I have a form and on a button click I automatically generate a date exactly a year in the future. I would like to know how to make sure that this date is not on a public holiday or on a weekend. Any help please ?
I would like to store the value of the date inside a variable in order to just place it inside the command.Parameters.AddWithValue("#DueDate",___________)
You can create your own logic for this. It's quite simple. Create a method that check if the date is a weekday or a holiday. But you have to hard-code the holidays since they differ per country/continent/culture etc.
public bool IsWeekday(DateTime date)
{
int dayOfWeek = (int)date.DayOfWeek;
//week starts on sunday
if (dayOfWeek == 0 || dayOfWeek == 6)
{
return false;
}
else
{
return true;
}
}
public bool IsHoliday(DateTime date)
{
int currentYear = DateTime.Now.Year;
//define your holidays here, they differ between cultures and continents etc
List<DateTime> holidays = new List<DateTime>()
{
new DateTime(currentYear, 1, 1), //new years day
new DateTime(currentYear, 1, 9), //for testing
new DateTime(currentYear, 4, 27), //kings day
new DateTime(currentYear, 6, 21), //longest day of the year
new DateTime(currentYear, 12, 25), //christmas
new DateTime(currentYear, 12, 26) //christmas
};
//check the date against the list of holidays
if (holidays.Any(x => x == date.Date))
{
return true;
}
else
{
return false;
}
}
Now you can check if it works.
//get a monday
DateTime monday = new DateTime(2019, 1, 7);
//loop all days of the week
for (int i = 0; i < 7; i++)
{
DateTime nextDay = monday.AddDays(i);
Label1.Text += string.Format("{0} - {1} - {2}<br>", nextDay.ToLongDateString(), IsWeekday(nextDay), IsHoliday(nextDay));
}
Result of the above loop
maandag 7 januari 2019 - True - False
dinsdag 8 januari 2019 - True - False
woensdag 9 januari 2019 - True - True
donderdag 10 januari 2019 - True - False
vrijdag 11 januari 2019 - True - False
zaterdag 12 januari 2019 - False - False
zondag 13 januari 2019 - False - False
Related
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
If you pass a non-existing/non-real date like: '20181364' (2018/13/64) into DateTime (constructor or parse-method), no exception is thrown. Instead a calculated DateTime is returned.
Example:
'20181364' --> 2019-03-05 00:00:00.000
How can I check if a given date really exists/is valid?
I tried to solve this using DartPad (without success), so no Flutter doctor output required here.
void main() {
var inputs = ['20180101', // -> 2018-01-01 00:00:00.000
'20181231', // -> 2018-12-31 00:00:00.000
'20180230', // -> 2018-03-02 00:00:00.000
'20181301', // -> 2019-01-01 00:00:00.000
'20181364'];// -> 2019-03-05 00:00:00.000
inputs.forEach((input) => print(convertToDate(input)));
}
String convertToDate(String input){
return DateTime.parse(input).toString();
}
It would be great if there exist some kind of method to check if a given date really exists/is valid, e.g.:
a validate function in DateTime
another lib that does not use DateTime.parse() for validation
How would you solve this?
You can convert parsed date to string with original format and then compare if it's matching the input.
void main() {
var inputs = ['20180101', // -> 2018-01-01 00:00:00.000
'20181231', // -> 2018-12-31 00:00:00.000
'20180230', // -> 2018-03-02 00:00:00.000
'20181301', // -> 2019-01-01 00:00:00.000
'20181364'];// -> 2019-03-05 00:00:00.000
inputs.forEach((input) {
print("$input is valid string: ${isValidDate(input)}");
});
}
bool isValidDate(String input) {
final date = DateTime.parse(input);
final originalFormatString = toOriginalFormatString(date);
return input == originalFormatString;
}
String toOriginalFormatString(DateTime dateTime) {
final y = dateTime.year.toString().padLeft(4, '0');
final m = dateTime.month.toString().padLeft(2, '0');
final d = dateTime.day.toString().padLeft(2, '0');
return "$y$m$d";
}
My solution to validate the birthday was this, we can see that it has the leap year calculation.
class DateHelper{
/*
* Is valid date and format
*
* Format: dd/MM/yyyy
* valid:
* 01/12/1996
* invalid:
* 01/13/1996
*
* Format: MM/dd/yyyy
* valid:
* 12/01/1996
* invalid
* 13/01/1996
* */
static bool isValidDateBirth(String date, String format) {
try {
int day, month, year;
//Get separator data 10/10/2020, 2020-10-10, 10.10.2020
String separator = RegExp("([-/.])").firstMatch(date).group(0)[0];
//Split by separator [mm, dd, yyyy]
var frSplit = format.split(separator);
//Split by separtor [10, 10, 2020]
var dtSplit = date.split(separator);
for (int i = 0; i < frSplit.length; i++) {
var frm = frSplit[i].toLowerCase();
var vl = dtSplit[i];
if (frm == "dd")
day = int.parse(vl);
else if (frm == "mm")
month = int.parse(vl);
else if (frm == "yyyy")
year = int.parse(vl);
}
//First date check
//The dart does not throw an exception for invalid date.
var now = DateTime.now();
if(month > 12 || month < 1 || day < 1 || day > daysInMonth(month, year) || year < 1810 || (year > now.year && day > now.day && month > now.month))
throw Exception("Date birth invalid.");
return true;
} catch (e) {
return false;
}
}
static int daysInMonth(int month, int year) {
int days = 28 + (month + (month/8).floor()) % 2 + 2 % month + 2 * (1/month).floor();
return (isLeapYear(year) && month == 2)? 29 : days;
}
static bool isLeapYear(int year)
=> (( year % 4 == 0 && year % 100 != 0 ) || year % 400 == 0 );
}
Support for validating dates was added for dart in December 2020: https://pub.dev/documentation/intl/latest/intl/DateFormat/parseStrict.html
I am trying to get date range for a particular week number.
Following is my code to get week number with respect to current date
final date = DateTime.now();
final startOfYear = new DateTime(date.year, 1, 1, 0, 0);
final firstMonday = startOfYear.weekday;
final daysInFirstWeek = 8 - firstMonday;
final diff = date.difference(startOfYear);
var weeks = ((diff.inDays - daysInFirstWeek) / 7).ceil();
if (daysInFirstWeek > 3) {
weeks += 1;
}
print("Week Range $weeks");
What I don't understand is how I get start and end date for a particular week number.
Any help would be appreciated.
Assuming your week is zero based, you need to do something like this:
final date = new DateTime.now();
final startOfYear = new DateTime(date.year, 1, 1, 0, 0);
final firstMonday = startOfYear.weekday;
final daysInFirstWeek = 8 - firstMonday;
final diff = date.difference(startOfYear);
var weeks = ((diff.inDays - daysInFirstWeek) / 7).ceil();
main(){
int week = weeks;
print("Start Date for week $week: ${startOfYear.add(Duration(days: 7*week))}");
print("End Date for week $week: ${startOfYear.add(Duration(days: 7*week+6))}");
}
To get the starting and day of week from week number and year.
DateTime getDateByWeekNumber({
int weeknumber,
int year,
bool start
}) {
//check if start == true retrun start date of week
//else return end date
var days = ((weeknumber - 1) * 7) + (start ? 0 : 6);
return DateTime.utc(year, 1, days);
}
This is in accordance to ISO 8601. You can very week number here
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 would like to subtract days from the current date in TypeScript.
For example, if the current date is October 1st, 2017, I would like to subtract 1 day to get September 30th, 2017, or if I want to subtract 3 days I would get September 28th etc.
This is what I have so far, the result is I received December 31st, 1969. Which I assume means that tempDate.getDate() is returning zero, as in the Epoch of January 1, 1970.
This is my code, the goal is to return the previous working day.
protected generateLastWorkingDay(): Date {
var tempDate = new Date(Date.now());
var day = tempDate.getDay();
//** if Monday, return Friday
if (day == 1) {
tempDate = new Date(tempDate.getDate() - 3);
} else if (1 < day && day <= 6) {
tempDate = new Date(tempDate.getDate() - 1);
}
return tempDate;
}
getDate returns the date of the month (1-31), so creating a new Date from it treats that number as "milliseconds since epoch".
What you probably want is to use setDate to change the date as it automatically handled going backwards through months/years.
protected generateLastWorkingDay(): Date {
const lastWorkingDay = new Date();
while(!this.isWorkingDay(lastWorkingDay)) {
lastWorkingDay.setDate(lastWorkingDay.getDate()-1);
}
return lastWorkingDay;
}
private isWorkingDay(date: Date) {
const day = date.getDay();
const isWeekday = (day > 0 && day < 6);
return isWeekday; // && !isPublicHoliday?
}
This is how I did
let yesterday=new Date(new Date().getTime() - (1 * 24 * 60 * 60 * 1000));
let last3days=new Date(new Date().getTime() - (3 * 24 * 60 * 60 * 1000));
We need to minus (no_of_days) * 24 * 60 * 60 * 1000 from current date.
You can just
const current = new Date()
and then
const numberOfDaysToSubstract= 3;
const prior = new Date().setDate(current.getDate) - numberOfDaysToSubstract);
you can see an example of this here
https://codepen.io/Jeysoon/pen/poNZRwd?editors=1112