In Laravel, I'm trying to check if the current date is before or after a specific date in any year. A client's fiscal year ends October 31st and I'm having trouble checking if today is before or after that.
I've tried using the built in comparison functions in Carbon but I can't figure out what to compare now() with, since all Carbon instances seems to include a specific year. How do I check if now() has passed that day in any calender year?
The code below works but using dayOfYear() will return different values on leap years and seems like a poor solution.
if (now()->dayOfYear >= 304) {
$current_year = now()->year;
} else {
$current_year = now()->year - 1;
}
How can I return true if now() is after October 31st in any year?
It looks like you already know about using getters with the Carbon object to find pieces of the date. If you're looking to see if the current date is after October 31, just use them some more:
if (now()->month > 10) {
$current_year = now()->year;
} else {
$current_year = now()->subYear()->year;
}
To check if the current date is before any given date
$isBefore31October = now()->lt(Carbon::parse('2020-10-31'));
//Will return false if now() occurs after 31st October 2020
Similarly to check if the current date occurs after a given date
$isAfter31October = now()->gt(Carbon::parse('2020-10-31'));
//Will return true if now() occurs after 31st October 2020
For setting the current year depending upon whether current date occurs after 31st October of current year
$fyEnd = Carbon::parse(now()->year . "-10-31");
$current_year = now()->gt($fyEnd) ? now()->year : now()->year - 1;
Related
I'm working on a project and I need to format date in the following : YYYYWW where WW is the week number in the year, for example : today is 202131.
There are several ways to do this, I can use isoFormat('YYYYWW') directly, or I can get weekOfYear attribute and append it to current year.
Both methods however, break for the following date : 01-01-2021, indeed, the 1st january is set in the last week of 2020, but the year is 2021. Both methods give : 202153 while the correct result should be 202053 .
I currently have a workaround by checking if the month is less than 3 and the week is bigger than 50 then there is a problem and I decrement the year by 1.
public static function formatTestWeek($d)
{
$current = Carbon::parse($d);
$currentWeek = $current->weekOfYear;
$currentYear = $current->year;
if($currentWeek > 50 && $current->month < 3){
$currentYear -= 1;
}
$formattedDate = strval($currentYear) . $current->isoFormat("WW");
return $formattedDate;
}
Is there a more elegant approach in Carbon to get YYYYWW format that works in all cases ?
YYYY is the year
GGGG is the ISO week-year
gggg is the week-year following current locale settings (first_day_of_week + day_of_first_week_of_year)
So you need ->isoFormat('GGGGWW')
Complete list of codes available in isoFormat() are in the documentation:
https://carbon.nesbot.com/docs/#iso-format-available-replacements
I need to know if a date is in the current month.
Examples:
If the date is 2018-06-30 and current month is June (06), then true.
If the date is 2018-07-30 and current month is June (06), then false.
I have a list of dates with more than 1000 dates and I want to show or colorize only the dates that belongs to a current month.
You can do it all on one line. Basically convert the date in question to a PHP time, and get the month.
date('m',strtotime('2018-06-30' )) == date('m');
Using the date() function, if you pass in only the format, it'll assume the current date/time. You can pass in a second optional variable of a time() object to use in lieu of the current date/time.
I hope this helps -
$date = "2018-07-31";
if(date("m", strtotime($date)) == date("m"))
{
//if they are the same it will come here
}
else
{
// they aren't the same
}
As an alternative you could use a DateTime and for the format use for example the n to get the numeric representation of a month without leading zeros and use Y to get the full numeric representation of a year in 4 digits.
$d = DateTime::createFromFormat('Y-m-d', '2018-06-30');
$today = new DateTime();
if($d->format('n') === $today->format('n') && $d->format('Y') === $today->format('Y')) {
echo "Months match and year match";
}
Test
PHP doesn't implement a date type. If you are starting with a date/time and you know that your you are only dealing with a single timezone, AND you mean you want the current month in the curent year
$testdate=strtotime('2018-06-31 12:00'); // this will be converted to 2018-07-01
if (date('Ym')==date('Ym', $testdate)) {
// current month
} else {
// not current month
}
I'm trying to use DateTime to check if a credit card expiry date has expired but I'm a bit lost.
I only want to compare the mm/yy date.
Here is my code so far
$expmonth = $_POST['expMonth']; //e.g 08
$expyear = $_POST['expYear']; //e.g 15
$rawExpiry = $expmonth . $expyear;
$expiryDateTime = \DateTime::createFromFormat('my', $rawExpiry);
$expiryDate = $expiryDateTime->format('m y');
$currentDateTime = new \DateTime();
$currentDate = $currentDateTime->format('m y');
if ($expiryDate < $currentDate) {
echo 'Expired';
} else {
echo 'Valid';
}
I feel i'm almost there but the if statement is producing incorrect results. Any help would be appreciated.
It's simpler than you think. The format of the datess you are working with is not important as PHP does the comparison internally.
$expires = \DateTime::createFromFormat('my', $_POST['expMonth'].$_POST['expYear']);
$now = new \DateTime();
if ($expires < $now) {
// expired
}
You can use the DateTime class to generate a DateTime object matching the format of your given date string using the DateTime::createFromFormat() constructor.
The format ('my') would match any date string with the string pattern 'mmyy', e.g. '0620'. Or for dates with 4 digit years use the format 'mY' which will match dates with the following string pattern 'mmyyyy', e.g. '062020'. It's also sensible to specify the timezone using the DateTimeZone class.
$expiryMonth = 06;
$expiryYear = 20;
$timezone = new DateTimeZone('Europe/London');
$expiryTime = \DateTime::createFromFormat('my', $expiryMonth.$expiryYear, $timezone);
See the DateTime::createFromFormat page for more formats.
However - for credit/debit card expiry dates you will also need to take into account the full expiry DATE and TIME - not just the month and year.
DateTime::createFromFormat will by default use todays day of the month (e.g. 17) if it is not specified. This means that a credit card could appear expired when it still has several days to go. If a card expires 06/20 (i.e. June 2020) then it actually stops working at 00:00:00 on 1st July 2020. The modify method fixes this. E.g.
$expiryTime = \DateTime::createFromFormat('my', $expiryMonth.$expiryYear, $timezone)->modify('+1 month first day of midnight');
The string '+1 month first day of midnight' does three things.
'+1 month' - add one month.
'first day of' - switch to the first day of the month
'midnight' - change the time to 00:00:00
The modify method is really useful for many date manipulations!
So to answer the op, this is what you need — with a slight adjustment to format to cater for single digit months:
$expiryMonth = 6;
$expiryYear = 20;
$timezone = new DateTimeZone('Europe/London');
$expiryTime = \DateTime::createFromFormat(
'm-y',
$expiryMonth.'-'.$expiryYear,
$timezone
)->modify('+1 month first day of midnight');
$currentTime = new \DateTime('now', $timezone);
if ($expiryTime < $currentTime) {
// Card has expired.
}
An addition to the above answers.
Be aware that by default the days will also be in the calculation.
For example today is 2019-10-31 and if you run this:
\DateTime::createFromFormat('Ym', '202111');
It will output 2021-12-01, because day 31 does not exist in November and it will add 1 extra day to your DateTime object with a side effect that you will be in the month December instead of the expected November.
My suggestion is always use the day in your code.
For op's question:
$y=15;
$m=05;
if(strtotime( substr(date('Y'), 0, 2)."{$y}-{$m}" ) < strtotime( date("Y-m") ))
{
echo 'card is expired';
}
For others with full year:
$y=2015;
$m=5;
if(strtotime("{$y}-{$m}") < strtotime( date("Y-m") ))
{
echo 'card is expired';
}
Would it not be simpler to just compare the string "201709" to the current year-month? Creating datetime objects will cost php some effort, I suppose.
if($_POST['expYear']. str_pad($_POST['expMonth'],2,'0', STR_PAD_LEFT ) < date('Ym')) {
echo 'expired';
}
edited as Adam states
The best answer is provided by John Conde above. It it does the minimum amount of processing: creates two correct DateTime objects, compares them and that's all it needs.
It could work also as you started but you must format the dates in a way that puts the year first.
Think a bit about it: as dates, 08/15 (August 2015) is after 12/14 (December 2014) but as strings, '08 15' is before '12 14'.
When the year is in front, even as strings the years are compared first and then, only when the years are equal the months are compared:
$expiryDate = $expiryDateTime->format('y m');
$currentDate = $currentDateTime->format('y m');
if ($expiryDate < $currentDate) {
echo 'Expired';
} else {
echo 'Valid';
}
Keep it simple, as the answer above me says except you need to string pad to the left:
isCardExpired($month, $year)
{
$expires = $year.str_pad($month, 2, '0', STR_PAD_LEFT);
$now = date('Ym');
return $expires < $now;
}
No need to add extra PHP load using DateTime
If you are using Carbon, which is a very popular Datetime extension library. Then this should be:
$expMonth = $_POST['month'];
$expYear = $_POST['year'];
$format_m_y = str_pad($expMonth,2,'0', STR_PAD_LEFT).'-'.substr($expYear, 2);
$date = \Carbon\Carbon::createFromFormat('m-y', $format_m_y)
->endOfMonth()
->startOfDay();
if ($date->isPast()) {
// this card is expired
}
Also take into consideration the exact date and time expiration:
Credit cards expire at the end of the month printed as its expiration date, not at the beginning. Many cards actually technically expire one day after the end of that month. In any case, unless they list a specific day of expiration along with month and year, they should work all the way through the end of their expiration month. Cardholders should not wait until the last moment to secure a replacement card. Source
I need to create functions in PHP that let me step up/down given datetime units. Specifically, I need to be able to move to the next/previous month from the current one.
I thought I could do this using DateTime::add/sub(P1M). However, when trying to get the previous month, it messes up if the date value = 31- looks like it's actually trying to count back 30 days instead of decrementing the month value!:
$prevMonth = new DateTime('2010-12-31');
Try to decrement the month:
$prevMonth->sub(new DateInterval('P1M')); // = '2010-12-01'
$prevMonth->add(DateInterval::createFromDateString('-1 month')); // = '2010-12-01'
$prevMonth->sub(DateInterval::createFromDateString('+1 month')); // = '2010-12-01'
$prevMonth->add(DateInterval::createFromDateString('previous month')); // = '2010-12-01'
This certainly seems like the wrong behavior. Anyone have any insight?
Thanks-
NOTE: PHP version 5.3.3
(Credit actually belongs to Alex for pointing this out in the comments)
The problem is not a PHP one but a GNU one, as outlined here:
Relative items in date strings
The key here is differentiating between the concept of 'this date last month', which, because months are 'fuzzy units' with different numbers of dates, is impossible to define for a date like Dec 31 (because Nov 31 doesn't exist), and the concept of 'last month, irrespective of date'.
If all we're interested in is the previous month, the only way to gaurantee a proper DateInterval calculation is to reset the date value to the 1st, or some other number that every month will have.
What really strikes me is how undocumented this issue is, in PHP and elsewhere- considering how much date-dependent software it's probably affecting.
Here's a safe way to handle it:
/*
Handles month/year increment calculations in a safe way,
avoiding the pitfall of 'fuzzy' month units.
Returns a DateTime object with incremented month/year values, and a date value == 1.
*/
function incrementDate($startDate, $monthIncrement = 0, $yearIncrement = 0) {
$startingTimeStamp = $startDate->getTimestamp();
// Get the month value of the given date:
$monthString = date('Y-m', $startingTimeStamp);
// Create a date string corresponding to the 1st of the give month,
// making it safe for monthly/yearly calculations:
$safeDateString = "first day of $monthString";
// Increment date by given month/year increments:
$incrementedDateString = "$safeDateString $monthIncrement month $yearIncrement year";
$newTimeStamp = strtotime($incrementedDateString);
$newDate = DateTime::createFromFormat('U', $newTimeStamp);
return $newDate;
}
Easiest way to achieve this in my opinion is using mktime.
Like this:
$date = mktime(0,0,0,date('m')-1,date('d'),date('Y'));
echo date('d-m-Y', $date);
Greetz Michael
p.s mktime documentation can be found here: http://nl2.php.net/mktime
You could go old school on it and just use the date and strtotime functions.
$date = '2010-12-31';
$monthOnly = date('Y-m', strtotime($date));
$previousMonth = date('Y-m-d', strtotime($monthOnly . ' -1 month'));
(This maybe should be a comment but it's to long for one)
Here is how it works on windows 7 Apache 2.2.15 with PHP 5.3.3:
<?php $dt = new DateTime('2010-12-31');
$dt->sub(new DateInterval('P1M'));
print $dt->format('Y-m-d').'<br>';
$dt->add(DateInterval::createFromDateString('-1 month'));
print $dt->format('Y-m-d').'<br>';
$dt->sub(DateInterval::createFromDateString('+1 month'));
print $dt->format('Y-m-d').'<br>';
$dt->add(DateInterval::createFromDateString('previous month'));
print $dt->format('Y-m-d').'<br>'; ?>
2010-12-01
2010-11-01
2010-10-01
2010-09-01
So this does seem to confirm it's related to the GNU above.
Note: IMO the code below works as expected.
$dt->sub(new DateInterval('P1M'));
Current month: 12
Last month: 11
Number of Days in 12th month: 31
Number of Days in 11th month: 30
Dec 31st - 31 days = Nov 31st
Nov 31st = Nov 1 + 31 Days = 1st of Dec (30+1)
I have a simple situation where I have a user supplied week number X, and I need to find out that week's monday's date (e.g. 12 December). How would I achieve this? I know year and week.
Some code based mainly on previous proposals:
$predefinedYear = 2009;
$predefinedWeeks = 47;
// find first mоnday of the year
$firstMon = strtotime("mon jan {$predefinedYear}");
// calculate how much weeks to add
$weeksOffset = $predefinedWeeks - date('W', $firstMon);
// calculate searched monday
$searchedMon = strtotime("+{$weeksOffset} week " . date('Y-m-d', $firstMon));
An idea to get you started:
take first day of year
add 7 * X days
use strtodate, passing in "last Monday" and the date calculated above.
May need to add one day to the above.
Depending on the way you are calculating week numbers and the start of the week this may sometimes be out. (i.e. if the monday in the first week of the year was actually in the previous year!)
TEST THIS THOROUGHLY - but I've used a similar approach for similar calcualtions in the past.
This will solve the problem for you. It mainly derives from Mihail Dimitrov's answer, but simplifies and condenses this somewhat. It can be a one-line solution if you really want it to be.
function getMondaysDate($year, $week) {
if (!is_numeric($year) || !is_numeric($week)) {
return null;
// or throw Exception, etc.
}
$timestamp = strtotime("+$week weeks Monday January $year");
$prettyDate = date('d M Y');
return $prettyDate;
}
A couple of notes:
As above, strtotime("Monday January $year") will give you the timestamp of the first Monday of the year.
As above +X weeks will increment a specified date by that many weeks.
You can validate this by trying:
date('c',strtotime('Sunday Jan 2018'));
// "2018-01-07T00:00:00+11:00" (or whatever your timezone is)
date('c',strtotime('+1 weeks Sunday Jan 2018'));
// "2018-01-14T00:00:00+11:00" (or whatever your timezone is)
date('c',strtotime('+52 weeks Sunday Jan 2018'));
// "2019-01-06T00:00:00+11:00"
Due to reputation restriction i can't post multiple links
for details check
http://php.net/manual/en/function.date.php and http://php.net/manual/en/function.mktime.php
you can use something like this :
use mktime to get a timestamp of the week : $stamp = mktime(0,0,0,0,<7*x>,) {used something similar a few years back, so i'm not sure it works like this}
and then use $wDay = date('N',$stamp). You now have the day of the week, the timestamp of the monday should be
mktime(0,0,0,0,<7*x>-$wDay+1,) {the 'N' parameter returns 1 for monday, 6 for sunday}
hope this helps
//To calculate 12 th Monday from this Monday(2014-04-07)
$n_monday=12;
$cur_mon=strtotime("next Monday");
for($i=1;$i<=$n_monday;$i++){
echo date('Y-m-d', $cur_mon);
$cur_mon=strtotime(date('Y-m-d', strtotime("next Monday",$cur_mon)));
}
Out Put
2014-04-07
2014-04-14
2014-04-21
2014-04-28
2014-05-05
2014-05-12
2014-05-19
2014-05-26
2014-06-02
2014-06-09
2014-06-16
2014-06-23