在PHP中寻找等效的JavaScript数学代码

Question

PHP中是否存在以下JavaScript代码？

我希望用PHP回显金额，以便每当访问者禁用JavaScript时，该金额仍可在页面上显示。

我对PHP不太好，所以如果有人能帮助我或指出正确的方向，我将不胜感激。

的JavaScript

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

Answer 1

试试这个：

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

Answer 2

Phew ......它看起来比它更容易！

好的，这真的很棘手，因为你必须考虑到JS客户端的偏移。如果将下一个代码复制并粘贴到foo.php中并使用活动控制台执行它，您将看到控制台中的两个（PHP和JS）是相同的。

您必须小心客户端OFFSET的值（在我的情况下，UTC + 1,60分钟），所以我在PHP中更正了。您应该发送您使用的客户端的OFFSET并使用它来将其更正到服务器中，或者更好，说服务器时间是您所在的时区以及数据与该时区相关（由于您需要它）在没有JS可用的客户端使用它，所以你不会得到客户端Offset）

<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>';

这里有我从浏览器中获得的内容：

Answer 3

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