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我在一個項目上，哪裏有那捲 1 3個套餐工作的PHP 3周旋轉軸線B 3：C菜單

所以爲了適應這個我試着寫一個腳本會知道這一週是什麼號碼（1-52）

$firstDayOfMonth = '2014-01-01';  

    $week1 = $firstDayOfMonth; 
    $week2 = date("Y-m-d" ,strtotime('next Sunday', strtotime($week1))); 
    $week3 = date("Y-m-d" ,strtotime('+1 week', strtotime($week2))); 
    $week4 = date("Y-m-d" ,strtotime('+1 week', strtotime($week3))); 
    $week5 = date("Y-m-d" ,strtotime('+1 week', strtotime($week4))); 
    $today = strtotime('today UTC');

然後我用一個簡單的回聲推出以下

echo ' 
    Meals for Week starting ' .$week1 .' : Week A(code for menu list here)<br> 
    Meals for Week starting ' .$week2 .' : Week B<br> 
    Meals for Week starting ' .$week3 .' : Week C<br>

然後我決定，我寧願沒有52個實例的htm l字符串。所以我試圖做這樣的事情

if ($today =='$week1' || '$week4' || '$week7' || '$week10' || '$week13' || '$week16' || '$week19' || '$week22' || '$week25' || '$week28' || '$week31' || '$week34' || '$week37' || '$week40' || '$week43' || '$week46' || '$week49'|| '$week52') { 
echo "Display Week A$"; 
} elseif ($today == '$week2' || '$week5' || '$week8' || '$week11' || '$week14' || '$week17' || '$week20' || '$week23' || '$week26' || '$week29' || '$week32' || '$week35' || '$week38' || '$week41' || '$week44' || '$week47' || '$week50') { 
echo "Display Week B$"; 
} else { 
echo "Display Week C$";

我想計算出一週看到今天星期在運行但是這始終顯示如果結果。

我想知道是否有人能爲我2個月的努力提供更好的解決方案。

來源

2014-01-09 Draftrabbit

$weekOfTheYear = date('W'); 

switch ($weekOfTheYear % 3) { 
    case 1 : 
     echo "It's the first week"; 
     break; 
    case 2 : 
     echo "It's the second week"; 
     break; 
    default : 
     echo "It's the third week"; 
}

來源

2014-01-09 12:01:19 deceze

我與它是一月的第二個星期但是嘗試過這種方法其實際的公司，這就是爲什麼我試圖不再是一個星期分配給每個52的計數方法，但也許是我的第一個星期很愚蠢，應該在第2周時使用A，然後以這種方式向前滾動。 – Draftrabbit

因此，這一週應該只是你的情況一年中的第一週，是對的嗎？ – rccoros

從php date documentation

您可以使用日期（ 'W'）來獲得一年的ISO-8601的週數

$currentWeekNumber = date('W');

然後可以使用MOD來計算，如果它是A周，B周或C周。

switch ($currentWeekNumber % 3) { 
    case 1 : 
     echo "Display Week A$"; 
     break; 
    case 2 : 
     echo "Display WeekB$"; 
     break; 
    default : 
     echo "Display Week A$"; 
     break; 
}

來源

2014-01-09 12:07:08 rccoros

這是我完成的代碼謝謝你yone爲您的輸入你們是驚人的。

$image1 = "<img src=img/weekc.png>"; 
$image2 = "<img src=img/weeka.png>"; 
$image3 = "<img src=img/weekb.png>"; 


$weekOfTheYear = date('W'); 

switch ($weekOfTheYear % 3) { 
     case 1 : 
    echo $image1; 
    break; 
case 2 : 
    echo $image2; 
    break; 
default : 
    echo $image3;

來源

2014-01-13 10:33:01 Draftrabbit

菜單

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