I have two dates in EPOCH value.
Open : 1579269496000
Close : 1579270005225
I want to get display different between two dates.
So difference = Close - Open = <ddd> Days <hh> Hours <mm> Mins <ss> Sec.
I'm using Moment.js to convert the date but I don't see to substract EPOCH date using that.
var c = new Date(close);
var o = new Date(open);
var seconds =enter code here (c.getTime() - o.getTime()) / 1000;
var ms = moment(close,"DD/MM/YYYY HH:mm:ss").diff(moment(open,"DD/MM/YYYY HH:mm:ss"));
Hope this solves your concern.
First convert epoch value to the value supported by momentJS. Then you could get the difference in years and add that to the initial date; then get the difference in weeks and add that to the initial date again.
var moment = require("moment");
var a = "1582113418";
var b = "1582113444";
var aa = moment.unix(a); // converted value
var bb = moment.unix(b); // converted value
var days = bb.diff(aa, "days");
aa.add(days, "years");
var hours = bb.diff(aa, "hour");
aa.add(hours, "hours");
var seconds = bb.diff(aa, "seconds");
console.log(days + " days " + hours + " hours " + seconds + " seconds");
Working sandbox: https://codesandbox.io/s/adoring-shamir-0yuxq
Related
I have a one time column in format 00:01:30 (HH:MM:SS). This column is a text format column.
this text column format convert to a time format column and create a new measure for the total time sum.
You can easily convert your Text column into Time using
Time = TIMEVALUE('Table'[Text])
But the problem is that the Time format doesn't support more than 24 hours, so your SUM will potentially lead to an overflow. Here's a workaround:
Create a calculated "Seconds" Column
Seconds =
VAR Time =
TIMEVALUE('Table'[Text])
RETURN
HOUR(Time) * 3600 + MINUTE(Time) * 60 + SECOND(Time)
Aggregate the Seconds with this Measure and convert back to a "Duration-like" format:
Total Duration =
VAR total_seconds =
SUM('Table'[Seconds])
VAR days =
QUOTIENT(total_seconds, 24 * 60 *60)
VAR rest1 =
MOD(total_seconds, 24 * 60 * 60)
VAR hours =
QUOTIENT(rest1, 60 * 60)
VAR rest2 =
MOD(total_seconds, 60 * 60)
VAR minutes =
QUOTIENT(rest2, 60)
VAR seconds =
MOD(rest2, 60)
RETURN
days & "." & FORMAT(hours, "0#") & ":" & FORMAT(minutes, "0#") & ":" & FORMAT(seconds, "0#")
I'm trying to build my first app using flutter framework. The app is about my "End of Year Challenge". It started from 1st Sept 2019 and will last till the end of this year.
What I'm trying to achieve is - I want to display the current day number of the challenge period. eg: 1st Sept is Day 1, 30th Sept is Day 30 and 1st Oct is Day 31 and so on.
I'm trying to get the first day of Sept and assign it to 1. Then using a loop I want the app to update the day to the current day. The loop will stop once the current day equals to 122 (as this would be the last day of the challenge)
Here's the screenshot of the UI
final firstSeptember = DateTime.utc(2019, DateTime.september, 1);
static const totalNumberOfDays = 122;
int noOfDay(){
int dayOne = firstSeptember.day; // I'm just trying codes, IDK the actual code/business logic
return dayOne;
}
In function you'd use
int noOfDay(){
var todayDate = DateTime.now();
final firstSeptember = DateTime.utc(2019, DateTime.september, 1);
var difference = todayDate.difference(firstSeptember);
return difference.inDays + 1;
}
Explanation:
Get today's date
var todayDate = DateTime.now();
You already have start date which is
final firstSeptember = DateTime.utc(2019, DateTime.september, 1);
All you need to do is subtraction.
var difference = todayDate.difference(firstSeptember);
int daysCompleted = difference.inDays + 1;
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.
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();
}
}
What I am trying to do here is this - I want to give index to only the workdays in each week.
So, if in a week, Monday and Wednesday are holidays, then Tuesday should get 1, Thursday should get 2, Friday should get the index 3. Otherwise, in a normal week without any holidays, Monday should get 1, Tuesday 2, Wednesday 3, and so on ...
Here is the code I have written (I haven't coded in years now, so please pardon the crude approach)
Sheet 'Holidays' contains a list of holidays in the column B starting from row 2
Variable date is the date for which I want to find out the index for
Variable dayOfTheWeek is the number of day of 'date' counted from last Sunday, so if date is a Monday, dayOfTheWeek is 1; if date is Tuesday, dayOfTheWeek is 2, and so on ...
function indexOfWorkdayOfTheWeek (date, dayOfTheWeek, lastSundayDate)
{
var activeSheet = SpreadsheetApp.getActiveSpreadsheet();
var activeCell = activeSheet.getActiveRange();
var activeRow = activeCell.getRowIndex();
var activeColumn = activeCell.getColumn();
var count = 1;
for (var j = 1; j < dayOfTheWeek; j++)
{
var date2 = lastSundayDate.valueOf() + j*86400;
Logger.log('Date ' + j + ' is:' + date2);
Logger.log('Last Sunday is:' + lastSundayDate);
if (holidayOrNot(date2) == true)
{
}
else
{
count = count + 1;
}
}
return count;
}
function holidayOrNot(date2)
{
var holidaysSheet = SpreadsheetApp.getActiveSpreadsheet().getSheetByName('Holidays');
var listOfHolidays = holidaysSheet.getSheetValues(2, 2, 95, 1);
var isDateMatch = false;
for (var k = 0; k < 90; k++)
{
if (date2 == listOfHolidays[k].valueOf())
{
isDateMatch = true;
break;
}
else
{
continue;
}
}
return isDateMatch;
}
I think the problem is two-fold here:
The date2 calculation isn't working for some reason (var date2 = lastSundayDate.valueOf() + j*86400;)
The function holidayOrNot is returning false, no matter what, even if it encounters a holiday ... the condition date2 == listOfHolidays[k] isn't working for some reason...
Help would be appreciated!
maybe this method below could help you in your calculations, it returns an integer corresponding to the day of the year so if you apply this to your holidays days and compare to the days of interest it could be a good way to find matches.
here it is, just add these lines outside of any function in your script (so you can use it anywhere) then use it like this :
var d = new Date().getDOY();
Logger.log(d)
Here the method :
Date.prototype.getDOY = function() {
var onejan = new Date(this.getFullYear(),0,1);
return Math.ceil((this - onejan) / 86400000);
}
Assuming that lastSundayDate is being passed around correctly, I see a glaring problem:
lastSundayDate.valueOf().
valueOf() on Date objects returns the primitive value... it looks like you're going for adding a day to the date (86400 seconds * j)? I can't tell what the logic is supposed to be here. But the valueOf() date2 is definitely giving you an integer something like: 1384628769399 (see here).
What you really want to accomplish is something like Date.getDay(), or something similar so that you can add hours, days, etc. to the original Date. This is likely the source of all your problems.
What you can do is read the Mozilla Developer Network documentation on Date objects to see all of the functions on Dates and their uses. You can greatly simplify what you're trying to do by using these functions, instead of doing abstract operations like j * 86400.
It should also be noted that you can do simple operations such as the following, to add 4 hours to the current Date (time):
var myDate = new Date();
Logger.log(myDate); // ~ console.write
var laterDate = new Date(myDate.setHours(myDate.getHours() + 4));
Logger.log(laterDate); // ~ console.write
which gives the following:
[13-11-16 14:13:38:947 EST] Sat Nov 16 14:13:38 GMT-05:00 2013
[13-11-16 14:13:38:954 EST] Sat Nov 16 18:13:38 GMT-05:00 2013
Working with dates can be tricky - but it's always best to use the simplest methods that are available, which are built into the Date objects themselves. There are also numerous other libraries that provide extended functionality for Dates such as Date js.
If you're still running into your problem after attempting to try using methods I displayed above, please run your script and post both the Execution Transcript and the content of the Logger so that I can help you narrow down the issue :)