How to calculate the time difference between 2 date time values - date

I am trying to calculate the time difference between 2 date time strings.
I have 2 inputs where the input string is something like this "1:00 PM" and the second one "3:15 PM". I want to know the time difference. So for the above example I want to display 3.15
What I have done:
Converted the time to a 24 hours format. So "1:00 PM" becomes "13:00:00"
Appended the new time to a date like so: new Date("1970-1-1 13:00:00")
Calculated the difference like so:
Code:
var total = Math.round(((new Date("1970-1-1 " + end_time) -
new Date("1970-1-1 " + start_time) ) / 1000 / 3600) , 2 )
But the total is always returning integers and not decimals, so the difference between "1:00 PM" and "3:15 PM" is 2 not 2.15.
I have also tried this (using jQuery, but that is irrelevant):
$('#to_ad,#from_ad').change(function(){
$('#total_ad').val( getDiffTime() );
});
function fixTimeString(time){
var hours = Number(time.match(/^(\d+)/)[1]);
var minutes = Number(time.match(/:(\d+)/)[1]);
var AMPM = time.match(/\s(.*)$/)[1];
if(AMPM == "PM" && hours<12) hours = hours+12;
if(AMPM == "AM" && hours==12) hours = hours-12;
var sHours = hours.toString();
var sMinutes = minutes.toString();
if(hours<10) sHours = "0" + sHours;
if(minutes<10) sMinutes = "0" + sMinutes;
return sHours + ':' + sMinutes + ':00';
}
function getDiffTime(){
var start_time = fixTimeString($('#from_ad').val());
var end_time = fixTimeString($('#to_ad').val());
var start = new Date("1970-1-1 " + end_time).getTime(),
end = new Date("1970-1-1 " + start_time).getTime();
return parseInt(((start - end) / 1000 / 3600, 10)*100) / 100;
}
But the total_ad input is displaying only integer values.
How can I fix this problem?

Math.round rounds to the nearest integer, multiply and divide instead
var start = new Date("1970-1-1 " + start_time).getTime(),
end = new Date("1970-1-1 " + end_time).getTime();
var total = (parseInt(((start-end) / 1000 / 3600)*100, 10)) / 100;
FIDDLE
When you take the time 15:15:00 and subtract 13:00:00, you're left with 2.15 hours, not 3.15, and this example would return 2.15 even without making sure there is only two decimals, but for other times that might not be the case.
You could also use toFixed(2), but that would leave you with 3.00 and not 3 etc.

This is how I calculate it:
calculateDiff();
function calculateDiff(){
_start = "7:00 AM";
_end = "1:00 PM";
_start_time = parseAMDate(_start);
_end_time = parseAMDate(_end);
if (_end_time < _start_time){
_end_time = parseAMDate(_end,1);
}
var difference= _end_time - _start_time;
var hours = Math.floor(difference / 36e5),
minutes = Math.floor(difference % 36e5 / 60000);
if (parseInt(hours) >= 0 ){
if (minutes == 0){
minutes = "00";
}
alert(hours+":"+minutes);
}
}
function parseAMDate(input, next_day) {
var dateReg = /(\d{1,2}):(\d{2})\s*(AM|PM)/;
var hour, minute, result = dateReg.exec(input);
if (result) {
hour = +result[1];
minute = +result[2];
if (result[3] === 'PM' && hour !== 12) {
hour += 12;
}
}
if (!next_day) {
return new Date(1970, 01, 01, hour, minute).getTime();
}else{
return new Date(1970, 01, 02, hour, minute).getTime();
}
}

Related

Formula of Hijri date

What is the formula of converting "Time since epoch" stamp such as 1525249213 to Hijri date (Year - Month - Day)?
I know there would be one day inaccuracy which is OK for me.
The algorithm I found is to convert the Date to a Julian Date then convert the julian date to Hijri.
P.S. I can not remember where I found it. I am not responsible for any error or miss-accuracy or an consequence bla bla bla of using this code.
unsigned long gregorian_to_julian(const int gyear, const int gmonth, const int gday) {
if (gmonth < 3) {
gyear -= 1;
gmonth += 12;
}
auto a = int(gyear / 100.0f);
auto b = (gyear == 1582 && (gmonth > 10 || (gmonth == 10 && gday > 4)) ? -10 :
(gyear == 1582 && gmonth == 10 ? 0 :
(gyear < 1583 ? 0 : 2 - a + int(a / 4.0f))));
return int(365.25f * (gyear + 4716)) + int(30.6001f * (gmonth + 1)) + gday + b - 1524;
}
std::array<int,3> julian_to_hijri(const unsigned long julian_datestamp) {
std::array<int,3> result;
auto y = 10631.0f / 30.0f;
auto epoch_astro = 1948084;
auto shift1 = 8.01f / 60.0f;
auto z = julian_day - epoch_astro;
auto cyc = int(z / 10631.0f);
z = z - 10631 * cyc;
auto j = int((z - shift1) / y);
z = z - int(j * y + shift1);
result[0] = 30 * cyc + j; //Hijri Year
result[1]= int((z + 28.5001f) / 29.5f); //Hijri Month
if (result[1] == 13) {
result[1]= 12;
}
result[2] = z - int(29.5001f * result[1]- 29);// Hijri day
return result;
}
P.S. there will be +-1 one day error in the result Hijri date.

Date difference in hours - Angular 2

How can I get difference in hours between two dates in angular2? I don't want to use external libs like moment.js.
Having, for example: incidentTime = '2017-03-05 11:26:16 AM' and creationTime = '2017-03-06 12:26:16 AM'
let time = +params.data.incidentTime - +params.data.creationTime;
console.log("time: " + time);
It should return 25 hours, but It returns NaN.
This should do the job:
const date1 = params.data.incidentTime;
const date2 = params.data.creationTime;
const diffInMs = Date.parse(date2) - Date.parse(date1);
const diffInHours = diffInMs / 1000 / 60 / 60;
console.log(diffInHours);
Use Math.floor to round the result down, or Math.ceil to round it up
parse the date type parameters to javascript's Date type.
let date1 = new Date(params.data.incidentTime).getTime();
let date2 = new Date(params.data.creationTime).getTime();
let time = date1 - date2; //msec
let hoursDiff = time / (3600 * 1000);
Try this:-
let time:any = new Date("2017-03-05 11:26:16").getHours();
let date2:any = new Date("2017-03-06 12:26:16").getHours();
console.log(time -date2, time, date2, "sdfsd");
but It returns NaN
this is because you are using + sign before date which is converting youtr date into number format so returning NAN(not a number)

Get today's date print to the console in Protractor

I'm new to Protractor. I need to select today's date from the date picker.
Is there a specific anyway to select the today's date from the date picker?
Thanks. :)
Please see my answer here
If you need to be more quick,
var pickerDue = element(by.model("supplier.enroll_date"));
var today = new Date();
var dd = today.getDate();
var mm = today.getMonth()+1; //January is 0!
var yyyy = today.getFullYear();
if(dd<10) {
dd='0'+dd
}
if(mm<10) {
mm='0'+mm
}
today = yyyy+'-'+mm+'-'+dd;
pickerDue.clear();
pickerDue.sendKeys(today);
Hope this helps. :)
This method will give you current date time with seconds:
getCurrentDatemmddyyyy: function () {
var today = new Date();
var dd = today.getDate();
var mm = today.getMonth() + 1; //January is 0!
var yyyy = today.getFullYear();
var hh = today.getHours();
var mins = today.getMinutes();
var ss = today.getSeconds();
if (dd < 10) {
dd = "0" + dd;
}
if (mm < 10) {
mm = "0" + mm;
}
if (hh < 10) {
hh = "0" + hh;
}
if (mins < 10) {
mins = "0" + mins;
}
if (ss < 10) {
ss = "0" + ss;
}
var today = mm + dd + yyyy + hh + mins + ss;
return today;
},

Dates (Duration) in Google Sheets Script

I have a Google Sheet with three columns:
- Date and time (timestamp)
- Duration
- Description
I have an script that when I write something in 'Description', inserts in 'Date' the date and time at this moment, and the 'Duration':
function onEdit(e) {
if(e.source.getActiveSheet().getName() == "Sheet2" ) {
var col = e.source.getActiveCell().getColumn();
if(col == 3 ) {
// I'm in column three
var cellTimeStamp = e.range.offset(0,-2); // First column of the same row
var cellTimeDiff = e.range.offset(0,-1); // Second column of the same row
var cellTimePrev = e.range.offset(-1,-2); // First column of the previous row
var timeTimeStamp = new Date();
var iniTime = cellTimePrev.getValue().getTime();
var finTime = timeTimeStamp.getTime() ;
var timeDiff = String(finTime - iniTime) ;
cellTimeStamp.setValue(timeTimeStamp);
cellTimeDiff.setValue(timeDiff); // [***]
}
}
}
When this executes (as an event) in the column of 'Duration' there is NOT something in the format of 'HH:mm:ss'.
But if I remove the last line in this script and adds this formulae in the sheet:
=A3-A2 (in row 3)
=A4-A3 (in row 4)
...
then it works ok.
I'd like to know how to meet the same result but with a script.
Thanks in advance.
timeDiff is the result of finTime - iniTime which are both native date object values, which means we have milliseconds .
converting that in hh:mm:ss is simple math... : 60 seconds in a minute and 60 minutes in an hour...
A simple code could be like this :
function msToTime(s) {
var ms = s % 1000;
s = (s - ms) / 1000;
var secs = s % 60;
s = (s - secs) / 60;
var mins = s % 60;
var hrs = (s - mins) / 60;
return hrs + ':' + mins + ':' + secs; // milliSecs are not shown but you can use ms if needed
}
If you prefer formating your string more conventionally (2 digits for each value) don't forget you can use Utilities.formatString() to do so.
example below :
return Utilities.formatString("%02d",hrs) + ':' + Utilities.formatString("%02d",mins) + ':' + Utilities.formatString("%02d",secs);
EDIT
Following your comment :
Spreadsheets are smarter than you think, you can try the code below and you will see that the result is actually a time value.(check by double clicking on it)
function test() {
var sh = SpreadsheetApp.getActiveSheet();
var t1 = sh.getRange('a1').getValue().getTime();
var t2 = sh.getRange('b1').getValue().getTime();
sh.getRange('c1').setValue(msToTime(t1-t2)).setNumberFormat('hh:mm:ss');
}
function msToTime(s) {
var ms = s % 1000;
s = (s - ms) / 1000;
var secs = s % 60;
s = (s - secs) / 60;
var mins = s % 60;
var hrs = (s - mins) / 60;
return hrs + ':' + mins + ':' + secs; // milliSecs are not shown but you can use ms if needed
}
note that setNumberFormat('hh:mm:ss') is optional, it's only there to force the spreadsheet to display hour:min:sec format but automatic mode works as well.

How to convert milliseconds into human readable form?

I need to convert an arbitrary amount of milliseconds into Days, Hours, Minutes Second.
For example: 10 Days, 5 hours, 13 minutes, 1 second.
Well, since nobody else has stepped up, I'll write the easy code to do this:
x = ms / 1000
seconds = x % 60
x /= 60
minutes = x % 60
x /= 60
hours = x % 24
x /= 24
days = x
I'm just glad you stopped at days and didn't ask for months. :)
Note that in the above, it is assumed that / represents truncating integer division. If you use this code in a language where / represents floating point division, you will need to manually truncate the results of the division as needed.
Let A be the amount of milliseconds. Then you have:
seconds=(A/1000)%60
minutes=(A/(1000*60))%60
hours=(A/(1000*60*60))%24
and so on (% is the modulus operator).
Hope this helps.
Both solutions below use javascript (I had no idea the solution was language agnostic!). Both solutions will need to be extended if capturing durations > 1 month.
Solution 1: Use the Date object
var date = new Date(536643021);
var str = '';
str += date.getUTCDate()-1 + " days, ";
str += date.getUTCHours() + " hours, ";
str += date.getUTCMinutes() + " minutes, ";
str += date.getUTCSeconds() + " seconds, ";
str += date.getUTCMilliseconds() + " millis";
console.log(str);
Gives:
"6 days, 5 hours, 4 minutes, 3 seconds, 21 millis"
Libraries are helpful, but why use a library when you can re-invent the wheel! :)
Solution 2: Write your own parser
var getDuration = function(millis){
var dur = {};
var units = [
{label:"millis", mod:1000},
{label:"seconds", mod:60},
{label:"minutes", mod:60},
{label:"hours", mod:24},
{label:"days", mod:31}
];
// calculate the individual unit values...
units.forEach(function(u){
millis = (millis - (dur[u.label] = (millis % u.mod))) / u.mod;
});
// convert object to a string representation...
var nonZero = function(u){ return dur[u.label]; };
dur.toString = function(){
return units
.reverse()
.filter(nonZero)
.map(function(u){
return dur[u.label] + " " + (dur[u.label]==1?u.label.slice(0,-1):u.label);
})
.join(', ');
};
return dur;
};
Creates a "duration" object, with whatever fields you require.
Formatting a timestamp then becomes simple...
console.log(getDuration(536643021).toString());
Gives:
"6 days, 5 hours, 4 minutes, 3 seconds, 21 millis"
Apache Commons Lang has a DurationFormatUtils that has very helpful methods like formatDurationWords.
You should use the datetime functions of whatever language you're using, but, just for fun here's the code:
int milliseconds = someNumber;
int seconds = milliseconds / 1000;
int minutes = seconds / 60;
seconds %= 60;
int hours = minutes / 60;
minutes %= 60;
int days = hours / 24;
hours %= 24;
This is a method I wrote. It takes an integer milliseconds value and returns a human-readable String:
public String convertMS(int ms) {
int seconds = (int) ((ms / 1000) % 60);
int minutes = (int) (((ms / 1000) / 60) % 60);
int hours = (int) ((((ms / 1000) / 60) / 60) % 24);
String sec, min, hrs;
if(seconds<10) sec="0"+seconds;
else sec= ""+seconds;
if(minutes<10) min="0"+minutes;
else min= ""+minutes;
if(hours<10) hrs="0"+hours;
else hrs= ""+hours;
if(hours == 0) return min+":"+sec;
else return hrs+":"+min+":"+sec;
}
function convertTime(time) {
var millis= time % 1000;
time = parseInt(time/1000);
var seconds = time % 60;
time = parseInt(time/60);
var minutes = time % 60;
time = parseInt(time/60);
var hours = time % 24;
var out = "";
if(hours && hours > 0) out += hours + " " + ((hours == 1)?"hr":"hrs") + " ";
if(minutes && minutes > 0) out += minutes + " " + ((minutes == 1)?"min":"mins") + " ";
if(seconds && seconds > 0) out += seconds + " " + ((seconds == 1)?"sec":"secs") + " ";
if(millis&& millis> 0) out += millis+ " " + ((millis== 1)?"msec":"msecs") + " ";
return out.trim();
}
In java
public static String formatMs(long millis) {
long hours = TimeUnit.MILLISECONDS.toHours(millis);
long mins = TimeUnit.MILLISECONDS.toMinutes(millis);
long secs = TimeUnit.MILLISECONDS.toSeconds(millis);
return String.format("%dh %d min, %d sec",
hours,
mins - TimeUnit.HOURS.toMinutes(hours),
secs - TimeUnit.MINUTES.toSeconds(mins)
);
}
Gives something like this:
12h 1 min, 34 sec
I would suggest using whatever date/time functions/libraries your language/framework of choice provides. Also check out string formatting functions as they often provide easy ways to pass date/timestamps and output a human readable string format.
Your choices are simple:
Write the code to do the conversion (ie, divide by milliSecondsPerDay to get days and use the modulus to divide by milliSecondsPerHour to get hours and use the modulus to divide by milliSecondsPerMinute and divide by 1000 for seconds. milliSecondsPerMinute = 60000, milliSecondsPerHour = 60 * milliSecondsPerMinute, milliSecondsPerDay = 24 * milliSecondsPerHour.
Use an operating routine of some kind. UNIX and Windows both have structures that you can get from a Ticks or seconds type value.
Long serverUptimeSeconds =
(System.currentTimeMillis() - SINCE_TIME_IN_MILLISECONDS) / 1000;
String serverUptimeText =
String.format("%d days %d hours %d minutes %d seconds",
serverUptimeSeconds / 86400,
( serverUptimeSeconds % 86400) / 3600 ,
((serverUptimeSeconds % 86400) % 3600 ) / 60,
((serverUptimeSeconds % 86400) % 3600 ) % 60
);
Why just don't do something like this:
var ms = 86400;
var seconds = ms / 1000; //86.4
var minutes = seconds / 60; //1.4400000000000002
var hours = minutes / 60; //0.024000000000000004
var days = hours / 24; //0.0010000000000000002
And dealing with float precision e.g. Number(minutes.toFixed(5)) //1.44
I'm not able to comment first answer to your question, but there's a small mistake. You should use parseInt or Math.floor to convert floating point numbers to integer, i
var days, hours, minutes, seconds, x;
x = ms / 1000;
seconds = Math.floor(x % 60);
x /= 60;
minutes = Math.floor(x % 60);
x /= 60;
hours = Math.floor(x % 24);
x /= 24;
days = Math.floor(x);
Personally, I use CoffeeScript in my projects and my code looks like that:
getFormattedTime : (ms)->
x = ms / 1000
seconds = Math.floor x % 60
x /= 60
minutes = Math.floor x % 60
x /= 60
hours = Math.floor x % 24
x /= 24
days = Math.floor x
formattedTime = "#{seconds}s"
if minutes then formattedTime = "#{minutes}m " + formattedTime
if hours then formattedTime = "#{hours}h " + formattedTime
formattedTime
This is a solution. Later you can split by ":" and take the values of the array
/**
* Converts milliseconds to human readeable language separated by ":"
* Example: 190980000 --> 2:05:3 --> 2days 5hours 3min
*/
function dhm(t){
var cd = 24 * 60 * 60 * 1000,
ch = 60 * 60 * 1000,
d = Math.floor(t / cd),
h = '0' + Math.floor( (t - d * cd) / ch),
m = '0' + Math.round( (t - d * cd - h * ch) / 60000);
return [d, h.substr(-2), m.substr(-2)].join(':');
}
var delay = 190980000;
var fullTime = dhm(delay);
console.log(fullTime);
Long expireTime = 69l;
Long tempParam = 0l;
Long seconds = math.mod(expireTime, 60);
tempParam = expireTime - seconds;
expireTime = tempParam/60;
Long minutes = math.mod(expireTime, 60);
tempParam = expireTime - minutes;
expireTime = expireTime/60;
Long hours = math.mod(expireTime, 24);
tempParam = expireTime - hours;
expireTime = expireTime/24;
Long days = math.mod(expireTime, 30);
system.debug(days + '.' + hours + ':' + minutes + ':' + seconds);
This should print: 0.0:1:9
Here's my solution using TimeUnit.
UPDATE: I should point out that this is written in groovy, but Java is almost identical.
def remainingStr = ""
/* Days */
int days = MILLISECONDS.toDays(remainingTime) as int
remainingStr += (days == 1) ? '1 Day : ' : "${days} Days : "
remainingTime -= DAYS.toMillis(days)
/* Hours */
int hours = MILLISECONDS.toHours(remainingTime) as int
remainingStr += (hours == 1) ? '1 Hour : ' : "${hours} Hours : "
remainingTime -= HOURS.toMillis(hours)
/* Minutes */
int minutes = MILLISECONDS.toMinutes(remainingTime) as int
remainingStr += (minutes == 1) ? '1 Minute : ' : "${minutes} Minutes : "
remainingTime -= MINUTES.toMillis(minutes)
/* Seconds */
int seconds = MILLISECONDS.toSeconds(remainingTime) as int
remainingStr += (seconds == 1) ? '1 Second' : "${seconds} Seconds"
A flexible way to do it :
(Not made for current date but good enough for durations)
/**
convert duration to a ms/sec/min/hour/day/week array
#param {int} msTime : time in milliseconds
#param {bool} fillEmpty(optional) : fill array values even when they are 0.
#param {string[]} suffixes(optional) : add suffixes to returned values.
values are filled with missings '0'
#return {int[]/string[]} : time values from higher to lower(ms) range.
*/
var msToTimeList=function(msTime,fillEmpty,suffixes){
suffixes=(suffixes instanceof Array)?suffixes:[]; //suffixes is optional
var timeSteps=[1000,60,60,24,7]; // time ranges : ms/sec/min/hour/day/week
timeSteps.push(1000000); //add very big time at the end to stop cutting
var result=[];
for(var i=0;(msTime>0||i<1||fillEmpty)&&i<timeSteps.length;i++){
var timerange = msTime%timeSteps[i];
if(typeof(suffixes[i])=="string"){
timerange+=suffixes[i]; // add suffix (converting )
// and fill zeros :
while( i<timeSteps.length-1 &&
timerange.length<((timeSteps[i]-1)+suffixes[i]).length )
timerange="0"+timerange;
}
result.unshift(timerange); // stack time range from higher to lower
msTime = Math.floor(msTime/timeSteps[i]);
}
return result;
};
NB : you could also set timeSteps as parameter if you want to control the time ranges.
how to use (copy an test):
var elsapsed = Math.floor(Math.random()*3000000000);
console.log( "elsapsed (labels) = "+
msToTimeList(elsapsed,false,["ms","sec","min","h","days","weeks"]).join("/") );
console.log( "half hour : "+msToTimeList(elsapsed,true)[3]<30?"first":"second" );
console.log( "elsapsed (classic) = "+
msToTimeList(elsapsed,false,["","","","","",""]).join(" : ") );
I suggest to use http://www.ocpsoft.org/prettytime/ library..
it's very simple to get time interval in human readable form like
PrettyTime p = new PrettyTime();
System.out.println(p.format(new Date()));
it will print like "moments from now"
other example
PrettyTime p = new PrettyTime());
Date d = new Date(System.currentTimeMillis());
d.setHours(d.getHours() - 1);
String ago = p.format(d);
then string ago = "1 hour ago"
In python 3 you could achieve your goal by using the following snippet:
from datetime import timedelta
ms = 536643021
td = timedelta(milliseconds=ms)
print(str(td))
# --> 6 days, 5:04:03.021000
Timedelta documentation: https://docs.python.org/3/library/datetime.html#datetime.timedelta
Source of the __str__ method of timedelta str: https://github.com/python/cpython/blob/33922cb0aa0c81ebff91ab4e938a58dfec2acf19/Lib/datetime.py#L607
Here is more precise method in JAVA , I have implemented this simple logic , hope this will help you:
public String getDuration(String _currentTimemilliSecond)
{
long _currentTimeMiles = 1;
int x = 0;
int seconds = 0;
int minutes = 0;
int hours = 0;
int days = 0;
int month = 0;
int year = 0;
try
{
_currentTimeMiles = Long.parseLong(_currentTimemilliSecond);
/** x in seconds **/
x = (int) (_currentTimeMiles / 1000) ;
seconds = x ;
if(seconds >59)
{
minutes = seconds/60 ;
if(minutes > 59)
{
hours = minutes/60;
if(hours > 23)
{
days = hours/24 ;
if(days > 30)
{
month = days/30;
if(month > 11)
{
year = month/12;
Log.d("Year", year);
Log.d("Month", month%12);
Log.d("Days", days % 30);
Log.d("hours ", hours % 24);
Log.d("Minutes ", minutes % 60);
Log.d("Seconds ", seconds % 60);
return "Year "+year + " Month "+month%12 +" Days " +days%30 +" hours "+hours%24 +" Minutes "+minutes %60+" Seconds "+seconds%60;
}
else
{
Log.d("Month", month);
Log.d("Days", days % 30);
Log.d("hours ", hours % 24);
Log.d("Minutes ", minutes % 60);
Log.d("Seconds ", seconds % 60);
return "Month "+month +" Days " +days%30 +" hours "+hours%24 +" Minutes "+minutes %60+" Seconds "+seconds%60;
}
}
else
{
Log.d("Days", days );
Log.d("hours ", hours % 24);
Log.d("Minutes ", minutes % 60);
Log.d("Seconds ", seconds % 60);
return "Days " +days +" hours "+hours%24 +" Minutes "+minutes %60+" Seconds "+seconds%60;
}
}
else
{
Log.d("hours ", hours);
Log.d("Minutes ", minutes % 60);
Log.d("Seconds ", seconds % 60);
return "hours "+hours+" Minutes "+minutes %60+" Seconds "+seconds%60;
}
}
else
{
Log.d("Minutes ", minutes);
Log.d("Seconds ", seconds % 60);
return "Minutes "+minutes +" Seconds "+seconds%60;
}
}
else
{
Log.d("Seconds ", x);
return " Seconds "+seconds;
}
}
catch (Exception e)
{
Log.e(getClass().getName().toString(), e.toString());
}
return "";
}
private Class Log
{
public static void d(String tag , int value)
{
System.out.println("##### [ Debug ] ## "+tag +" :: "+value);
}
}
A solution using awk:
$ ms=10000001; awk -v ms=$ms 'BEGIN {x=ms/1000;
s=x%60; x/=60;
m=x%60; x/=60;
h=x%60;
printf("%02d:%02d:%02d.%03d\n", h, m, s, ms%1000)}'
02:46:40.001
This one leaves out 0 values. With tests.
const toTimeString = (value, singularName) =>
`${value} ${singularName}${value !== 1 ? 's' : ''}`;
const readableTime = (ms) => {
const days = Math.floor(ms / (24 * 60 * 60 * 1000));
const daysMs = ms % (24 * 60 * 60 * 1000);
const hours = Math.floor(daysMs / (60 * 60 * 1000));
const hoursMs = ms % (60 * 60 * 1000);
const minutes = Math.floor(hoursMs / (60 * 1000));
const minutesMs = ms % (60 * 1000);
const seconds = Math.round(minutesMs / 1000);
const data = [
[days, 'day'],
[hours, 'hour'],
[minutes, 'minute'],
[seconds, 'second'],
];
return data
.filter(([value]) => value > 0)
.map(([value, name]) => toTimeString(value, name))
.join(', ');
};
// Tests
const hundredDaysTwentyHoursFiftyMinutesThirtySeconds = 8715030000;
const oneDayTwoHoursEightMinutesTwelveSeconds = 94092000;
const twoHoursFiftyMinutes = 10200000;
const oneMinute = 60000;
const fortySeconds = 40000;
const oneSecond = 1000;
const oneDayTwelveSeconds = 86412000;
const test = (result, expected) => {
console.log(expected, '- ' + (result === expected));
};
test(readableTime(
hundredDaysTwentyHoursFiftyMinutesThirtySeconds
), '100 days, 20 hours, 50 minutes, 30 seconds');
test(readableTime(
oneDayTwoHoursEightMinutesTwelveSeconds
), '1 day, 2 hours, 8 minutes, 12 seconds');
test(readableTime(
twoHoursFiftyMinutes
), '2 hours, 50 minutes');
test(readableTime(
oneMinute
), '1 minute');
test(readableTime(
fortySeconds
), '40 seconds');
test(readableTime(
oneSecond
), '1 second');
test(readableTime(
oneDayTwelveSeconds
), '1 day, 12 seconds');