在PHP中的年份和星期

Question

我从MySQL数据库中提取了两条信息,即年份(2009年,2010年,等)和周(1-52).我需要将其转换为日期开始和日期结束..

例如:

Year=2010, Week=1 would be (Friday, Jan 1st, 2010) - (Sunday, Jan 3rd, 2010)
Year=2010, Week=33 would be (Monday, Aug 16th, 2010) - (Sunday, Aug 22nd, 2010)
Year=2010, Week=34 would be (Monday, Aug 23rd, 2010) - (Sunday, Aug 29th, 2010)

我将如何在PHP中执行此操作？

Answer 1

$year = "2010"; // Year 2010
$week = "01"; // Week 1

$date1 = date( "l, M jS, Y", strtotime($year."W".$week."1") ); // First day of week
$date2 = date( "l, M jS, Y", strtotime($year."W".$week."7") ); // Last day of week
echo $date1 . " - " . $date2;

如果周数小于10,则在数字前附加0.1将无法正常工作,应为01.

Answer 2

由于发布了这个问题和接受的答案,DateTime课程使这更容易: -

function daysInWeek($weekNum)
{
    $result = array();
    $datetime = new DateTime();
    $datetime->setISODate((int)$datetime->format('o'), $weekNum, 1);
    $interval = new DateInterval('P1D');
    $week = new DatePeriod($datetime, $interval, 6);

    foreach($week as $day){
        $result[] = $day->format('d/m/Y');
    }
    return $result;
}

var_dump(daysInWeek(24));

输出: -

array (size=7)
  0 => string '10/06/2013' (length=10)
  1 => string '11/06/2013' (length=10)
  2 => string '12/06/2013' (length=10)
  3 => string '13/06/2013' (length=10)
  4 => string '14/06/2013' (length=10)
  5 => string '15/06/2013' (length=10)
  6 => string '16/06/2013' (length=10)

这具有照顾闰年等的额外优势.