Fairly new to PS, I want to batch file on Sunday between the 1st and the 2nd Tuesday of the month.
I know how to find the the 1st and 2nd Tuesdays I am looking for, cannot figure out the rest.
$FindNthDay = 1
$WeekDay = 'Tuesday'
[datetime]$Today = [datetime]::NOW
$todayM = $Today.Month.ToString()
$todayY = $Today.Year.ToString()
[datetime]$StrtMonth1 = $todayM + '/1/' + $todayY
while ($StrtMonth1.DayofWeek -ine $WeekDay ) { $StrtMonth1 = $StrtMonth1.AddDays(1) }
$StrtMonth1.AddDays(7 * ($FindNthDay - 1))
#
$FindNthDay = 2
$WeekDay = 'Tuesday'
[datetime]$Today = [datetime]::NOW
$todayM = $Today.Month.ToString()
$todayY = $Today.Year.ToString()
[datetime]$StrtMonth = $todayM + '/1/' + $todayY
while ($StrtMonth.DayofWeek -ine $WeekDay ) { $StrtMonth = $StrtMonth.AddDays(1) }
$StrtMonth.AddDays(7 * ($FindNthDay - 1))
I know how to find the the 1st and 2nd Tuesdays
Since there's only one Sunday in between, you only need to find the first one:
# Get-Date -Day 1 will give us the 1st of the current month
$firstTuesday = Get-Date -Day 1
while($firstTuesday.DayOfWeek -ne 'Tuesday') {
$firstTuesday = $firstTuesday.AddDays(1)
}
And then add another 5 days:
$sundayAfterFirstTuesday = $firstTuesday.AddDays(5).Date
Which (in January 2021) gives us:
PS ~> $sundayAfterFirstTuesday
Sunday, January 10, 2021 12:00:00 AM
In addition to Mathias's fine answer, I got hung up on this. Given that .DayOfWeek is an [enum] that's easily converted to a [Int] I was looking for a mathematically concise way to derive the first Tuesday. Since it's fairly obvious how to then find the following Sunday by just doing .AddDays(5) .
Honestly, I was stumbling a bit because While .DayOfWeek is 0 - 6 how many days to add depends on if the current day, in this case the first of the month, is less than or greater than Tuesday (2). It was worth playing around; here are 2 alternate approaches I came up with:
Example 1: A switch statement that's not at all concise but is very readable:
$Day1 = Get-Date "1/1/2021"
Switch ( $Day1.DayOfWeek )
{
'Sunday' { $Sunday = $Day1.Adddays(2).AddDays( 5 ); Break }
'Monday' { $Sunday = $Day1.Adddays(1).AddDays( 5 ); Break }
'Tuesday' { $Sunday = $Day1.AddDays( 5 ); Break }
'Wednesday' { $Sunday = $Day1.Adddays(6).AddDays( 5 ); Break }
'Thursday' { $Sunday = $Day1.Adddays(5).AddDays( 5 ); Break }
'Friday' { $Sunday = $Day1.Adddays(4).AddDays( 5 ); Break }
'Saturday' { $Sunday = $Day1.Adddays(3).AddDays( 5 ); Break }
}
$Sunday
This is also easy to adjust. For example, if you wanted to switch to 1st Sunday between 1st & 2nd Monday etc...
For the mathematical / logic approach it came out a little more crude:
$Day1 = Get-Date "1/1/2021"
If( [Int]$Day1.DayOfWeek -gt 2 ) { $Interval = 7 - ([Int]$Day1.DayOfWeek - 2) }
Else { $Interval = 2 - [Int]$Day1.DayOfWeek }
$Sunday = $Day1.AddDays( $Interval ).AddDays( 5 )
$Sunday
The above example can be more concise in PowerShell 7+ using the ternary operator:
$Day1 = Get-Date "1/1/2021"
$Interval = $Day1.DayOfWeek -gt 2 ? 7 - ([Int]$Day1.DayOfWeek - 2) : 2 - [Int]$Day1.DayOfWeek
$Sunday = $Day1.AddDays( $Interval ).AddDays( 5 )
$Sunday
Note: all of the examples user 1/1/2021 as the starting date, but I did test across multiple 1st days of the month, For example, if the first day was a Monday, Sunday, Thursday etc...
Note: Some of the [Int] casting can probably be removed if one is careful about PowerShell type conversion system, but I wanted to get this out there. If I have time I'll cleanup further.
Or a slightly different approach:
$BaseDate = (Get-Date 3/1/2021)
$BaseDate.AddDays( (& {if ([int]$BaseDate.DayOfWeek -gt 2) {14}
else { 7 } }) - [int]($BaseDate).DayOfWeek)
Sunday, March 7, 2021 12:00:00 AM
Related
I need to find the date of the third saturday of the month.
I've got the following function, which finds me the last saturday of the month
function Get-3rdSaturdayOfMonth {
$Date = Get-Date
$lastDay = new-object DateTime($Date.Year, $Date.Month, [DateTime]::DaysInMonth($Date.Year, $Date.Month))
$diff = ([int] [DayOfWeek]::Saturday) - ([int] $lastDay.DayOfWeek)
if ($diff -ge 0) {
return $lastDay.AddDays(- (7-$diff))
}
else {
return $lastDay.AddDays($diff)
}
}
However, I don't know how to edit it to find the third saturday of the month.
Right now the function goes from last date of the month to the saturday that comes before it. But if I want the third saturday i think i need to calculate from the first day of the month, not the last.
Based on the solution from Brandon, since the third Saturday can't possibly occur before the 15th of the month, we can increment from the 15. until we find a saturday:
$thirdSaturday = [DateTime]::new((Get-Date).Year,(Get-Date).Month,15)
while ($thirdSaturday.DayOfWeek -ne [DayOfWeek]::Saturday)
{
$thirdSaturday = $thirdSaturday.AddDays(1)
}
I would like Powershell to determine the current week & print monday - friday in the format:
mm/dd/yyyy - mm/dd/yyyy
I saw a way to get the day of the week which is nice but I would like to just have it show the dates.
What I have so far, it works but if I run any day but Monday the dates would be off:
$bar = "------------------------------------"
$today = (Get-Date)
$dates = #($today.AddDays(0).ToString('MM-dd-yyyy'),
$today.AddDays(1).ToString('MM-dd-yyyy'),
$today.AddDays(2).ToString('MM-dd-yyyy'),
$today.AddDays(3).ToString('MM-dd-yyyy'),
$today.AddDays(4).ToString('MM-dd-yyyy'))
$result = "`n{0}`n{1}`n`n`n{2}`n{3}`n`n`n{4}`n{5}`n`n`n{6}`n{7}`n`n`n{8}`n{9}`n" -f $dates[0], $bar, $dates[1], $bar, $dates[2], $bar, $dates[3], $bar, $dates[4], $bar
echo $result
this seems to do what you want. [grin] it uses the builtin weekday list, indexes into that with the current weekday name, calculates the 1st day of the current week, generates an array of dates for the current week, and finally prints it out with the lines & vertical spacing that you seem to want.
# for my locale, the 1st day is "Sunday"
$WeekDayList = [System.DayOfWeek].GetEnumNames()
$Line = '-' * 40
$Newline = [environment]::NewLine
$BlankLineCount = 3
# the ".Date" property gives you midnite, not "now"
$Today = (Get-Date).Date
$TodayNumber = $WeekDayList.IndexOf($Today.DayOfWeek.ToString())
$WeekStartDate = $Today.AddDays(-$TodayNumber)
$CurrentWeek = foreach ($Offset in 0..6)
{
$WeekStartDate.AddDays($Offset).ToString('yyyy-MM-dd')
}
-join ($CurrentWeek -join "$Newline$Line$($Newline * $BlankLineCount)"), "$Line$($Newline * $BlankLineCount)"
output ...
2019-01-20
----------------------------------------
2019-01-21
----------------------------------------
2019-01-22
----------------------------------------
2019-01-23
----------------------------------------
2019-01-24
----------------------------------------
2019-01-25
----------------------------------------
2019-01-26
----------------------------------------
[not part of the output - needed to show that there are two blank lines above this one. [*grin*]
I have come across the requirement to get a date from a week day, and wrote a function to return a [DateTime].
Using this function and ToString() to format the date to your requirements, gives an output of:
2019/01/21 - 2019/01/25
Code:
Function Get-DateFromDay {
param(
[Parameter(Mandatory = $true)]
[ValidateSet('Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday')]
[string]$WeekDay,
[Int]$AddWeeks = 0
)
$DayNumber = #{
'Saturday' = 1;
'Sunday' = 0;
'Monday' = -1;
'Tuesday' = -2;
'Wednesday' = -3;
'Thursday' = -4
'Friday' = -5
}
[System.Datetime]$Today = Get-Date
$NumDaysSinceDateFromDay = $Today.DayOfWeek.value__ + $DayNumber[$WeekDay]
[System.Datetime]$DateFromDayThisWeek = $Today.AddDays(-$NumDaysSinceDateFromDay)
$DateFromDayThisWeek.AddDays( + ($AddWeeks * 7))
}
$Monday = (Get-DateFromDay -WeekDay Monday).ToString('yyyy/MM/dd')
$Friday = (Get-DateFromDay -WeekDay Friday).ToString('yyyy/MM/dd')
Write-Output "$Monday - $Friday"
And a few examples of use for those who come across this post in the future:
Thursday next week:
Get-DateFromDay -WeekDay Thursday -AddWeeks 1
Tuesday last week:
Get-DateFromDay -WeekDay Tuesday -AddWeeks -1
I have the need to retrieve the number of days in the current month, and despite my research, I have yet to find anything in powershell that does it. The following is what I have currently built to get the result I want. is there a better way to do it?
Please note that I am limited to Powershell
#check to see if this is a leap year
function LYC
{
if ([System.DateTime]::isleapyear((Get-Date).Year))
{
$Global:LY = True
}
}
#Get the number of days in current month
function fNOD
{
$MNum = (Get-Date).Month
switch ($MNum)
{
1 {$Global:DaysInMonth=31}
2 {
LYC
if (LY)
{
$Global:DaysInMonth=29
} else {
$Global:DaysInMonth=28
}
}
3 {$Global:DaysInMonth=31}
4 {$Global:DaysInMonth=30}
5 {$Global:DaysInMonth=31}
6 {$Global:DaysInMonth=30}
7 {$Global:DaysInMonth=31}
8 {$Global:DaysInMonth=31}
9 {$Global:DaysInMonth=30}
10 {$Global:DaysInMonth=31}
11 {$Global:DaysInMonth=30}
12 {$Global:DaysInMonth=31}
}
}
http://msdn.microsoft.com/en-us/library/system.datetime.daysinmonth.aspx
# static int DaysInMonth(int year, int month)
[DateTime]::DaysInMonth(2013, 3)
Assuming you want it for any date stored in $date:
((get-date $date -Day 1 -hour 0 -Minute 0 -Second 0).AddMonths(1).AddSeconds(-1)).Day
This seems to me like it might be deceptively easy:
$numDays = ((Get-Date -Month 3) - (Get-Date -Month 2))
28 days in February, July minus June gives me 30 days.
As mentioned below, that apparently sometimes does some weird integer rounding that gives off 29d, 23h, 59m or something else not right.
This seems to be consistent:
$monthLess = Get-Date -Month 4
$NumDays = (Get-Date -Month 5).Subtract($monthLess)
I have been using the following code for the last few months, which loops through a period of months from a predefined date until it gets to today's date.
use Date::Pcalc qw(:all);
$startDay = 1;
$startMonth = '4';
$startYear = '2009';
$dateToday = `date +%Y-%m-%d`;
($yt,$mt,$dt) = split(/\-/,$dateToday);
while ($endMonth <= $mt || $startYear < $yt ) {
if ($startMonth eq '12') {
$endMonth = 1;
$endYear = $startYear + 1;
} else {
$endMonth = $startMonth + 1;
$endYear = $startYear;
}
if ($startMonth eq '12') {
$endYear = $startYear + 1;
}
($meYear,$meMonth,$meDay) = Add_Delta_Days($endYear,$endMonth,$startDay,-1);
$endOfMonth = "$meYear-$meMonth-$meDay";
$monthText = Month_to_Text($startMonth);
$startDate = "$startYear-$startMonth-1";
$endDate = "$endYear-$endMonth-1";
print "$startDate - $endDate\n";
if ($startMonth eq '12') {
$startMonth = 1;
$startYear++;
} else {
$startMonth++
}
}
This has been working great for the last few months, but I've realised that now in December, as $endmonth will never be greater $mt (12), this causes an infinite loop.
I've not been able to figure out any alternate way of doing this. I feel like I should be able to fix this relatively easily but I seem to be having severe 'developer's block'
Thanks in advance to anyone who can assist.
my $date = DateTime->new(
time_zone => 'local',
year => $startYear,
month => $startMonth,
day => 1,
);
my $today = DateTime->today(time_zone => 'local');
while ($date <= $today) {
say $date->ymd('-');
$date->add( months => 1 );
}
I think you have a couple of problems with your code. But lets get to the first problem which is the enddate in month 12 which causes a loop in this statement:
while ($endMonth <= $mt || $startYear < $yt ) {
OK what you should do is something like this, once you have the current date, year month and day. You will notice that others have suggested different way to get the current date You should take up this suggestion. However once you have the date this code below should be adopted:
($yt,$mt,$dt) = split(/\-/,$dateToday);
# the line below will create a date like 201212 (yyyy mm) but if the month is a 1 digit month it will place a 0 in front of it to ensure your yymm variable always holds 6 characters in the format of yyyy mm - ok
my $yymm = $yt . ${\(length($mt) == 1 ? '0' : '')} . $mt;
# Now lets check the end date against the yymm
# initialise end date as end_yymm - again it inserts a 0 for single digit month
my $end_yymm = $startyear . ${\(length($startMonth) == 1 ? '0' : '')} . $startMonth;
# the above should get the date as '200904' from your code provide
# the while will check end_yymm like 200904 < 201212 - yes it is...
## the end_yymm will keep getting incremented each month and so will the year component at the end of each year until it reaches 201212
## then the question 201212 < 201212 will cause the while to end
## If you want it go into 201301 then say while ($end_yymm <= $yymm) {
## Hope you get the picture
while ($end_yymm < $yymm) {
if ($startMonth eq '12') {
$endMonth = 1;
$endYear = $startYear + 1;
} else {
$endMonth = $startMonth + 1;
$endYear = $startYear;
}
## Now this one seems to be repeating the endYear calculation as above - to me it seems redundant - maybe get rid of it
if ($startMonth eq '12') {
$endYear = $startYear + 1;
}
## Now that you have the end year and month incremented setup the end_yymm variable again to be picked up in the while statement:
$end_yymm = $startyear . ${\(length($startMonth) == 1 ? '0' : '')} . $startMonth;
# ...... carry on with the rest of your code
} # end the while loop
And that should do it.
All the best
I need to get the end date of the given month for some calculation purpose,
how can I do that in PHP, I tried using date() function, but It didn't work.
I used this:
date($year.'-'.$month.'-t');
But this gives the current month's end date.
I think I'm wrong somewhere, I couldn't find where I'm going wrong here.
If I give year as 2012 & month as 03, then it must show me as 2012-03-31.
This code will give you last day for a specific month.
$datetocheck = "2012-03-01";
$lastday = date('t',strtotime($datetocheck));
You want to replace your date() call with:
date('Y-m-t', strtotime($year.'-'.$month.'-01'));
The first parameter to date() is the format you want to be returned, and the second parameter has to be a unix timestamp (or not passed to use the current timestamp). In your case, you can generate a timestamp with the function strtotime(), passing it a date string with the year, the month, and 01 for the day. It will return that same year and month, but the -t in the format will be replaced by the last day of the month.
If you want to return only the last day of the month without year and month:
date('t', strtotime($year.'-'.$month.'-01'));
Just use 't' as your format string.
Current month:
echo date('Y-m-t');
Any month:
echo date('Y-m-t', strtotime("$year-$month-1"));
Try below code.
$m = '03';//
$y = '2012'; //
$first_date = date('Y-m-d',mktime(0, 0, 0, $m , 1, $y));
$last_day = date('t',strtotime($first_date));
$last_date = date('Y-m-d',mktime(0, 0, 0, $m ,$last_day, $y));
function lastday($month = '', $year = '') {
if (empty($month)) {
$month = date('m');
}
if (empty($year)) {
$year = date('Y');
}
$result = strtotime("{$year}-{$month}-01");
$result = strtotime('-1 second', strtotime('+1 month', $result));
return date('Y-m-d', $result);
}
function firstOfMonth() {
return date("Y-m-d", strtotime(date('m').'/01/'.date('Y').' 00:00:00')). 'T00:00:00';}
function lastOfMonth() {
return date("Y-m-d", strtotime('-1 second',strtotime('+1 month',strtotime(date('m').'/01/'.date('Y').' 00:00:00')))). 'T23:59:59';}
$date1 = firstOfMonth();
$date2 = lastOfMonth();
try this ,this give you a current month's starting and ending date.
date("Y-m-d",strtotime("-1 day" ,strtotime("+1 month",strtotime(date("m")."-01-".date("Y")))));
function getEndDate($year, $month)
{
$day = array(1=>31,2=>28,3=>31,4=>30,5=>31,6=>30,7=>31,8=>31,9=>30,10=>31,11=>30,12=>31);
if($year%100 == 0)
{
if($year%400 == 0)
$day[$month] = 29;
}
else if($year%4 == 0)
$day[$month] = 29;
return "{$year}-{$month}-{$day[$month]}";
}
If you are using PHP >= 5.2 I strongly suggest you use the new DateTime object. For example like below:
$a_date = "2012-03-23";
$date = new DateTime($a_date);
$date->modify('last day of this month');
echo $date->format('Y-m-d');