PHP中的年龄计算leap年问题 [英] Age calculation leap year issue in php

问题描述

我正在使用以下功能从给定的出生日期开始计算年龄，但是如果使用a年日(即29)，则该功能未显示正确的差值.请帮助我修复此代码.

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")
?>

如果有人可以提出更好的代码，请对其进行更新.

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

如果一个人不到1岁，我需要以几个月和几天为单位查找出生日期.

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

我的服务器上安装了最新的5.4 php版本.

I am having latest 5.4 php version on my server.

2012年2月29日，它返回2年，而应该是3年.请帮忙.

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

推荐答案

为什么不一直使用date_diff()函数?它会给您想要的结果:

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);
}

另一种方法是通过计算以秒为单位的时间差，然后从那里获取时间差:

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);

然后一天是24小时，每个60分钟，又是60秒:

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

$sday = 60 * 60 * 24;

然后计算年差为:

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

但是我只会坚持使用date_diff()

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