根据今天的日期的 PHP 日期范围

Question

任何人都可以想出一种更好的方法来将其写在循环中并获得相同的结果吗？

$today = date('l');

    if($today == 'Wednesday'){
        $min = date('l-m-d-y');
        $max = date('l-m-d-y', strtotime('+4 days'));
    }else if($today == 'Thursday'){
        $min = date('l-m-d-y', strtotime('-1 days'));
        $max = date('l-m-d-y', strtotime('+3 days'));
    }else if($today == 'Friday'){
        $min = date('l-m-d-y', strtotime('-2 days'));
        $max = date('l-m-d-y', strtotime('+2 days'));
    }else if($today == 'Saturday'){
        $min = date('l-m-d-y', strtotime('-2 days'));
        $max = date('l-m-d-y', strtotime('+1 days'));
    }else if($today == 'Sunday'){
        $min = date('l-m-d-y', strtotime('-3 days'));
        $max = date('l-m-d-y');
    }

    echo $min . ' - ' . $max;

Answer 1

我假设您希望周六分钟内为 -3，周日分钟内为 -4。无论如何，这就是这个想法：

$weekday = date("w");
if ($weekday == 0)
    $weekday = 7;

if ($weekday >= 3) {
    $min = date('l-m-d-y',
        strtotime(($weekday==3?"+0":(3-$weekday))." days");
    $max = date('l-m-d-y',
        strtotime("+".(7-$weekday)." days");
}

Answer 2

可以将其存储在一个数组中，以日期为键，+/-x 天作为值。