函数返回没有星期六、星期日和节假日的交货日期 - Function to return the date of delivery without Saturdays, Sundays and holidays

Function to return the date of delivery without Saturdays, Sundays and holidays

如何在PHP中编写一个函数，返回没有星期六，星期日和节假日的交付日期?

例如:相同产品3天交货，其他产品5天交货。

如果我今天执行命令，我将在3天内收到我的命令，没有星期六，星期日和节假日。

如果我现在对产品"3j"执行"29/12/2011"的命令，我将在"4/1/2011"收到您的命令

在数据库中，我只知道每种产品的日销售额。

例如:

prod1: 3day  
prod2: 5day  
prod3: 7day

我试图做一个函数，但失败了。

以下代码(稍作修改)用于我的一个站点的生产环境:

// $dt = date of shipping, $numdays = expected number of days in transit
function realDeliveryDate($dt, $numdays)
{
    $holidays = array("05/30/2011","07/04/2011","09/05/2011","11/24/2011","11/25/2011","12/25/2011","12/31/2011","01/01/2012","05/28/2012","07/04/2012","09/03/2012","11/22/2012","11/23/2012","12/25/2012");
    $checkday = strtotime($dt." +".$numdays." days");
    // check if it's a holiday
    while(in_array(date("m/d/Y",$checkday), $holidays)) {
        $checkday = strtotime(date("m/d/Y",$checkday)." +1 day");
    }
    // make sure it's not Saturday
    if (date("w",$checkday) == 6) {
        $checkday = strtotime(date("m/d/Y",$checkday)." +2 days");
    }
    // make sure it's not Sunday
    if (date("w",$checkday) == 0) {
        $checkday = strtotime(date("m/d/Y",$checkday)." +1 day");
    }
    // make sure it's not another holiday
    while(in_array(date("m/d/Y",$checkday), $holidays)) {
        $checkday = strtotime(date("m/d/Y",$checkday)." +1 day");
    }
    return $checkday;
}

这是一个非常复杂的问题，你不能用一个简单的函数来解决。你在谈论工作日的计算策略，它们涉及很多思考。

如果您只想处理工作日而不是工作时间，那么它会变得稍微容易一些。

第一步是创建工作日数组或非工作日数组。例如:

//Build the days based of weekends
$nonWorkingDays = array();
foreach($iDate = 0; $iDate < 365; $iDate++){
    $date = strtotime('today +'.$iDate.' day');
    if(date('w', $date) == 0 || date('w', $date) == 6){
        $nonWorkingDays = date('Y-m-d', $date);
    }
}
//Add the holidays
$nonWorkingDays[] = '2011-12-25';
$nonWorkingDays[] = '2012-01-01';
//Determine the date of delivery
$daysToDelivery = 6;
$deliveryDate = time();
while($daysToDelivery > 0){
    $deliveryDate = time() + (24*60*60);
    if(!in_array(date('Y-m-d', $deliveryDate), $nonWorkingDays)){
        $daysToDelivery--;
    }
}

看一下:

http://php.net/manual/en/function.localtime.php

这将根据您提供的时间返回一个工作日。如果你现在的时间是X，你需要在2天内交货，你会做localtime(X + 2*24*3600)。生成的结构体将包含工作日，您可以使用它根据需要向前或向后移动时间。