (_secondsAreNatural)
{
_secondsAreShadowed = FALSE;
double dayDuration = 12 * 60 * 60 + 12 * 60 * 60 * sin(currentHeight);
double midday = fmod(24 * 60 * 60 * _longitudeAngle / (2 * M_PI) + 12 * 60 * 60, 24 * 60 * 60);
double sunrise = midday - dayDuration / 2;
double sunset = midday + dayDuration / 2;
double seconds = fmod([now timeIntervalSinceReferenceDate], 24 * 60 * 60);
double proportionAlong = 0;
if (seconds < sunrise)
{
_naturalSeconds = (seconds - sunset - 24 * 60 * 60) / (sunrise - sunset - 24 * 60 * 60);
}
else if (seconds > sunset)
{
_naturalSeconds = 12 * 60 * 60 * (seconds - sunset) / (sunrise + 24 * 60 * 60 - sunset) + 18 * 60 * 60;
}
else
{
_naturalSeconds = 12 * 60 * 60 * (seconds - sunrise) / (sunset - sunrise) + 6 * 60 * 60;
}
}
Are there any problems (given that this approximation can probably be refined to any extent) you can pinpoint in this code?
Thanks,
--EDIT--
The code I wrote above was fairly demanding in terms of the loose ends presented to someone reading it. I tried to take another pass, and rewrite it in simpler terms and with a purer mathematical model. I wrote, comments added:
NSDate *now = [[NSDate alloc] init];
NSDate *summerSolstice2013 = [[NSDate alloc] initWithTimeIntervalSinceReferenceDate:_referenceSummerSolstice];
double distanceAlong = [now timeIntervalSinceDate:summerSolstice2013];
// How far along are we, in seconds, since the reference date?
double angleAlong = M_PI * 2 * distanceAlong / (2 * (_referenceWinterSolstice - _referenceSummerSolstice));
// What's the angle if 2 π radians corresponds to a whole year?
double currentHeight = cos(angleAlong) * _latitudeAngle + _tiltAngle;
// _latitudeAngle is the angle represented by our latitude; _tiltAngle is the angle of the earth's tilt.
NSInteger day = 24 * 60 * 60;
// 'day' could have been called secondsInADay, but it was mean to reduce the number of multiplicands represented in the code.
// If we are in the endless day or endless night around the poles, leave the user with standard clock hours.
if (currentHeight > M_PI / 2)
{
_secondsAreShadowed = TRUE;
}
else if (currentHeight < - M_PI / 2)
{
_secondsAreShadowed = TRUE;
}
// Otherwise, calculate the time this routine is meant to calculate. (This is the main intended use case.)
else if (_secondsAreNatural)
{
_secondsAreShadowed = FALSE;
// closestDay is intended to be the nearest midnight (or, in another hemisphere, midday), not exactly in hours offset from UTC, but in longitude offset from Greenwich.
double closestDay;
if (fmod(distanceAlong, day) < .5 * day)
{
closestDay = distanceAlong - fmod(distanceAlong, day);
}
else
{
closestDay = day + distanceAlong - fmod(distanceAlong, day);
}
// As we go through the calculations, for the most part we keep up information on the previous and next days, which will to some degree be consulted at the end.
double previousDay = closestDay - day;
double nextDay = closestDay + day;
// For the three days, what proportion of the way along are they from the solstices?
double closestDayAngleAlong = M_PI * 2 * closestDay / (2 * (_referenceWinterSolstice - _referenceSummerSolstice));
double previousDayAngleAlong = M_PI * 2 * previousDay / (2 * (_referenceWinterSolstice - _referenceSummerSolstice));
double nextDayAngleAlong = M_PI * 2 * nextDay / (2 * (_referenceSummerSolstice - _referenceSummerSolstice));
// What angle are we placed by on the year's cycle, between _latitudeAngle + _tiltAngle and -latitudeAngle + _tiltAngle?
double closestDayHeight = cos(closestDayAngleAlong) * _latitudeAngle + _tiltAngle;
double previousDayHeight = cos(previousDayAngleAlong) * _latitudeAngle + _tiltAngle;
double nextDayHeight = cos(nextDayAngleAlong) * _latitudeAngle + _tiltAngle;
// Based on that, what are the daylight durations for the three twenty-four hour days?
double closestDayDuration = day / 2 + (day / 2) * sin(closestDayHeight);
double previousDayDuration = day / 2 + (day / 2) * sin(previousDayHeight);
double nextDayDuration = day / 2 + (day / 2) * sin(nextDayHeight);
// Here we use both morning and evening for the closest day, and the previous day's morning and the next day's evening.
double closestDayMorning = closestDay + (day / 2) - (closestDayDuration / 2);
double closestDayEvening = closestDay + (day / 2) + (closestDayDuration / 2);
double previousDayEvening = previousDay + (day / 2) + (previousDayDuration / 2);
double nextDayMorning = nextDay + (day / 2) + (nextDayDuration / 2);
// We calculate the proportion along the day that we are between evening and morning (or morning and evening), along with the sooner endpoint of that interval.
double proportion;
double referenceTime;
if (distanceAlong < closestDayMorning)
{
proportion = (distanceAlong - previousDayEvening) / (closestDayMorning - previousDayEvening);
referenceTime = previousDay + day * 3 / 4;
}
else if (distanceAlong > closestDayEvening)
{
proportion = (distanceAlong - closestDayEvening) / (nextDayMorning - closestDayEvening);
referenceTime = closestDay + day * 3 / 4;
}
else
{
proportion = (distanceAlong - closestDayMorning) / (closestDayEvening - closestDayMorning);
referenceTime = closestDay + day * 1 / 4;
}
// Lastly, we take both that endpoint and the proportion of it, and we get the number of seconds according to the daylight / nighttime calculation intended.
_naturalSeconds = referenceTime + proportion * day / 2;
I was hoping to make the code clearer and easier to grasp, and I think I did that, but it is displaying similar behavior to my previous attempt: the clock hands spin by at about ten times natural time when they should be within a factor of .8 to 1.2 of standard hours/minutes/seconds.
Any advice? Has my edited code been any clearer either about what is intended or what is wrong?
Thanks,
Your code is hard to follow, but I'll try to get you some tips:
There are existing libraries out there that compute solar angle/azimuth and sunrise/sunset for a given date. Use google as a help, here's some relevant resources: http://www.esrl.noaa.gov/gmd/grad/solcalc/ If you don't find any useful source code, I could post some.
Do not use double to calculate with dates and times. That's confusing and results in errors. Use a data type that is intended to store dates.
For your code, you say that the time is running to fast. Since referenceTime and day in the last line are constant (at least for half a day), the error must be in proportion. I think you're mixing to many cases there. The interpolation should go from the start of the range to the end, so in the case
proportion = (distanceAlong - previousDayEvening) / (closestDayMorning - previousDayEvening);
referenceTime = previousDay + day * 3 / 4;
proportion should run from (previousDay + day * 3 / 4) to (closestDay + day * 3 / 4), or, described differently, from the dusk to dawn of closestDay. But it's completely unclear how this interpolation should work.
Try to draw a diagram of the different cases (I believe there should only be two, one for day and one for night) and the corresponding interpolation.
But: What are you trying to achieve after all? The resulting time is just a forward running time, it is actually independent of latitude or longitude or time of day. So to make the time run, you don't need to know where the sun is.

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