用于时间和日期的PHP开关语句

Question

我有一个非常复杂的交货情况，有一个转换语句，它检查一周中的某天，每天的截止时间为16:00，并且回声是星期一，星期二和星期三是今天的交货日+ 1天（如果已过帐）截止日期应与今天的日期+ 2相呼应，由于周末不提供服务，因此周四和周五的交付日期完全不同

<?php
    $today = date("D");
        switch($today){
            case "Mon":
            if(mktime(16, 0, 0) <= time()) {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 2 days')) . "</strong><p>";
                } else {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 1 days')). "</strong><p>";
                }
                break;
            case "Tue":
                if(mktime(16, 0, 0) <= time()) {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 2 days')). "</strong><p>";
                } else {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 1 days')). "</strong><p>";
                }
                break;
            case "Wed":
                if(mktime(16, 0, 0) <= time()) {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 2 days')). "</strong><p>";
                } else {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 1 days')). "</strong><p>";
                }
                break;
            case "Thu":
            if(mktime(16, 0, 0) <= time()) {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 4 days')). "</strong><p>";
                } else {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 1 days')). "</strong><p>";
                }
                break;
            case "Fri":
                if(mktime(16, 0, 0) <= time()) {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 5 days')). "</strong><p>";
                } else {
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 4 days')). "</strong><p>";
                }
                break;
            case "Sat":
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 4 days')). "</strong><p>";
                break;
            case "Sun":
                    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. ' + 3 days')). "</strong><p>";
                break;
            default:
                echo "No information available for that day.";
                break;
        }
        ?>

当然，任何人都应该知道采用更优雅的方式来实现相同的结果，它们必须是实现这一目标的更优雅的方式。此代码有效，但可能会更好，更短]

Answer 1

一个快速的尝试，就像我在评论中提到的，删除重复的代码，这既是用于回显的字符串，又是能够组织切换的函数，因此任何具有共同值的日子都将由同一代码处理。 ..

$today = date("D");
switch($today){
    case "Mon":
    case "Tue":
    case "Wed":
        $days = (mktime(16, 0, 0) <= time())? 2 : 1;
        break;
    case "Thu":
        $days = (mktime(16, 0, 0) <= time())? 4 : 1;
        break;
    case "Fri":
        $days = (mktime(16, 0, 0) <= time())? 5 : 4;
        break;
    case "Sat":
        $days = 4;
        break;
    case "Sun":
        $days = 3;;
        break;
    default:
        break;
}
if ( isset($days) ) {
    echo "<p>For Delivery on <strong> " . date('D jS', strtotime($Date. " + {$days} days")). "</strong><p>";
}
else    {
    echo "No information available for that day.";
}

Answer 2

注意，您可以堆叠具有相同功能的case匹配项：

switch ($today) {
    case 'Mon':
    case 'Tue':
    case 'Wed':
        // code here
        break;
}

但是最好只制作一个包含当日所需值的小数组，然后通过将日期名称用作数组索引来查找今天的值：

$leadtimes = [
    'Mon' => [1, 2], // first value is for before 16:00, second is for after
    'Tue' => [1, 2],
    'Wed' => [1, 2],
    'Thu' => [1, 4],
    'Fri' => [4, 5],
    'Sat' => [4, 4],
    'Sun' => [3, 3],
];

然后，用于计算提前期的代码仅一行：

$leadtime = $leadtimes[date('D')][date('H') < 16 ? 0 : 1];

然后像以前一样将其插入输出中：

echo
    "<p>For Delivery on <strong> " .
    date('D jS', strtotime("$Date + $leadtime days")) .
    "</strong><p>";

现在，当您要更改提前期时，无需编写任何代码，只需调整数组中的值即可。

而且因为我做过任何代码高尔夫已经有一段时间了，所以整件事都在一行上：

echo date('D jS',strtotime('+'.['Mon'=>[1,2],'Tue'=>[1,2],'Wed'=>[1,2],'Thu'=>[1,4],'Fri'=>[4,5],'Sat'=>[4,4],'Sun'=>[3,3]][date('D')][date('H')<16].'day'));

Answer 3

怎么样：

$today = date("D");
$cutoff = mktime(16, 0, 0) <= time();

$offset = $cutoff ? 2 : 1;
$offset = $cutoff and $today === 'Thu' ? 4 : 1;
$offset = $cutoff and $today === 'Fri' ? 5 : 4;
$offset = $today === 'Sat' ? 4 : $offset;
$offset = $today === 'Sun' ? 3 : $offset;

echo "<p>For Delivery on <strong> " . date('D jS', strtotime("$Date $offset days")) . "</strong><p>";

此外，我建议使用Carbon处理日期。