1

I want to calculate average of time-in for a user for particular duration , i have timestamp values for each time-in . To calculate average i want to add all timestamps and divide by no of days . But sum of all timestamps gives wrong input so i want convert timestamps to seconds so i can add them and calculate average . I am using following code .

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

but

date("H",$punchintime)

gives me proper value but 

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

giving me 0

Thanks in advance .

4

3 回答 3

2

你的问题不是很清楚，但我想我明白你想从一系列打卡时间中计算平均打卡时间。

日期对此没有好处，您需要隔离每个午夜后的秒数$punchintime并计算其平均值。下面的代码就是这样做的。我创建了一个时间数组来说明我的观点，我对您的系统一无所知，因此生成输入数组取决于您。

$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);

输出：-

平均时钟时间为 09h 00m 10s

这不适用于跨越午夜的时间，例如这样的数组：-

$aroundMidnight = array(
    '2013-08-01 23:59',
    '2013-08-02 00:02',
);
于 2013-08-29T13:46:49.320 回答
1

你说的是unixtime。Unixtime 是从 unix 纪元（1970 年 1 月 1 日）开始的秒数。要在两个时间戳中获得不同的时间，您可以简单地从第二个时间戳中减去第一个时间戳。

$timestamp1 = date('U');

然后过了一段时间：

$timestamp2 = date('U');

存储这些变量，并在需要时获得差异：

$difference = $timestamp2 - $timestamp1;

然后，您可以使用基本数学来格式化时间：

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

希望这可以帮助！

于 2013-08-27T14:03:43.923 回答
0

试试这个代码

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;
}
于 2013-09-26T13:10:20.277 回答