Php - 计算学年

Question

我正在努力编写一个可以解决当前学年的脚本.

学年从每年的8月1日开始.

我如何根据当前日期确定我们所处的学年.

即2012年7月31日(20120631)将作为2011/2012

2012年8月13日(20120801)将成为2012/2013

目前这是我所拥有的,但它不是很好,因为我不想定义日期,它不会返回正确的学年只是最初定义的$ academic_start_date.

function check_in_range($start_date, $end_date, $date_from_user)
{
  // Convert to timestamp
  $start_ts = strtotime($start_date);
  $end_ts = strtotime($end_date);
  $user_ts = strtotime($date_from_user);

  // Check that user date is between start & end
  return (($user_ts >= $start_ts) && ($user_ts <= $end_ts));
}


$academic_start_date = '20110801';
$academic_end_date = '20120731';
$startdate = '20120813';

$acyear_check = check_in_range($academic_start_date, $academic_end_date, $startdate);
if ($acyear_check == 1) { $acyear = $academic_start_date;}
else { $acyear = '';}

Answer 1

$time = ??;// here you put timestamp, it also may be strtotime(smth);

$year = date('Y', $time);
if(date('n', $time) < 8)
     $ayear = ($year - 1).'/'.$year;
else
    $ayear = ($year).'/'.($year + 1);

对于您的格式:

$datestr = 'YYYYmmdd';
$year = substr($datestr, 0, 4);
if(intval(substr($datestr,4,2)) < 8)
     $ayear = ($year - 1).'/'.$year;
else
    $ayear = ($year).'/'.($year + 1);

Answer 2

我会改用Datetime对象。最好在代码中传递DateTime对象，停止与日期格式（例如，美国/英国）等的混淆。

function academicYear(DateTime $userDate) {
    $currentYear = $userDate->format('Y');
    $cutoff = new DateTime($userDate->format('Y') . '/07/31 23:59:59');
    if ($userDate < $cutoff) {
        return ($currentYear-1) . '/' . $currentYear;
    }
    return $currentYear . '/' . ($currentYear+1);
}