I need help with some quick coding with google apps script, linking to my googlesheets spreadsheet.
In the googlespreadsheets, I have a cell with the value “26-Jun-2020”. It is a date.
I want to use google apps script to calculate the number of days difference between that date (“26-Jun-2020”) and today’s day, but it won’t do the calculation for me somehow.
If I print only “expiry_date[i]” using Logger.log(expiry_date[i]), it will provide the output “Fri Dec 17 2021 01:00:00 GMT-0500 (Eastern Standard Time) “
function Put_Options_Expiry_Alert() {
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheet = ss.getSheetByName("Long equity (sell puts)");
//var timeZone = AdsApp.currentAccount().getTimeZone(); //Get timezone of current spreadsheet
var status = sheet.getRange("F:F").getValues();
var expiry_date = sheet.getRange("M:M").getValues();
var potential_capitaloutlay_USD = sheet.getRange("Z:Z").getValues();
Logger.log("Length of status = " + status.length);
Logger.log("Length of expiry_date = " + expiry_date.length);
Logger.log("Length of potential_capitaloutlay_USD = " + potential_capitaloutlay_USD.length);
for (var i = 0; i < status.length; i++) {
if (status[i] == "Entered") { //Evaluate if this is a live Put position
//Calculate the time difference of two dates using date2. getTime() – date1. getTime();
//Calculate the no. of days between two dates, divide the time difference of both the dates by no. of milliseconds in a day (1000*60*60*24)
Logger.log("expiry date is = "+expiry_date[i]);
Logger.log("today's date is = "+Date());
var days_to_expiry = dateDiffInDays(expiry_date[i],Date());
Logger.log(days_to_expiry);
}
}
}
// Function that returns the number of days difference between DateA and DateB
// DateA and DateB are javascript Date objects
function dateDiffInDays(DateA, DateB) {
var milliseconds_per_day = 1000 * 24 * 60; // number of milliseconds in a day
const utcA = Date.UTC(2021, DateA.getMonth(), DateA.getDate());
const utcB = Date.UTC(2020, DateB.getMonth(), DateB.getDate());
return Math.floor((utc2 - utc1) / milliseconds_per_day);
}
function timeDiffDays(Start, End) {
var day = 86400000;
var t1 = new Date(Start).valueOf();
var t2 = new Date(End).valueOf();
var d = t2 - t1;
return Math.floor(d / day);
}
I'm looking for a way to use DateTime to parse two dates, to show the difference.
I want to have it on the format: "X years, Y months, Z days".
For JS, we have momentjs library and following code::
var a = moment([2015, 11, 29]);
var b = moment([2007, 06, 27]);
var years = a.diff(b, 'year');
b.add(years, 'years');
var months = a.diff(b, 'months');
b.add(months, 'months');
var days = a.diff(b, 'days');
console.log(years + ' years ' + months + ' months ' + days + ' days');
// 8 years 5 months 2 days
Is there similar library available for dart that can help achieve this usecase?
I think it is not possible to do exactly what you want easily with DateTime. Therefore you can use https://pub.dev/packages/time_machine package that is quite powerful with date time handling:
import 'package:time_machine/time_machine.dart';
void main() {
LocalDate a = LocalDate.today();
LocalDate b = LocalDate.dateTime(DateTime(2022, 1, 2));
Period diff = b.periodSince(a);
print("years: ${diff.years}; months: ${diff.months}; days: ${diff.days}");
}
for hours/minutes/seconds precision:
import 'package:time_machine/time_machine.dart';
void main() {
LocalDateTime a = LocalDateTime.now();
LocalDateTime b = LocalDateTime.dateTime(DateTime(2022, 1, 2, 10, 15, 47));
Period diff = b.periodSince(a);
print("years: ${diff.years}; months: ${diff.months}; days: ${diff.days}; hours: ${diff.hours}; minutes: ${diff.minutes}; seconds: ${diff.seconds}");
}
What you are looking for is the Dart DateTime class
You can get close to what you want in moment.js with
main() {
var a = DateTime.utc(2015, 11, 29);
var b = DateTime.utc(2007, 06, 27);
var years = a.difference(b);
print(years.inDays ~/365);
}
There is no inYears or inMonths option for DateTime though that's why the year is divided in the print.
the difference function returns the difference in seconds so you have to process it yourself to days.
You could write an extension on duration class to format it:
extension DurationExtensions on Duration {
String toYearsMonthsDaysString() {
final years = this.inDays ~/ 365
// You will need a custom logic for the months part, since not every month has 30 days
final months = (this.inDays ~% 365) ~/ 30
final days = (this.inDays ~% 365) ~% 30
return "$years years $months months $days days";
}
}
The usage will be:
final date1 = DateTime()
final date2 = DateTime()
date1.difference(date2).toYearsMonthsDaysString()
You can use Jiffy Package for this like this
var jiffy1 = Jiffy("2008-10", "yyyy-MM");
var jiffy2 = Jiffy("2007-1", "yyyy-MM");
jiff1.diff(jiffy2, Units.YEAR); // 1
jiff1.diff(jiffy2, Units.YEAR, true);
You can calculate from the total number of days:
void main() {
DateTime a = DateTime(2015, 11, 29);
DateTime b = DateTime(2007, 06, 27);
int totalDays = a.difference(b).inDays;
int years = totalDays ~/ 365;
int months = (totalDays-years*365) ~/ 30;
int days = totalDays-years*365-months*30;
print("$years $months $days $totalDays");
}
Result is: 8 5 7 3077
I created my own class for Gregorian Dates, and I created a method which handle this issue, it calculates "logically" the difference between two dates in years, months, and days...
i actually created the class from scratch without using any other packages (including DateTime package) but here I used DateTime package to illustrate how this method works.. until now it works fine for me...
method to determine if it's a leap year or no:
static bool leapYear(DateTime date) {
if(date.year%4 == 0) {
if(date.year%100 == 0){
return date.year%400 == 0;
}
return true;
}
return false;
}
this is the method which calculates the difference between two dates in years, months, and days. it puts the result in a list of integers:
static List<int> differenceInYearsMonthsDays(DateTime dt1, DateTime dt2) {
List<int> simpleYear = [31,28,31,30,31,30,31,31,30,31,30,31];
if(dt1.isAfter(dt2)) {
DateTime temp = dt1;
dt1 = dt2;
dt2 = temp;
}
int totalMonthsDifference = ((dt2.year*12) + (dt2.month - 1)) - ((dt1.year*12) + (dt1.month - 1));
int years = (totalMonthsDifference/12).floor();
int months = totalMonthsDifference%12;
late int days;
if(dt2.day >= dt1.day) {days = dt2.day - dt1.day;}
else {
int monthDays = dt2.month == 3
? (leapYear(dt2)? 29: 28)
: (dt2.month - 2 == -1? simpleYear[11]: simpleYear[dt2.month - 2]);
int day = dt1.day;
if(day > monthDays) day = monthDays;
days = monthDays - (day - dt2.day);
months--;
}
if(months < 0) {
months = 11;
years--;
}
return [years, months, days];
}
the method which calculates the difference between two dates in months, and days:
static List<int> differenceInMonths(DateTime dt1, DateTime dt2){
List<int> inYears = differenceInYearsMonthsDays(dt1, dt2);
int difMonths = (inYears[0]*12) + inYears[1];
return [difMonths, inYears[2]];
}
the method which calculates the difference between two dates in days:
static int differenceInDays(DateTime dt1, DateTime dt2) {
if(dt1.isAfter(dt2)) {
DateTime temp = dt1;
dt1 = dt2;
dt2 = temp;
}
return dt2.difference(dt1).inDays;
}
usage example:
void main() {
DateTime date1 = DateTime(2005, 10, 3);
DateTime date2 = DateTime(2022, 1, 12);
List<int> diffYMD = GregorianDate.differenceInYearsMonthsDays(date1, date2);
List<int> diffMD = GregorianDate.differenceInMonths(date1, date2);
int diffD = GregorianDate.differenceInDays(date1, date2);
print("The difference in years, months and days: ${diffYMD[0]} years, ${diffYMD[1]} months, and ${diffYMD[2]} days.");
print("The difference in months and days: ${diffMD[0]} months, and ${diffMD[1]} days.");
print("The difference in days: $diffD days.");
}
output:
The difference in years, months and days: 16 years, 3 months, and 9 days.
The difference in months and days: 195 months, and 9 days.
The difference in days: 5945 days.
the answer is yes, you can easilly achieve it with DateTime class in Dart. See: https://api.dart.dev/stable/2.8.3/dart-core/DateTime-class.html
Example
void main() {
var moonLanding = DateTime(1969,07,20)
var marsLanding = DateTime(2024,06,10);
var diff = moonLanding.difference(marsLanding);
print(diff.inDays.abs());
print(diff.inMinutes.abs());
print(diff.inHours.abs());
}
outputs:
20049
28870560
481176
final firstDate = DateTime.now();
final secondDate = DateTime(firstDate.year, firstDate.month - 20);
final yearsDifference = firstDate.year - secondDate.year;
final monthsDifference = (firstDate.year - secondDate.year) * 12 +
firstDate.month - secondDate.month;
final totalDays = firstDate.difference(secondDate).inDays;
Simple approach, no packages needed.
try intl package with the following code:
import 'package:intl/intl.dart';
String startDate = '01/01/2021';
String endDate = '01/01/2022';
final start = DateFormat('dd/MM/yyyy').parse(startDate);
final end = DateFormat('dd/MM/yyyy').parse(endDate);
Then, you can calculate the duration between the two dates with the following code:
final duration = end.difference(start);
To obtain the number of years, months and days, you can do the following:
final years = duration.inDays / 365;
final months = duration.inDays % 365 / 30;
final days = duration.inDays % 365 % 30;
Finally, you can use these variables to display the result in the desired format:
final result = '${years.toInt()} years ${months.toInt()} months y ${days.toInt()} days';
DateTime difference in years is a specific function, like this:
static int getDateDiffInYear(DateTime dateFrom, DateTime dateTo) {
int sign = 1;
if (dateFrom.isAfter(dateTo)) {
DateTime temp = dateFrom;
dateFrom = dateTo;
dateTo = temp;
sign = -1;
}
int years = dateTo.year - dateFrom.year;
int months = dateTo.month - dateFrom.month;
if (months < 0) {
years--;
} else {
int days = dateTo.day - dateFrom.day;
if (days < 0) {
years--;
}
}
return years * sign;
}
difHour = someDateTime.difference(DateTime.now()).inHours;
difMin = (someDateTime.difference(DateTime.now()).inMinutes)-(difHour*60);
and same for years and days
in flutter we can get current month using this
var now = new DateTime.now();
var formatter = new DateFormat('MM');
String month = formatter.format(now);
But how to get the last month date? Especially if current date is January (01). we can't get the right month when we use operand minus (-) , like month - 1.
You can just use
var prevMonth = new DateTime(date.year, date.month - 1, date.day);
with
var date = new DateTime(2018, 1, 13);
you get
2017-12-13
It's usually a good idea to convert to UTC and then back to local date/time before doing date calculations to avoid issues with daylight saving and time zones.
We can calculate both first day of the month and the last day of the month:
DateTime firstDayCurrentMonth = DateTime.utc(DateTime.now().year, DateTime.now().month, 1);
DateTime lastDayCurrentMonth = DateTime.utc(DateTime.now().year, DateTime.now().month + 1).subtract(Duration(days: 1));
DateTime.utc takes in integer values as parameters: int year, int month, int day and so on.
Try this package, Jiffy, it used momentjs syntax. See below
Jiffy().subtract(months: 1);
Where Jiffy() returns date now. You can also do the following, the same result
var now = DateTime.now();
Jiffy(now).subtract(months: 1);
We can use the subtract method to get past month date.
DateTime pastMonth = DateTime.now().subtract(Duration(days: 30));
Dates are pretty hard to calculate. There is an open proposal to add support for adding years and months here https://github.com/dart-lang/sdk/issues/27245.
There is a semantic problem with adding months and years in that "a
month" and "a year" isn't a specific amount of time. Years vary by one
day, months by up to three days. Adding "one month" to the 30th of
January is ambiguous. We can do it, we just have to pick some
arbitrary day between the 27th of February and the 2nd of March.
That's why we haven't added month and year to Duration - they do not
describe durations.
You can use the below code to add months in a arbitrary fashion (I presume its not completely accurate. Taken from the issue)
const _daysInMonth = const [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
bool isLeapYear(int value) =>
value % 400 == 0 || (value % 4 == 0 && value % 100 != 0);
int daysInMonth(int year, int month) {
var result = _daysInMonth[month];
if (month == 2 && isLeapYear(year)) result++;
return result;
}
DateTime addMonths(DateTime dt, int value) {
var r = value % 12;
var q = (value - r) ~/ 12;
var newYear = dt.year + q;
var newMonth = dt.month + r;
if (newMonth > 12) {
newYear++;
newMonth -= 12;
}
var newDay = min(dt.day, daysInMonth(newYear, newMonth));
if (dt.isUtc) {
return new DateTime.utc(
newYear,
newMonth,
newDay,
dt.hour,
dt.minute,
dt.second,
dt.millisecond,
dt.microsecond);
} else {
return new DateTime(
newYear,
newMonth,
newDay,
dt.hour,
dt.minute,
dt.second,
dt.millisecond,
dt.microsecond);
}
}
To get a set starting point at the start of a month, you can use DateTime along with the Jiffy package.
DateTime firstOfPreviousMonth
= DateTime.parse(
Jiffy().startOf(Units.MONTH)
.subtract(months: 1)
.format('yyyy-MM-dd'). //--> Jan 1 '2021-01-01 00:00:00.000'
);
var fifthOfMonth
= firstOfPreviousMonth.add(Duration(days: 4)); //--> Jan 5 '2021-01-05 00:00:00.000'
or
DateTime endOfPreviousMonth
= DateTime.parse(
Jiffy().endOf(Units.MONTH)
.subtract(months: 2)
.format('yyyy-MM-dd'). //--> Dec 30 '2020-12-31 00:00:00.000'
// endOf always goes to 30th
);
var previousMonth
= endOfPreviousMonth.add(Duration(days: 2)); //--> Jan 1 '2021-01-01 00:00:00.000'
DateFormat('MMMM yyyy')
.format(DateTime(DateTime.now().year, DateTime.now().month - 2)),
List<DateTime> newList = [];
DateFormat format = DateFormat("yyyy-MM-dd");
for (var i = 0; i < recents.length; i++) {
newList.add(format.parse(recents[i]['date'].toString()));
}
newList.sort(((a, b) => a.compareTo(b)));
var total = 0;
for (var i = 0; i < newList.length; i++) {
if (DateTime.now().difference(newList[i]).inDays < 30) {
print(newList[i]);
total++;
}
}
print(total);
You can use this to fetch the last 30 days.
In addition to Günter Zöchbauer Answer
var now = new DateTime.now();
String g = ('${now.year}/ ${now.month}/ ${now.day}');
print(g);
I am using Asp:calendar. I need to find the day of week of the date selected.
For example for the month of February,2014 if 27th is selected it should give value as 4th Thursday. Please help in this
You can get the Day Of Week through DateTime.DayOfWeek property. like:
DayOfWeek daOfWeek = youCalendar.SelectedDate.DayOfWeek;
To calculate the occurrence of the day in that particular month, Create a new DateTime starting from 1st of the month, Loop through to SelectedDate check for DayOfWeek equals to SelectedDate.DayOfWeek and increment a counter.
DateTime selecteDate = yourCalendar.SelectedDate;
DateTime startDate = new DateTime(selecteDate.Year, selecteDate.Month, 1);
int counter = 0;
while (startDate <= selecteDate)
{
if (startDate.DayOfWeek == selecteDate.DayOfWeek)
counter++;
startDate = startDate.AddDays(1);
}
I need to setup last week , last month periods on changing dropdownlist
I'm making
switch (DDL.SelectedIndex)
{
case 0:
{
// last week
this.TextBox3.Text = DateTime. //Previos week first day
this.TextBox4.Text = DateTime. //Previos week last day
} break;
case 1:
{
// last mouth
this.TextBox3.Text = DateTime.// Previos month first day
this.TextBox4.Text = DateTime.// Previos month last day
} break;
}
So is there some ways how can I select date values like I want ?
also , I've got AJAX calendar extender on text boxes
thank you.
Something like this, I think:
int dayOfWeekNumber = (int)DateTime.Today.DayOfWeek - (int)CultureInfo.CurrentCulture.DateTimeFormat.FirstDayOfWeek;
var previosWeekFirstDay = DateTime.Today.AddDays(-7 - dayOfWeekNumber);
var previosWeekLastDay = previosWeekFirstDay.AddDays(6);
var previosMonthFirstDay = DateTime.Today.AddMonths(-1);
previosMonthFirstDay = previosMonthFirstDay.AddDays(-previosMonthFirstDay.Day + 1);
var previosMonthLastDay = previosMonthFirstDay.AddDays(DateTime.DaysInMonth(previosMonthFirstDay.Year, previosMonthFirstDay.Month) - 1);
Edited: see Fredrik Mörk comment.