使用array()PHP的城市交货天数

Question

我在两个不同城市的网店有一个交货日期安排：

$city1 = array("Monday", "Friday");
$city2 = array("Monday", "Thursday", "Saturday");

这家商店每2天送货一次，但根据城市的不同，只在特定的日子送货。

如果今天是星期四，有人在城市1购买，那么订单将在2天后收到，在这种情况下是在星期一（见上面的数组）。

如果今天是星期四，有人在城市2购买，订单将在两天后收到，星期六在这里。

基于以上内容，我如何使用PHP来了解您在哪一天收到产品？

编程的逻辑应该是什么？

Answer 1

为了获得接收日，你可以做这样的

<?php
$days = array("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday");
$city1 = array("Monday","Thursday");
$city2 = array("Monday", "Thursday", "Saturday");
$cities = array("city1" => $city1,"city2" => $city2);
//returns the receive day actually the index for $days array
function get_day($city,$order_date){
    global $days;
    $key = array_keys($days,$order_date);
    $key = $key[0];
    $receive_day = getnexttwo($key+2,$city);
    return $receive_day;
}
//Utility to iterate by 2
function getnexttwo($key,$city){
    global $cities,$days;
    $next_day_key = $key%7;
    if(in_array($days[$next_day_key],$cities[$city])){
        return $next_day_key;
    }
    return getnexttwo($next_day_key+2,$city);
}
$order_day =  $days[get_day("city2","Saturday")]; 
echo $order_day;

Answer 2

如果我正确理解你的问题，商店至少需要两天才能发货，对吧？

如果是这样的话，你可以试试这样的东西：

function guess_delivery_day($order_city, $order_day) {
    $delivery_days = array(
        //index of the week days are based on date('N'), see http://php.net/manual/en/function.date.php
        'city1' => array(1 => "Monday", 5 => "Friday"),
        'city2' => array(1 => "Monday", 4 =>"Thursday", 6 => "Saturday")
    );
    if (isset($delivery_days[$order_city])) {
        $guessed_day = $order_day + 2; //Assuming the store take at least two days to ship a product;
        if ($guessed_day > 7) {
            $guessed_day = $guessed_day - 7; //If bigger then 7, it will be shipped on the next week
        }
        if (isset($delivery_days[$order_city][$guessed_day])) {
            return $delivery_days[$order_city][$guessed_day]; //we got lucky and the guessed day matches a delivery day
        }
        end($delivery_days[$order_city]); //set the pointer at the end of the array
        if ($guessed_day > key($delivery_days[$order_city])) {
            reset($delivery_days[$order_city]); //set the pointer at begining of the array
            return current($delivery_days[$order_city]); //the delivery day will be the first day available on that city
        }
        //We search for the following delivery days
        for($i = $guessed_day; $i <= 7; $i++) {
            if (isset($delivery_days[$order_city][$i])) {
                return $delivery_days[$order_city][$i];
            }
        }
        return false; //Something is wrong, it shouldn't be reached
    } else {
        return false; //Something is wrong with this order;
    }
}

Answer 3

不太容易解释，但

我已经用数字替换了日期名称，这样我就可以计算下一个了。

0是星期日，1是星期一，依此类推。

function run()
{
    $delivery = 2; // 2 days
    $city1 = array(1, 5); // Monday, Friday
    $city2 = array(1, 4, 6); // Monday, Thursday, Saturday
    // 0 = Sunday
    $today = 4; // Thursday
    $delivery1 = get_delivery_day($city1, $today, $delivery);
    echo "Order in city1 gets delivered $delivery1n";
    $delivery2 = get_delivery_day($city2, $today, $delivery);
    echo "Order in city2 gets delivered $delivery2n";
}
function get_delivery_day($city, $today, $delivery)
{
    echo "Trying to delivern";
    echo "Today is $todayn";
    echo "city " . implode(',', $city) . "n";
    // delivery must be zero or lower, and city must deliver today
    // looping before it happens
    while ($delivery > 0 || !in_array($today, $city))
    {
        $today++; // time is increasing by day
        $today %= 7; // if 7 then 0 - looping from Saturday to Sunday
        // remaining days are decreasing
        $delivery--; // can be lower than 0 because it would break >0 condition
        echo "today $today, remaining days $delivery, city " . implode('-', $city) ."n";
    }
    return $today; // today it will get delivered
}
run();