Question

我想计算特定持续时间内用户的平均时间，我有每个时间的时间戳值。要计算平均值，我想添加所有时间戳并除以天数。但所有时间戳的总和给出了错误的输入，所以我想将时间戳转换为秒，以便我可以添加它们并计算平均值。我正在使用以下代码。

$timeInTotalSec = 0;
$timeInTotalSec += intval(date("H",$punchintime)) * 60 * 60;
$timeInTotalSec += intval(date("i",$punchintime)) * 60;
$timeInTotalSec += intval(date("s",$punchintime));`

但

date("H",$punchintime)

给了我适当的价值但是

intval(date("H",$punchintime))

给我 0

提前致谢。

Answer 1

你所说的是unixtime。 Unixtime是unix时代（1970年1月1日）的秒数。要在两个时间戳中获得不同的时间，您可以简单地减去第二个时间戳的第一个时间戳。

$timestamp1 = date('U');

然后一段时间后：

$timestamp2 = date('U');

存储这些变量，何时需要获得差异：

$difference = $timestamp2 - $timestamp1;

然后，您可以使用基本数学格式化时间：

$seconds = $difference;
$minutes = $seconds/60; 
$hours = $minutes/60;
$days = $hours/24;

希望这有帮助！

Answer 2

你的问题不是很清楚，但我想我明白你想要计算一系列时间的平均时间。

日期对此无益，您需要隔离每个$punchintime午夜后的秒数并计算其平均值。以下代码执行此操作。我创建了一系列时间来说明我的观点，我对你的系统一无所知，所以生成输入数组归你所有。

$punchInTimes = array(
    '2013-08-01 09:00',
    '2013-08-02 09:06',
    '2013-08-03 08:50',
    '2013-08-04 09:20',
    '2013-08-05 09:01',
    '2013-08-06 08:56',
);

function getAverageTime(array $times)
{
    $seconds = $average = 0;
    $result = null;
    //get seconds after midnight
    foreach($times as $dateString){
        $date = new \DateTime($dateString);
        list($datePart) = explode(' ', $dateString);
        $midnight = new \DateTime($datePart);
        $seconds += $date->getTimestamp() - $midnight->getTimestamp();
    }

    if($seconds > 0){
        $average = $seconds/count($times);
        $hours = floor($average/3600);
        $average -= ($hours * 3600);
        $minutes = floor($average/60);
        $average -= ($minutes * 60);
        $result = new \DateInterval("PT{$hours}H{$minutes}M{$average}S");
    } else $result = new \DateInterval('PT0S');
    return $result->format("%Hh %Mm %Ss");
}

echo "Average clock in time is " . getAverageTime($punchInTimes);

<强>输出： -

平均时钟是09h 00m 10s

这对于跨越午夜的时间不起作用，例如像这样的数组： -

$aroundMidnight = array(
    '2013-08-01 23:59',
    '2013-08-02 00:02',
);

Answer 3

试试这段代码

public static function sec2hms($sec, $padHours = false) {

    // start with a blank string
    $hms = "";

    // do the hours first: there are 3600 seconds in an hour, so if we divide
    // the total number of seconds by 3600 and throw away the remainder, we're
    // left with the number of hours in those seconds
    $hours = intval(intval($sec) / 3600);

    // add hours to $hms (with a leading 0 if asked for)
    $hms .= ($padHours) ? str_pad($hours, 2, "0", STR_PAD_LEFT) . ":" : $hours . ":";

    // dividing the total seconds by 60 will give us the number of minutes
    // in total, but we're interested in *minutes past the hour* and to get
    // this, we have to divide by 60 again and then use the remainder
    $minutes = intval(($sec / 60) % 60);

    // add minutes to $hms (with a leading 0 if needed)
    $hms .= str_pad($minutes, 2, "0", STR_PAD_LEFT) . ":";

    // seconds past the minute are found by dividing the total number of seconds
    // by 60 and using the remainder
    $seconds = intval($sec % 60);

    // add seconds to $hms (with a leading 0 if needed)
    $hms .= str_pad($seconds, 2, "0", STR_PAD_LEFT);

    // done!
    return $hms;
}

php平均时间戳

3 个答案: