php return date/time issue [duplicate] - php

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How to get time difference in minutes in PHP
(21 answers)
Closed 9 years ago.
I have a simple function below that gets seconds and returns it like "2 minutes ago", "9 hours ago" etc...
if I pass 1 on 'relativedate()' function it will just return 1 sec and so on. As we all know 60*60*24*7*30 or 18144000sec = 1 month. Therefore if I pass a value of 18144000 on the parameter, it should return 1 month.However, if I pass a value on the parameter lets say 231440000 on relativedate() which should be more than a year, it is returning '13 months' instead of 1 year.
function relativedate($secs) {
$second = 1;
$minute = 60;
$hour = 60*60;
$day = 60*60*24;
$week = 60*60*24*7;
$month = 60*60*24*7*30;
$year = 60*60*24*7*30*365;
if ($secs <= 0) { $output = "now";
}elseif ($secs > $second && $secs < $minute) { $output = round($secs/$second)." second";
}elseif ($secs >= $minute && $secs < $hour) { $output = round($secs/$minute)." minute";
}elseif ($secs >= $hour && $secs < $day) { $output = round($secs/$hour)." hour";
}elseif ($secs >= $day && $secs < $week) { $output = round($secs/$day)." day";
}elseif ($secs >= $week && $secs < $month) { $output = round($secs/$week)." week";
}elseif ($secs >= $month && $secs < $year) { $output = round($secs/$month)." month";
}elseif ($secs >= $year && $secs < $year*10) { $output = round($secs/$year)." year";
}else{ $output = " more than a decade ago"; }
if ($output <> "now"){
$output = (substr($output,0,2)<>"1 ") ? $output."s" : $output;
}
return $output;
}
echo relativedate(60); // 1 minute

Answer 1

Your year calculation is for 365 Months, not 12 months