1

谁能想到一个更好的方法来循环写出这个并得到相同的结果？

$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;
4

2 回答 2

3

我假设你想要在周六的分钟内 -3 和周日的 -4 分钟。无论如何，这是一个想法：

$weekday = date("w");
if ($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");
}
于 2010-08-01T00:31:03.280 回答
1

可以将它存储在一个数组中，以天为键，+/-x 天为值。

于 2010-08-01T00:30:02.970 回答