php时间戳的平均值 [英] php average of timestamp

问题描述

我想计算用户在特定持续时间内的平均时间，我有每个时间戳的时间戳值。
要计算平均值，我想添加所有时间戳，除以天数。
但是所有时间戳的总和给出错误的输入，所以我想将时间戳转换为秒，所以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））;`
  

但是

  date（H，$ punchintime）
  

给我正确的值，但是

  intval（date（H ，$ punchintime））
  

给我 0

提前感谢

解决方案

你的问题不是很清楚，但我认为我了解你想从一连串的时间中计算一下平均时间。

日期是不好的，你需要隔离秒数午夜为每个 $ punchintime ，并计算它的平均值。以下代码这样做。我创建了一系列时间来说明我的观点，我不知道你的系统，所以生成输入数组是由你而定。

 code> $ 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' 
）; 
 
函数getAverageTime（array $ times）
 {
 $ seconds = $ average = 0; 
 $ result = null; 
 //午夜后得到
 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）; 
} 
 
回声平均时钟是。 getAverageTime（$ punchInTimes）; 
  

输出： -

平均时钟为09h 00m 10s

这不适用于跨越午夜，像这样的阵列： -

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

如果我稍后再回来，我会看看应对这种情况。 >

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 .

解决方案

Your question isn't very clear, but I think I understand that you want to calculate the average punch in time from a series of punch in times.

Dates are no good for this, you need to isolate the number of seconds after midnight for each $punchintime and calculate the average of that. The following code does that. I created an array of times to illustrate my point, I don't know anything about your system, so generating the input array is down to you.

$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);

Output:-

Average clock in time is 09h 00m 10s

This does not work for times that span midnight, such as an array like this:-

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

If I get time later on I'll have a look at coping with that case.

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