Age calculation leap year issue in php

问题:

I am using following function to calculate age from given date of birth, but its not showing the correct difference if a leap year day i.e 29 is used. Please help me fix this code.

<?php
function getAbsAge($birthday)
    {
        list($year,$month,$day) = explode("-", $birthday);
        $year_diff  = date("Y") - $year;
        $month_diff = date("m") - $month;
        $day_diff   = date("d") - $day;

        if ($day_diff < 0 || $month_diff < 0)
        {
            $year_diff--;
        }

        if ($year_diff == 0)
        {
            $interval = date_diff(date_create(), date_create($birthday));
            $months = $interval->format("%M");
            $days = $interval->format("%d");

            if ($months > 0)
            {
                return $interval->format("%M Months %d Days");
            }
            else if ($months == 0 && $days > 1)
            {
                return $interval->format("%d Days");
            }
            else
            {
                return $interval->format("%d Day");
            }
        }
        else if ($year_diff == 1)
        {
        return "$year_diff Year";
    }
        else if ($year_diff > 1)
        {
        return "$year_diff Years";
    }
    }
echo getAbsAge("2012-02-29")
?>

Also if anyone can suggest a better code then please update it.

I need to find date of birth in months and days if a person is less than 1 year old.

I am having latest 5.4 php version on my server.

With 2012-02-29, its returning 2 Years whereas it should be 3 Years. Please help.

回答1:

Why are you not using the date_diff() function all the way through? it will give you the desired result:

function getAbsAge($birthday) {

    $age = '';
    $diff = date_diff(date_create(), date_create($birthday));
    $years = $diff->format("%y");
    $months = $diff->format("%m");
    $days = $diff->format("%d");

    if ($years) {
        $age = ($years < 2) ? '1 Year' : "$years Years";
    } else {
        $age = '';
        if ($months) $age .= ($months < 2) ? '1 Month ' : "$months Months ";
        if ($days) $age .= ($days < 2) ? '1 Day' : "$months Days";
    }
    return trim($age);
}

Another way would be by calculating the time difference in seconds and taking it from there:

list($year,$month,$day) = explode("-", $birthday);
$diff = mktime(0,0,0,date('n'),date('j'),date('Y')) - mktime(0,0,0,$month,$day,$year);

Then a day consists of 24 hours each with 60 minutes each with 60 seconds:

$sday = 60 * 60 * 24;

And then calculating the years difference would be:

$years = floor($diff / (365.2425 * $sday));     

But i would just stick to the first version i presented you using date_diff()