把我的时间变成时间戳php

Question

我有一个当前时间格式化方法是：

$time = date('mdY');

我需要把它转回正常的 linux 时间戳。我需要即时执行此操作。基本上，一个函数将 $time 传递给它，然后它需要将其转换为正常的时间戳（我假设是 time();）？

谢谢

编辑： 当我尝试每个人明显的建议时，我得到一个 警告：第 75 行 /home/content/50/5975650/html/checkEntry.php 中除以零

function RelativeTime($timeToSend){
    $difference = time() - $timestamp;
    $periods = array("sec", "min", "hour", "day", "week", "month", "years", "decade");
    $lengths = array("60","60","24","7","4.35","12","10");

    if ($difference > 0) { // this was in the past
        $ending = "ago";
    } else { // this was in the future
        $difference = -$difference;
        $ending = "to go";
    }
    for($j = 0; $difference >= $lengths[$j]; $j++)
        $difference /= $lengths[$j];
    $difference = round($difference);
    if($difference != 1) $periods[$j].= "s";
    $text = "$difference $periods[$j] $ending";
    return $text;
}

// 如果发现未付费，请检查服务呼叫状态和邮件

$query = "SELECT id, account, status, dateEntered FROM service WHERE status = 'Unpaid'";
        $result = mysql_query($query) or die(mysql_error());

    while($row = mysql_fetch_row($result)){

                $account = $row[1];
                $status = $row[2];
                $dateEntered = $row[3];

                $timestamp = strtotime($dateEntered);   
                            $timeToSend = RelativeTime($timestmap);

              // mailStatusUpdate($account, $status, $timeToSend);

        }   

Answer 1

开始，函数$difference = time() - $timestamp; 的第一行引用$timestamp 未设置，因为$timeToSend 是传递给函数的内容。这会导致 $difference 始终等于 time()。

其次，除以零错误是从 for 循环 for($j = 0; $difference >= $lengths[$j]; $j++) 到达 $lengths[j] 不存在的点（因为第一个错误，$difference==0），所以循环继续直到 @ 987654327@ 超出了$lengths 的范围。我建议在 for 循环中检查 $lengths[$j]，例如 for($j = 0; array_key_exists($j,$lengths)&&$difference >= $lengths[$j]; $j++)。

最终：

function RelativeTime($timeToSend){
$difference = time() - $timestamp;
$periods = array("sec", "min", "hour", "day", "week",
"month", "years", "decade");
$lengths = array("60","60","24","7","4.35","12","10");

if ($difference > 0) { // this was in the past
$ending = "ago";
} else { // this was in the future
$difference = -$difference;
$ending = "to go";
}
for($j = 0; $difference >= $lengths[$j]; $j++)
$difference /= $lengths[$j];
$difference = round($difference);
if($difference != 1) $periods[$j].= "s";
$text = "$difference $periods[$j] $ending";
return $text;
}
echo RelativeTime(strtotime('-3 months'))."\n";
echo RelativeTime(strtotime(0));

And example working.

Answer 2

像这样使用strptime：

$tm = strptime($time, '%m%d%Y');
$timestamp = strtotime(sprintf("%04d-%02d-%02d", $tm['tm_year'], $tm['tm_mon']+1, $tm['tm_mday']));

如果你使用PHP 5.3或以上，也可以使用date_parse_from_format

至于您问题的另一部分，如果时差大于 10 年，您的代码就有错误。

除法前的初始差值是多少？我的猜测是超过 10 年，所以你用完了“长度”，即系统试图达到未定义的 $lengths[8]，它算作零，所以它试图除以那个零。