如何创建从周三到周五的 7 天数组以在周五上午 10 点更改

Question

即：

如果现在低于本周周五 10:00，则创建包含前一周周六的数组，然后是本周周日、周一、周二、周三和周五。

如果 now 超过本周星期五 10:00，则将之前的数组替换为包含本周星期六的另一个数组，然后是星期日星期一星期二星期三和下周星期五。

我无法用脚本表达这种情况的逻辑。

谢谢

在阅读了很多很多帖子后，我发现这是行不通的

$current_time = strtotime('now');
$week_start_day = "friday";
$start_time = "10:00";

if ($current_time <= strtotime('$week_start_day this week $start_time')) {
    $day_1 = date('d/m/Y', strtotime('previous week friday', strtotime(date('d-m-Y'))));
    $day_2 = date('d/m/Y', strtotime('previous week saturday ', strtotime(date('d-m-Y'))));
    $day_3 = date('d/m/Y', strtotime('previous week sunday  ', strtotime(date('d-m-Y'))));
    $day_4 = date('d/m/Y', strtotime('previous week monday', strtotime(date('d-m-Y'))));
    $day_5 = date('d/m/Y', strtotime('previous week tuesday', strtotime(date('d-m-Y'))));
    $day_6 = date('d/m/Y', strtotime('this week wednesday  ', strtotime(date('d-m-Y'))));   
    $day_7 = date('d/m/Y', strtotime('this week thursday ', strtotime(date('d-m-Y')))); 

} 
else{

    $day_1 = date('d/m/Y', strtotime('this week saturday  ', strtotime(date('d-m-Y'))));
    $day_2 = date('d/m/Y', strtotime('this week sunday', strtotime(date('d-m-Y'))));
    $day_3 = date('d/m/Y', strtotime('this week monday', strtotime(date('d-m-Y'))));
    $day_4 = date('d/m/Y', strtotime('this week tuesday  ', strtotime(date('d-m-Y')))); 
    $day_5 = date('d/m/Y', strtotime('this week wednesday ', strtotime(date('d-m-Y'))));    
    $day_6 = date('d/m/Y', strtotime('this week thursday ', strtotime(date('d-m-Y'))));
    $day_7 = date('d/m/Y', strtotime('this week friday ', strtotime(date('d-m-Y'))));
}


echo "
    $day_1 <br> 
    $day_2 <br>
    $day_3 <br>
    $day_4 <br> 
    $day_5 <br>
    $day_6 <br> 
    $day_7 <br>     
    ";

Answer 1

如果我对您的理解正确，这就是您想要的，但尚未对其进行很好的测试，因此可能存在错误，但这就是想法。

$fri=strtotime(" Friday this week + 4 hours ");// 10:00 GMT 
$now=time();

$week=array();

if($now<=$fri)
{
    $day = "last Saturday";
}
else
{
    $day = "next Saturday";
}

for($i=0;$i<7;$i++)
{
    $week[] = date("d/m/Y", strtotime($day." +".$i." day"));
}

foreach($week as $day)
{
    echo $day;
}