指定的日期/時間範圍內的當前日期PHP

Question

，我創建了一個網站，只允許一定範圍內交付時間框架交付。

下面是一個例子正是我在尋找：

FakeCompany提供在週三和允許客戶將週五和週二之間的訂單，11點的週二晚截止時間。

我需要弄清楚客戶登錄時是否允許訂購（週五至週二晚上11點之間）。我還需要知道他們需要多長時間訂購。

我知道PHP date('N')函數，週五爲5：

date('N', strtotime('Friday')); 

和週二爲1：

date('N', strtotime('Tuesday')); 

這些時間段可能會改變，所以我需要一個簡單的解決方案。

這裏是我開始用，現在我失去了對如何做到這一點。

//Set today and get from database start/end days and end time 
$today = (int)date('N'); 
$startDay = (int)date('N', strtotime('Friday')); 
$endDay = (int)date('N', strtotime('Tuesday')); 
$endDayTime = '11:00:00'; 
//If today is before start date 
if($today >= $startDay && $today <= $endDay){ 
    //This works only if the end date is not the following week 
    //It also needs to be before end day time! 
} 

我想我需要獲得（星期五）根據當天的一週的日期，轉換至該週週五，如果週五沒有通過或下週週五和做與結束日期相同。

然後，我需要知道，如果今天的日期/時間之間。

Answer 1

這正是我所期待的。

所提供的日期可能不是本週甚至本月，所以我們需要根據日期確定一週中的哪一天，並根據今天將本週或下週的日期設置爲同一天（有點混亂）。

See It In Action

//IF USING A DATABASE TO STORE DATE/TIME 
//Get route times 
//When Using Database: $query = "SELECT * FROM routes WHERE id = '".$user['route_one']."'"; 
//When Using Database: $result = $this->query($query); 
//When Using Database: $route = $this->fetchArray($result); 
//Set date vaiables 
$thisWeek = new DateTime(); 
$routeStart = new DateTime(date('Y-m-d H:i:s', strtotime('2013-04-21 00:00:00'))); 
//When Using Database: $routeStart = new DateTime(date('Y-m-d H:i:s', strtotime($route['start_time']))); 
$routeEnd = new DateTime(date('Y-m-d H:i:s', strtotime('2013-04-24 00:00:00'))); 
//When Using Database: $routeEnd = new DateTime(date('Y-m-d H:i:s', strtotime($route['end_time']))); 
$interval = $routeStart->diff($routeEnd); 
$numDays = abs($interval->format('%d')); 
//Check if today is past or on the start date, else start date is next week, and set day of week 
if($thisWeek->format('N') >= $routeStart->format('N')){ 
    $startDate = $thisWeek->modify('last '.$routeStart->format('l')); 
} 
else{ 
    $startDate = $thisWeek->modify($routeStart->format('l')); 
} 
//Now that we know the start date add the amount of days to the start date to create the end date 
$endDate = new DateTime($startDate->format('Y-m-d H:s:i')); 
$endDate->modify('+'.$numDays.' days '.$routeEnd->format('H').' hours'); 
//Check to see if user is within the time range to order or not 
$today = new DateTime(); 
if($startDate <= $today && $today <= $endDate){ 
    //Find out how much longer ordering can take place 
    $interval = $endDate->diff($today); 
    $output = 'Allowed to order!<br>'; 
    $output .= '<div id="orderTimeCounter">'.$interval->format('%d days %h hours %i minutes left to order').'</div>'; 
} 
else{ 
    //If today is before start date set start date to THIS week otherwise NEXT week 
    if($startDate >= $today){ 
     $startDate = $startDate->modify('this '.$routeStart->format('l')); 
    } 
    else{ 
    $startDate = $startDate->modify('next '.$routeStart->format('l')); 
} 
    //Find out how much longer until ordering is allowed 
    $interval = $today->diff($startDate); 
    $output = 'Not allowed to order!'; 
    $output .= '<div id="orderTimeCounter">'.$interval->format('%d days %h hours %i minutes until you can order').'</div>'; 
} 
echo $output; 

Answer 2

這裏是一個函數來檢查，今天是經過批准的一天，然後如果週二也確保它是晚上11點前：

/* 
Acceptable days: 

5 - friday 
6 - saturday 
7 - sunday 
1 - monday 
2 - tuesday 
*/ 

//Can they order today? 
if(in_array(date('N'),array(1,2,5,6,7))){ 
    //if today is tuesday is it before 11pm? 
    if(date('N') == 2){ 
     if(date('H')<23){ 
      //23 = 11pm in 24 hour time 
      //Then can order 
     } 
     else{ 
      //Then CANT order 
     } 
    } 
    //Its not tuesday so we dont care what time it is they can order 
} 

Answer 3

爲末一天，我想你可以做這樣的：

$endDay = (int)date('N', strtotime('Friday') + 3 * 24 * 3600 + 23 * 3600); 

strtotime（'Friday'）獲得星期五並添加3天24小時到它，它將是星期二上午0點。然後你在晚上11點完成23小時的時間。

$today = (int)date('N'); 
$startDay = (int)date('N', strtotime('Friday')); 
$endDay = (int)date('N', strtotime('Friday') + 3 * 24 * 3600 + 23 * 3600); 

//If today is before start date 
if($today >= $startDay && $today <= $endDay){ 
    //now it works 
} 

Answer 4

$now  = new DateTime(); 
$tuesday = new DateTime('last Tuesday'); 
$friday = new DateTime('Friday 11pm'); 

if ($tuesday < $now && $now < $friday) { 
    $interval = $friday->diff($now); 
    echo $interval->format('%d day %h hours %i minutes left to order'); 
} 
else { 
    echo "you can't order now"; 
} 

See it in action