php 获取当前季度天数的简单方法？
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Easy way to get day number of current quarter?
提问by Chad Johnson
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 number of the current day of the current quarter?
PHP 提供了获取当月当前天数 (date('j')) 以及一年中当前天数 (date('z')) 的方法。有没有办法获取当前季度的当前天数？
So right now, August 5, it is day 36 of the third quarter.
所以现在，8 月 5 日，是第三季度的第 36 天。
If there is no standard way of calculating this, does anyone have a (prefereably PHP-based) algorithm handy?
如果没有标准的计算方法，是否有人手头有（最好是基于 PHP 的）算法？
采纳答案by Chad Johnson
I wrote a class with the following methods. Enjoy.
我用以下方法编写了一个类。享受。
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'));
}
回答by Michiel
How about:
怎么样：
$curMonth = date("m", time());
$curQuarter = ceil($curMonth/3);
回答by Mike
function date_quarter()
{
return ceil(date('n', time()) / 3);
}
or
或者
function date_quarter()
{
$month = date('n');
if ($month <= 3) return 1;
if ($month <= 6) return 2;
if ($month <= 9) return 3;
return 4;
}
回答by Eelke van den Bos
You can use Carbonit has easy modifiers for getFirstOf{Month,Year,Quarter}()
您可以使用Carbon，它对 getFirstOf{Month,Year,Quarter}() 有简单的修饰符
<?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;
回答by Edward Dale
<?php
function day_of_quarter($ts=null) {
if( is_null($ts) ) $ts=time();
$d=date('d', $ts);
$m=date('m', $ts)-1;
while($m%3!=0) {
$lastmonth=mktime(0, 0, 0, $m, date("d", $ts), date("Y",$ts));
$d += date('t', $lastmonth);
$m--;
}
return $d;
}
echo day_of_quarter(mktime(0, 0, 0, 1, 1,2009));
echo "\n";
echo day_of_quarter(time());
echo "\n";
?>
回答by Jason Cohen
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:
假设您的意思是日历季度（因为公司财政年度可以从一年中的任何月份开始），您可以依靠日期（'z'）来确定一年中的哪一天，然后保留一个简单的数组每个季度开始的那一天：
$quarterStartDays = array( 1 /* Jan 1 */, 90 /* Mar 1, non leap-year */, ... );
Then with the current day-of-year you can first locate the largest start-day that's less than or equal to the day-of-year, then subtract.
然后使用当前的年份，您可以首先找到小于或等于年份的最大开始日，然后减去。
Note that you need different numbers depending on the leap year.
请注意，您需要根据闰年不同的数字。
回答by Simo P. M.
<?php
function quarter_day($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.
回答by Clemens Tolboom
We need to calculate the date of the first quarter first
我们需要先计算第一季度的日期
$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
接下来我们计算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');
回答by Robert
I've noticed that this thread went a bit beyond the question, and it's the first response to many google searches with "Quarter" & "PHP" in them.
我注意到这个线程有点超出了问题，它是对许多包含“Quarter”和“PHP”的谷歌搜索的第一个回应。
If you're working with the ISO standards of organization, which you should if you're doing a business app, then
如果你正在使用 ISO 组织标准，如果你正在做一个商业应用程序，那么你应该这样做
$curMonth = date("m", time());
$curQuarter = ceil($curMonth/3);
Is NOT correct, because the first day of a year in the ISO standards, can be 30, or 31 December.
不正确，因为 ISO 标准中一年的第一天可以是 30 或 12 月 31。
Instead, you should use this :
相反，你应该使用这个：
$current_yearly_cycle_year_number = 2019;
$current_yearly_cycle_start->setISODate( $current_yearly_cycle_year_number, 1, 1 );
$current_yearly_cycle_end->setISODate( $current_yearly_cycle_year_number, 53, 1 );
if( $current_yearly_cycle_end->format("W") !== "53" )
$current_yearly_cycle_end->setISODate( $current_yearly_cycle_year_number, 52, 1 );
$week_number_start = intval( $current_yearly_cycle_start->format( "W" ) );
$timestamp_start_quarter = ( $week_number_start === 1 ? 1 : intval( ceil( $current_yearly_cycle_start->format( "m" ) / 3 ) ) );
var_dump( $timestamp_start_quarter );
