时间和日期的 PHP switch 语句

Question

我有一个非常复杂的交付情况，其中一个 switch 语句正在检查星期几和每天 16:00 的截止时间，并回显交付日，周一、周二和周三是今天 + 1 天，如果切断后它应该与今天的日期 + 2 相呼应，由于周末没有交货，因此星期四和星期五的交货日期完全不同

<?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，它们一定是实现这一目标的更优雅的方式，任何人都知道实现相同结果的更清洁的方式吗？此代码有效，但可能会更好更短

Answer 1

请注意，您可以堆叠 case 执行相同操作的匹配项：

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

但你最好只创建一个包含当天所需值的小数组，然后使用日期名称作为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],
];

那么，计算提前期的代码就只有一行了：

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

然后像以前一样将其插入到您的输出中：

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

现在，当您要更改交货时间时，无需编辑编码，只需调整数组中的值即可。

因为我已经有一段时间没有做任何代码高尔夫了，所以整件事都在一行中：

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 2

正如我在评论中提到的那样，快速尝试删除重复代码，这既是用于回显的字符串，也能够组织开关，以便任何具有共同值的日子都由相同的代码处理...

$today = date("D");
switch($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 3

这个怎么样：

$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>";

我还建议使用 Carbon 处理日期。