在PHP中尋找等效的JavaScript數學代碼

Question

在PHP中，是否有相當於下面的JavaScript代碼的代碼？

我想用PHP回顯大小，這樣當訪問者禁用JavaScript時，頁面上的數量仍然可見。

我對PHP並不擅長，所以如果有人能幫助我或指出我正確的方向，我將不勝感激。

的JavaScript

var start = new Date(2014, 2, 1); 
var rate = 1/(30 * 1000); 
var amount = Math.floor((new Date() - start) * rate); 

Answer 1

呼......它看起來比它更容易了！

好的，這真的很棘手，因爲你必須考慮到JS客戶端偏移量。如果您將下一個代碼複製並粘貼到foo.php中，並且您使用活動控制檯執行它，則會在控制檯中看到兩者（PHP和JS）相同。

你必須小心客戶端OFFSET的值（在我的情況下，UTC + 1，60分鐘），所以我糾正了我的PHP。您應該發送您使用的客戶端的OFFSET並使用它將其更正到服務器中，或者更好地說，服務器時間是您所在的時區以及數據與該時區相關的時間（由於您需要它可用無JS與客戶使用，這樣你就不會得到客戶端偏移）

<script> 
    // Creates a JavaScript Date instance that represents a single moment in time. 
    // Date objects are based on a time value that is the number of milliseconds since 1 January, 1970 UTC. 
    // new Date(year, month [, day, hour, minute, second, millisecond]); 

    // REALLY TRICKY!! Month starts in 0! 
    var start = new Date(2014, 2, 1); 

    console.log('OFFSET (minutes): ' + start.getTimezoneOffset()); 

    console.log('Date Start: ' + start.getTime()); 
    console.log('Date NOW: ' + new Date().getTime()); 

    var rate = 1/(30 * 1000); 

    console.log('Date substract: ' + (new Date() - start)); 

    var amount = Math.floor((new Date() - start) * rate); 

    console.log('Amount: ' + amount); 
</script> 

<?php 
// Returns the Unix timestamp corresponding to the arguments given. 
// This timestamp is a long integer containing the number of seconds between the 
// Unix Epoch (January 1 1970 00:00:00 GMT) and the time specified. 
// int mktime ([ int $hour = date("H") [, int $minute = date("i") [, int $second = date("s") [, int $month = date("n") [, int $day = date("j") [, int $year = date("Y") [, int $is_dst = -1 ]]]]]]]) 
// As month starts in 0 in js, here is month 3 
$start = mktime(0, 0, 0, 3, 1, 2014); 
echo 'Date Start MkTime: ' . $start . '<br>'; 
echo 'Date Start MkTime + JS OFFSET (-60 min * 60 sec): ' . ($start - 3600) . '<br>'; 

$start = $start - 3600; 
echo 'Date Start: ' . $start . '<br>'; 
echo 'Date NOW: ' . time() . '<br>'; 

// You don't need divide between 1000 because PHP is seconds, in JS you get microsecs 
$rate = 1/30; 

echo 'Date substract: ' . (time() - $start) . '<br>'; 

$amount = floor((time() - $start) * $rate); 

echo 'Amount: ' . $amount . '<br>'; 

在這裏，你有什麼我從我的瀏覽器中得到：

Answer 2

試試這個：

$start = mktime(0, 0, 0, 2014, 2, 1); 
$rate = 1/(30 * 1000); 
$amount = ((time() - $start) * $rate) | 0; 

Answer 3

<?php 
$start = strtotime("2011-02-01"); 
$rate = 1/(30 * 1000); 
$amount = floor((time() - $start) * $rate); 
echo $amount; 
?>