Convert mysql datetime into simple text - php

Question

I have a MySql datetime value like "2012-04-17 20:48:29". I want to convert this to a simple text like "10 days ago". I want to do this in either php or javascript! I tried to create my own algorithm to do this. But is there an already available solution for doing this?

Answer 1

you could use this pattern
$date = "2012-04-17 20:48:29";
$seconds = time() - strtotime($date);
$days = floor($seconds / 86400);
$seconds -= $days * 86400;
$hours = floor($seconds / 3600);
$seconds -= $hours * 3600;
$minutes = floor($seconds / 60);
$seconds -= $minutes * 60;
echo "$days days, $hours hours, $minutes minutes, $seconds seconds ago";
you should of course add some conditions before echoing the result. to only show 1 minute ago, or 3 hours ago, or 10 days ago...

Answer 2

With this function you'll get outputs like:
1 minute
5 minutes
15 hours
4 days
2 months
1.5 years
function time_ago_in_words($time) {
$from_time = strtotime($time);
$to_time = strtotime(gmd());
$distance_in_minutes = round((($to_time - $from_time))/60);
if ($distance_in_minutes < 0)
return (string)$distance_in_minutes.'E';
if (between($distance_in_minutes, 0, 1))
return '1 minute';
elseif (between($distance_in_minutes, 2, 44))
return $distance_in_minutes.' minutes';
elseif (between($distance_in_minutes, 45, 89))
return '1 hour';
elseif (between($distance_in_minutes, 90, 1439))
return round($distance_in_minutes/60).' hours';
elseif (between($distance_in_minutes, 1440, 2879))
return '1 day';
elseif (between($distance_in_minutes, 2880, 43199))
return round($distance_in_minutes/1440).' days';
elseif (between($distance_in_minutes, 43200, 86399))
return '1 month';
elseif (between($distance_in_minutes, 86400, 525959))
return round($distance_in_minutes/43200).' months';
elseif ($distance_in_minutes > 525959)
return number_format(round(($distance_in_minutes/525960), 1), 1).' years';
}
So you could do:
// Last time you logged in: 15 days ago.
Last time you logged in: <?php echo time_ago_in_words($user['last_logged_in']) ?> ago.
// We haven't seen you for 15 days!
We haven't seen you for <?php echo time_ago_in_words($user['last_logged_in']) ?>!

Answer 3

With PHP, you can call strftime to get different out put. Look here for more details.
There is also a jQuery plugin that you might look at as well: jQuery-dateFormat
Have fun!

Answer 4

I use this function:
function duration($integer)
{
$seconds=$integer;
$minutes = 0;
$hours = 0;
$days = 0;
$weeks = 0;
$return = "";
if ($seconds/60 >=1)
{
$minutes=floor($seconds/60);
if ($minutes/60 >= 1)
{ # Hours
$hours=floor($minutes/60);
if ($hours/24 >= 1)
{ #days
$days=floor($hours/24);
if ($days/7 >=1)
{ #weeks
$weeks=floor($days/7);
if ($weeks>=2) $return="$weeks Weeks";
else $return="$weeks Week";
} #end of weeks
$days=$days-(floor($days/7))*7;
if ($weeks>=1 && $days >=1) $return="$return, ";
if ($days >=2) $return="$return $days days";
if ($days ==1) $return="$return $days day";
} #end of days
$hours=$hours-(floor($hours/24))*24;
if ($days>=1 && $hours >=1) $return="$return, ";
if ($hours >=2) $return="$return $hours hours";
if ($hours ==1) $return="$return $hours hour";
} #end of Hours
$minutes=$minutes-(floor($minutes/60))*60;
if ($hours>=1 && $minutes >=1) $return="$return, ";
if ($minutes >=2) $return="$return $minutes minutes";
if ($minutes ==1) $return="$return $minutes minute";
} #end of minutes
$seconds=$integer-(floor($integer/60))*60;
if ($minutes>=1 && $seconds >=1) $return="$return, ";
if ($seconds >=2) $return="$return $seconds seconds";
if ($seconds ==1) $return="$return $seconds second";
$return="$return.";
return $return;
}
echo duration(time() - strtotime($date));