I have the following method that I use to get the date for Monday and Friday of next week.
For example, today is 1/6/2017. If I ran it, I would hope to get the following results:
monday = 1/9/2017
friday = 1/13/2017
The method works fine if I run it earlier in the week, but if I run it later like a friday or saturday, it returns the monday and friday dates 2 weeks from now(not next week).
For example, running it today(Friday the 6th), I get the following results:
monday = 1/16/2017
friday = 1/20/2017
Here is the method:
public static DateTime NextWeekRange(DateTime start, DayOfWeek day)
{
var add_days = ((int)day - (int)start.DayOfWeek + 7) % 7;
return start.AddDays(add_days);
}
And I call it like this:
var monday = NextWeekRange(DateTime.Today.AddDays(i_today), DayOfWeek.Monday);
var friday = NextWeekRange(DateTime.Today.AddDays(i_today + 4), DayOfWeek.Friday);
I'm not quite sure what I got wrong, so another pair of eyes would help!
Thanks!
DateTime MyDate = DateTime.Now;
DateTime NextMondayDate;
DateTime NextFridayDate;
Boolean Test = true;
while (Test)
{
if (MyDate.DayOfWeek.ToString().ToUpper() == "MONDAY")
{
NextMondayDate = MyDate;
NextFridayDate = MyDate.AddDays(4);
Test = false;
}
else if (MyDate.DayOfWeek.ToString().ToUpper() == "FRIDAY")
{
NextFridayDate = MyDate;
NextMondayDate = MyDate.AddDays(3);
Test = false;
}
else
{
MyDate = MyDate.AddDays(1);
}
}
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
I have a date returned by a json, it is in the following variable as string:
val dateEvent = "2019-12-28 21:00:00"
The calculation I need is to know how many days hours minutes are left with the current date.
I have found some solutions but these use as input "2019-12-28" and I have my format with the time included.
java.time
Since Java 9 you can do (sorry that I can write only Java code):
DateTimeFormatter jsonFormatter = DateTimeFormatter.ofPattern("u-M-d H:mm:ss");
String dateEvent = "2019-12-28 21:00:00";
Instant eventTime = LocalDateTime.parse(dateEvent, jsonFormatter)
.atOffset(ZoneOffset.UTC)
.toInstant();
Duration timeLeft = Duration.between(Instant.now(), eventTime);
System.out.format("%d days %d hours %d minutes%n",
timeLeft.toDays(), timeLeft.toHoursPart(), timeLeft.toMinutesPart());
When I ran the code just now, the output was:
145 days 4 hours 19 minutes
In Java 6, 7 and 8 the formatting of the duration is a bit more wordy, search for how.
Avoid SimpleDateFormat and friends
The SimpleDateFormat and Date classes used in the other answer are poorly designed and long outdated. In my most honest opinion no one should use them in 2019. java.time, the modern Java date and time API, is so much nicer to work with.
Use the following function:
fun counterTime(eventtime: String): String {
var day = 0
var hh = 0
var mm = 0
try {
val dateFormat = SimpleDateFormat("yyyy-MM-dd HH:mm:ss")
val eventDate = dateFormat.parse(eventtime)
val cDate = Date()
val timeDiff = eventDate.time - cDate.time
day = TimeUnit.MILLISECONDS.toDays(timeDiff).toInt()
hh = (TimeUnit.MILLISECONDS.toHours(timeDiff) - TimeUnit.DAYS.toHours(day.toLong())).toInt()
mm =
(TimeUnit.MILLISECONDS.toMinutes(timeDiff) - TimeUnit.HOURS.toMinutes(TimeUnit.MILLISECONDS.toHours(timeDiff))).toInt()
} catch (e: ParseException) {
e.printStackTrace()
}
return if (day == 0) {
"$hh hour $mm min"
} else if (hh == 0) {
"$mm min"
} else {
"$day days $hh hour $mm min"
}
}
counterTime(2019-08-27 20:00:00)
This returns 24 days 6 hour 57 min
Note: The event date should always be a future date to the current date.
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 want to populate the weeks in a dropdown list when selecting month like
when I select
January
Then it will show only the weeks which starts from monday like
1st week:
Monday(02/01/2012)
2nd week:
Monday(09/01/2012)
3rd week:
Monday(16/01/2012)
4th week:
Monday(23/01/2012)
5th week:
Monday(30/01/2012)
So it sounds like you need to find the first Monday in the week, then just keep adding one week until you're not in the same month any more:
using System;
using System.Collections.Generic;
class Test
{
static void Main(string[] args)
{
foreach (DateTime date in GetMondays(2012, 1))
{
Console.WriteLine(date);
}
}
static IEnumerable<DateTime> GetMondays(int year, int month)
{
DateTime startOfMonth = new DateTime(year, month, 1);
// Get to the first Monday
int daysToMonday = DayOfWeek.Monday - startOfMonth.DayOfWeek;
// Now make sure it's non-negative...
int daysToNextMonday = (daysToMonday + 7) % 7;
// Add it to the start of the month to get to the first Monday
DateTime firstMonday = startOfMonth.AddDays(daysToNextMonday);
// Now yield and iterate until we're done
for (DateTime date = firstMonday;
date.Month == month;
date = date.AddDays(7))
{
yield return date;
}
}
}
(You could add the dates to a list and return that instead if you wanted... it doesn't make much difference really.)
simply parse everything from the starting date of the month to the ending date of the month.
checking DayOfWeek for every day and print out the results