Converting timestamp to time ago in PHP e.g 1 day

Question

I am trying to convert a timestamp of the format 2009-09-12 20:57:19 and turn it into something like 3 minutes ago with PHP.

I found a useful script to do this, but I think it's looking for a different format to be used as the time variable. The script I'm wanting to modify to work with this format is:

function _ago($tm,$rcs = 0) {
    $cur_tm = time(); 
    $dif = $cur_tm-$tm;
    $pds = array('second','minute','hour','day','week','month','year','decade');
    $lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);

    for($v = sizeof($lngh)-1; ($v >= 0)&&(($no = $dif/$lngh[$v])<=1); $v--); if($v < 0) $v = 0; $_tm = $cur_tm-($dif%$lngh[$v]);
        $no = floor($no);
        if($no <> 1)
            $pds[$v] .='s';
        $x = sprintf("%d %s ",$no,$pds[$v]);
        if(($rcs == 1)&&($v >= 1)&&(($cur_tm-$_tm) > 0))
            $x .= time_ago($_tm);
        return $x;
    }

I think on those first few lines the script is trying to do something that looks like this (different date format math):

$dif = 1252809479 - 2009-09-12 20:57:19;

How would I go about converting my timestamp into that (unix?) format?

Answer 1

I tried this and works fine for me

$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-10');
$difference = $datetime1->diff($datetime2);
echo formatOutput($difference);

function formatOutput($diff){
    /* function to return the highrst defference fount */
    if(!is_object($diff)){
        return;
    }

    if($diff->y > 0){
        return $diff->y .(" year".($diff->y > 1?"s":"")." ago");
    }

    if($diff->m > 0){
        return $diff->m .(" month".($diff->m > 1?"s":"")." ago");
    }

    if($diff->d > 0){
        return $diff->d .(" day".($diff->d > 1?"s":"")." ago");
    }

    if($diff->h > 0){
        return $diff->h .(" hour".($diff->h > 1?"s":"")." ago");
    }

    if($diff->i > 0){
        return $diff->i .(" minute".($diff->i > 1?"s":"")." ago");
    }

    if($diff->s > 0){
        return $diff->s .(" second".($diff->s > 1?"s":"")." ago");
    }
}

Check this link for reference here

Thanks! and have fun.

Answer 2

I'm aware that there are several answers here, but this is what I came up with. This only handles MySQL DATETIME values as per the original question I was responding to. The array $a needs some work. I welcome comments on how to improve. Call as:

echo time_elapsed_string('2014-11-14 09:42:28');

function time_elapsed_string($ptime)
{
    // Past time as MySQL DATETIME value
    $ptime = strtotime($ptime);

    // Current time as MySQL DATETIME value
    $csqltime = date('Y-m-d H:i:s');

    // Current time as Unix timestamp
    $ctime = strtotime($csqltime); 

    // Elapsed time
    $etime = $ctime - $ptime;

    // If no elapsed time, return 0
    if ($etime < 1){
        return '0 seconds';
    }

    $a = array( 365 * 24 * 60 * 60  =>  'year',
                 30 * 24 * 60 * 60  =>  'month',
                      24 * 60 * 60  =>  'day',
                           60 * 60  =>  'hour',
                                60  =>  'minute',
                                 1  =>  'second'
    );

    $a_plural = array( 'year'   => 'years',
                       'month'  => 'months',
                       'day'    => 'days',
                       'hour'   => 'hours',
                       'minute' => 'minutes',
                       'second' => 'seconds'
    );

    foreach ($a as $secs => $str){
        // Divide elapsed time by seconds
        $d = $etime / $secs;
        if ($d >= 1){
            // Round to the next lowest integer 
            $r = floor($d);
            // Calculate time to remove from elapsed time
            $rtime = $r * $secs;
            // Recalculate and store elapsed time for next loop
            if(($etime - $rtime)  < 0){
                $etime -= ($r - 1) * $secs;
            }
            else{
                $etime -= $rtime;
            }
            // Create string to return
            $estring = $estring . $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ';
        }
    }
    return $estring . ' ago';
}

Answer 3

You'll have to take each individual piece of your timestamp, and convert it into Unix time. For example for the timestamp, 2009-09-12 20:57:19.

(((2008-1970)*365)+(8*30)+12)*24+20 would give you a ROUGH estimate of the hours since January 1st, 1970.

Take that number, multiply by 60 and add 57 to get the minutes.

Take that, multiply by 60 and add 19.

That would convert it very roughly and inaccurately however.

Is there any reason you can't just take the normal Unix time to begin with?

Answer 4

I am using following function for several years. And it is working fine:

function timeDifference($timestamp)
{
    $otherDate=$timestamp;
    $now=@date("Y-m-d H:i:s");

    $secondDifference=@strtotime($now)-@strtotime($otherDate);
    $extra="";
    if ($secondDifference == 2592000) { 
    // months 
    $difference = $secondDifference/2592000; 
    $difference = round($difference,0); 
    if ($difference>1) { $extra="s"; } 
    $difference = $difference." month".$extra." ago"; 
}else if($secondDifference > 2592000)
    {$difference=timestamp($timestamp);} 
elseif ($secondDifference >= 604800) { 
    // weeks 
    $difference = $secondDifference/604800; 
    $difference = round($difference,0); 
    if ($difference>1) { $extra="s"; } 
    $difference = $difference." week".$extra." ago"; 
} 
elseif ($secondDifference >= 86400) { 
    // days 
    $difference = $secondDifference/86400; 
    $difference = round($difference,0); 
    if ($difference>1) { $extra="s"; } 
    $difference = $difference." day".$extra." ago"; 
} 
elseif ($secondDifference >= 3600) { 
    // hours 

    $difference = $secondDifference/3600; 
    $difference = round($difference,0); 
    if ($difference>1) { $extra="s"; } 
    $difference = $difference." hour".$extra." ago"; 
} 
elseif ($secondDifference < 3600) { 
    // hours 
    // for seconds (less than minute)
    if($secondDifference<=60)
    {       
        if($secondDifference==0)
        {
            $secondDifference=1;
        }
        if ($secondDifference>1) { $extra="s"; }
        $difference = $secondDifference." second".$extra." ago"; 

    }
    else
    {

$difference = $secondDifference/60; 
        if ($difference>1) { $extra="s"; }else{$extra="";}
        $difference = round($difference,0); 
        $difference = $difference." minute".$extra." ago"; 
    }
} 

$FinalDifference = $difference; 
return $FinalDifference;
}

Answer 5

This function is not made to be used for the English language. I translated the words in English. This needs more fixing before using for English.

function ago($d) {
$ts = time() - strtotime(str_replace("-","/",$d));

        if($ts>315360000) $val = round($ts/31536000,0).' year';
        else if($ts>94608000) $val = round($ts/31536000,0).' years';
        else if($ts>63072000) $val = ' two years';
        else if($ts>31536000) $val = ' a year';

        else if($ts>24192000) $val = round($ts/2419200,0).' month';
        else if($ts>7257600) $val = round($ts/2419200,0).' months';
        else if($ts>4838400) $val = ' two months';
        else if($ts>2419200) $val = ' a month';


        else if($ts>6048000) $val = round($ts/604800,0).' week';
        else if($ts>1814400) $val = round($ts/604800,0).' weeks';
        else if($ts>1209600) $val = ' two weeks';
        else if($ts>604800) $val = ' a week';

        else if($ts>864000) $val = round($ts/86400,0).' day';
        else if($ts>259200) $val = round($ts/86400,0).' days';
        else if($ts>172800) $val = ' two days';
        else if($ts>86400) $val = ' a day';

        else if($ts>36000) $val = round($ts/3600,0).' year';
        else if($ts>10800) $val = round($ts/3600,0).' years';
        else if($ts>7200) $val = ' two years';
        else if($ts>3600) $val = ' a year';

        else if($ts>600) $val = round($ts/60,0).' minute';
        else if($ts>180) $val = round($ts/60,0).' minutes';
        else if($ts>120) $val = ' two minutes';
        else if($ts>60) $val = ' a minute';

        else if($ts>10) $val = round($ts,0).' second';
        else if($ts>2) $val = round($ts,0).' seconds';
        else if($ts>1) $val = ' two seconds';
        else $val = $ts.' a second';


        return $val;
    }

Answer 6

The following is a very simple and extremely efficient solution.

function timeElapsed($originalTime){

        $timeElapsed=time()-$originalTime;

        /*
          You can change the values of the following 2 variables 
          based on your opinion. For 100% accuracy, you can call
          php's cal_days_in_month() and do some additional coding
          using the values you get for each month. After all the
          coding, your final answer will be approximately equal to
          mine. That is why it is okay to simply use the average
          values below.
        */
        $averageNumbDaysPerMonth=(365.242/12);
        $averageNumbWeeksPerMonth=($averageNumbDaysPerMonth/7);

        $time1=(((($timeElapsed/60)/60)/24)/365.242);
        $time2=floor($time1);//Years
        $time3=($time1-$time2)*(365.242);
        $time4=($time3/$averageNumbDaysPerMonth);
        $time5=floor($time4);//Months
        $time6=($time4-$time5)*$averageNumbWeeksPerMonth;
        $time7=floor($time6);//Weeks
        $time8=($time6-$time7)*7;
        $time9=floor($time8);//Days
        $time10=($time8-$time9)*24;
        $time11=floor($time10);//Hours
        $time12=($time10-$time11)*60;
        $time13=floor($time12);//Minutes
        $time14=($time12-$time13)*60;
        $time15=round($time14);//Seconds

        $timeElapsed=$time2 . 'yrs ' . $time5 . 'months ' . $time7 . 
                     'weeks ' . $time9 .  'days ' . $time11 . 'hrs '
                     . $time13 . 'mins and ' . $time15 . 'secs.';

        return $timeElapsed;

}

echo timeElapsed(1201570814);

Sample output:

6yrs 4months 3weeks 4days 12hrs 40mins and 36secs.