PHP provides ways to get the number of the current day of the month (date(\'j\')) as well as the number of the current day of the year (date(\'z\')). Is there a way to get the n
Assuming you mean a calendar-quarter (because a company fiscal year can start in any month of the year), you could rely on the date('z') to determine the day-of-year, and then keep a simple array of the day each quarter starts on:
$quarterStartDays = array( 1 /* Jan 1 */, 90 /* Mar 1, non leap-year */, ... );
Then with the current day-of-year you can first locate the largest start-day that's less than or equal to the day-of-year, then subtract.
Note that you need different numbers depending on the leap year.
<?php
function quarter_day($time = "") {
$time = $time ? strtotime($time) : time();
$date = intval(date("j", $time));
$month = intval(date("n", $time));
$year = intval(date("Y", $time));
// get selected quarter as number between 1 and 4
$quarter = ceil($month / 3);
// get first month of current quarter as number between 1 and 12
$fmonth = $quarter + (($quarter - 1) * 2);
// map days in a year by month
$map = [31,28,31,30,31,30,31,31,30,31,30,31];
// check if year is leap
if (((($year % 4) == 0) && ((($year % 100) != 0) || (($year % 400) == 0)))) $map[1] = 29;
// get total number of days in selected quarter, by summing the relative portion of $map array
$total = array_sum(array_slice($map, ($fmonth - 1), 3));
// get number of days passed in selected quarter, by summing the relative portion of $map array
$map[$month-1] = $date;
$day = array_sum(array_slice($map, ($fmonth - 1), ($month - $fmonth + 1)));
return "Day $day on $total of quarter $quarter, $year.";
}
print(quarter_day("2017-01-01")) . "\n"; // prints Day 1 on 90 of quarter 1, 2017.
print(quarter_day("2017-04-01")) . "\n"; // prints Day 1 on 91 of quarter 2, 2017.
print(quarter_day("2017-08-15")) . "\n"; // prints Day 46 on 92 of quarter 3, 2017.
print(quarter_day("2017-12-31")) . "\n"; // prints Day 92 on 92 of quarter 4, 2017.
I wrote a class with the following methods. Enjoy.
public static function getQuarterByMonth($monthNumber) {
return floor(($monthNumber - 1) / 3) + 1;
}
public static function getQuarterDay($monthNumber, $dayNumber, $yearNumber) {
$quarterDayNumber = 0;
$dayCountByMonth = array();
$startMonthNumber = ((self::getQuarterByMonth($monthNumber) - 1) * 3) + 1;
// Calculate the number of days in each month.
for ($i=1; $i<=12; $i++) {
$dayCountByMonth[$i] = date("t", strtotime($yearNumber . "-" . $i . "-01"));
}
for ($i=$startMonthNumber; $i<=$monthNumber-1; $i++) {
$quarterDayNumber += $dayCountByMonth[$i];
}
$quarterDayNumber += $dayNumber;
return $quarterDayNumber;
}
public static function getCurrentQuarterDay() {
return self::getQuarterDay(date('n'), date('j'), date('Y'));
}
function date_quarter()
{
return ceil(date('n', time()) / 3);
}
or
function date_quarter()
{
$month = date('n');
if ($month <= 3) return 1;
if ($month <= 6) return 2;
if ($month <= 9) return 3;
return 4;
}
We need to calculate the date of the first quarter first
$current_month = date('m');
// Get first month of quarter
$new_month = (3 * floor(($current_month - 1 ) / 3)) + 1;
// Add prefix zero if needed
$new_month = substr('0' . $new_month, -2);
$first_quarter_day_date = date('Y') . '-' . $new_month . '-01';
next we calculate the http://php.net/manual/en/datetime.diff.php
$datetime1 = new DateTime($first_quarter_day_date);
$datetime2 = new DateTime();
$interval = $datetime1->diff($datetime2);
echo $interval->format('%a days');
You can use Carbon it has easy modifiers for getFirstOf{Month,Year,Quarter}()
<?php
//take current date
$now = Carbon\Carbon::now();
//modify a copy of it to the first day of the current quarter
$firstOfQuarter = $now->copy()->firstOfQuarter();
//calculate the difference in days and add 1 to correct the index
$dayOfQuarter = $now->diffInDays($firstOfQuarter) + 1;