获取当前季度 php 的开始日期和结束日期

php date codeigniter

2022-08-30 21:24:10

我正在尝试按季度设置开始日期和结束日期。

例如，我正在开发一个报告系统，我需要报告第1季度，第2季度，第3季度和第4季度的数据。

第一季度 - 1月至3月

第二季度 - 4月至6月

第三季度 - 7月至9月

第四季度 - 10月至12月

例如，我有一些本月的情况，以及上个月的情况，如下所示。

   case 'this_month':
      $start_date = date(DATE_FORMAT, mktime(0, 0, 0, date("m"), 1, date("Y")));
      $end_date = date(DATE_FORMAT, mktime(0, 0, 0, date("m"), date("d"), date("Y")));
    break;
    case 'last_month':
      $start_date = date(DATE_FORMAT, mktime(0, 0, 0, date("m") - 1, 1, date("Y")));
      $end_date = date(DATE_FORMAT, mktime(0, 0, 0, date("m"), 0, date("Y")));
    break;

但现在我需要为这个季度和上个季度添加案例，我不确定如何实际做到这一点，以便它反映适当的季度范围。

有什么想法吗？

答案 1

/**
* Compute the start and end date of some fixed o relative quarter in a specific year.
* @param mixed $quarter  Integer from 1 to 4 or relative string value:
*                        'this', 'current', 'previous', 'first' or 'last'.
*                        'this' is equivalent to 'current'. Any other value
*                        will be ignored and instead current quarter will be used.
*                        Default value 'current'. Particulary, 'previous' value
*                        only make sense with current year so if you use it with
*                        other year like: get_dates_of_quarter('previous', 1990)
*                        the year will be ignored and instead the current year
*                        will be used.
* @param int $year       Year of the quarter. Any wrong value will be ignored and
*                        instead the current year will be used.
*                        Default value null (current year).
* @param string $format  String to format returned dates
* @return array          Array with two elements (keys): start and end date.
*/
public static function get_dates_of_quarter($quarter = 'current', $year = null, $format = null)
{
    if ( !is_int($year) ) {        
       $year = (new DateTime)->format('Y');
    }
    $current_quarter = ceil((new DateTime)->format('n') / 3);
    switch (  strtolower($quarter) ) {
    case 'this':
    case 'current':
       $quarter = ceil((new DateTime)->format('n') / 3);
       break;

    case 'previous':
       $year = (new DateTime)->format('Y');
       if ($current_quarter == 1) {
          $quarter = 4;
          $year--;
        } else {
          $quarter =  $current_quarter - 1;
        }
        break;

    case 'first':
        $quarter = 1;
        break;

    case 'last':
        $quarter = 4;
        break;

    default:
        $quarter = (!is_int($quarter) || $quarter < 1 || $quarter > 4) ? $current_quarter : $quarter;
        break;
    }
    if ( $quarter === 'this' ) {
        $quarter = ceil((new DateTime)->format('n') / 3);
    }
    $start = new DateTime($year.'-'.(3*$quarter-2).'-1 00:00:00');
    $end = new DateTime($year.'-'.(3*$quarter).'-'.($quarter == 1 || $quarter == 4 ? 31 : 30) .' 23:59:59');

    return array(
        'start' => $format ? $start->format($format) : $start,
        'end' => $format ? $end->format($format) : $end,
    );
}

我开发这个函数来以任何方式处理四分之一：相对（这个，上一个，第一个，最后一个）和固定的。

例子：

get_dates_of_quarter();
//return current quarter start and end dates

get_dates_of_quarter(2);
//return 2nd quarter start and end dates of current year

get_dates_of_quarter('first', 2010, 'Y-m-d');
//return start='2010-01-01' and end='2014-03-31'

get_dates_of_quarter('current', 2009, 'Y-m-d');
//Supposing today is '2014-08-22' (3rd quarter), this will return
//3rd quarter but of year 2009.
//return start='2009-07-01' and end='2009-09-30'

get_dates_of_quarter('previous');
//Supposing today is '2014-02-18' (1st quarter), this will return
//return start='2013-10-01' and end='2013-12-31'

等待此帮助某人;)

答案 2

简单代码：

$current_quarter = ceil(date('n') / 3);
$first_date = date('Y-m-d', strtotime(date('Y') . '-' . (($current_quarter * 3) - 2) . '-1'));
$last_date = date('Y-m-t', strtotime(date('Y') . '-' . (($current_quarter * 3)) . '-1'));