Php getdate[“weekday”如何从周一开始而不是从周日开始？

我需要将

getdate（）

的“工作日”从周日改为周一，有什么办法吗

这是我的代码，不要担心

validatePlanningVendeur

中的参数。问题是当我尝试获取当前日期

arret

时，它可以从周一到

arret

进行验证

你可能会注意到我写了

$arret=$today[“wday”]-1，这是我试图绕过
wkday
Sunday=0，使星期一为0，但现在它对星期天不起作用，因为它等于-1

简而言之，现在只从周一到周六工作

function validatePlanningVendeur($week, $vendeur, $boutique, $type_connecte, $user_connecte){
        $connexion = connexion();
        // Getting the current date in order to get the end
        $today = getdate();
        $arret = $today["wday"]-1; // attempt to bypass sunday = 0
        echo($arret);
        // echo("Aujourdhui : " . $today["wday"]);
        // echo("arret: " . $arret);

        // We get every day of the week until today ($arret)
        for($i = 0; $i <= $arret ; $i++){
            // On crée le year-month-day pour chaque jour de la semaine
            // echo($week[$i]);
            $dateCurrent = getdate($week[$i]);
            $day = $dateCurrent['mday'];
            $month = $dateCurrent['mon'];
            $year = $dateCurrent['year'];

            // Used to fetch from DB
            $jour = $year. '-' . $month . '-' . $day;
            // echo("Jour : " . $jour . " ");

            if($type_connecte == 1){ //Si c'est un RPV qui est connecté
                // On assigne id_validation = 1
                $request_validate = $connexion->prepare("UPDATE plannings SET id_validation = ? WHERE date_planning = ? AND id_employe = ? AND id_validation = ? AND id_boutique = ?");
                $request_validate->execute(array(1, $jour, $vendeur, 0, $boutique));
            }else{// On ne prend pas en compte la valeur boutique pour le RC
                $request_validate = $connexion->prepare("UPDATE plannings SET id_validation = ? WHERE date_planning = ? AND id_employe = ? AND id_validation = ?");
                $request_validate->execute(array(1, $jour, $vendeur, 0,));
            }
        }
}

功能验证计划供应商（$week、$vendur、$精品店、$type\u connecte、$user\u connecte）{
$connexion=connexion（）；
//获取当前日期以获取结束日期
$today=getdate（）；
$arret=$today[“wday”]-1；//尝试绕过sunday=0
echo（$arret）；
//echo（“Aujourdhui:”.$today[“wday”]）；
//echo（“arret:.$arret”）；
//直到今天，我们一周中的每一天都是如此（$arret）
对于（$i=0；$i请尝试使用此选项：

$today=new DateTime（）；
//按ISO8601获取一周中某一天的序号。
//从1日（星期一）到7日（星期日）开始。
$arret=$today->format（'N'）；//今天是第七天
对于（$i=0；$i请检查此项：


$week_days[saturday]=getdate();
$week_days[sunday]=getdate();
$week_days[monday]=getdate();
$week_days[tuesday]=getdate();
$week_days[wednesday]=getdate();
$week_days[thursday]=getdate();
$week_days[friday]=getdate();

$day_order[1]=$week_days[monday];
$day_order[2]=$week_days[tuesday];
$day_order[3]=$week_days[wednesday];
$day_order[4]=$week_days[thursday];
$day_order[5]=$week_days[friday];
$day_order[6]=$week_days[saturday];
$day_order[4]=$week_days[sunday];

请先告诉我们更多有关预期计算的信息。您不能更改不提供选项的函数签名。请听Ro的建议，并描述您希望代码执行的操作。
Sunday 7
Monday 0
Tuesday 1
Wednesday 2
Thursday 3
Friday 4
Saturday 5
Sunday 7


$week_days[saturday]=getdate();
$week_days[sunday]=getdate();
$week_days[monday]=getdate();
$week_days[tuesday]=getdate();
$week_days[wednesday]=getdate();
$week_days[thursday]=getdate();
$week_days[friday]=getdate();

$day_order[1]=$week_days[monday];
$day_order[2]=$week_days[tuesday];
$day_order[3]=$week_days[wednesday];
$day_order[4]=$week_days[thursday];
$day_order[5]=$week_days[friday];
$day_order[6]=$week_days[saturday];
$day_order[4]=$week_days[sunday];