Question

我正在建造一个小型网站并陷入困境。你如何使用他们的出生日期来获得一个人的年龄？

所以这应该是这样的：

$age=curr_date - YEAR($birth_date)

我希望这个问题很清楚。

提前致谢

Answer 1

使用date和strtotime来确定出生日期的人的年龄。

$birth_date = '1991-12-10';
echo $age= date("Y") - date("Y", strtotime($birth_date)); //25

Answer 2

以下是代码：

$dob = new DateTime('2015-10-02');
$today = new DateTime;
$age = $today->diff($dob);

你可以像这样回应年龄：

echo $age->format('%y Years, %m Months and %d Days');

Answer 3

首先在strtotime中转换你的日期

$userDob  = '18/01/2000'; //Try to convert your date format like this
$userDob  =  strtotime($userDob);
$currDate =  time();
$dateDiff =  $currDate  - $userDob;
echo Date('y',$dateDiff);

Answer 4

假设$birth_date格式为YYYY-MM-DD。

您可以通过以下方式完成此操作。首先创建BirthDate的 DateTime 对象，

$date=date("Y-m-d",strtotime($birth_date));
$bDObj=new DateTime($date);
$cDate=new DateTime();

$age=$cDate->format("Y")-$bDObj->format("Y");

Answer 5

您可以使用：

<?php
$from = new DateTime('1993-09-19');
$to   = new DateTime('today');
echo $from->diff($to)->y;
?>

Answer 6

使用DateInterval Class？

<?php
$date1=date_create("2013-01-01");
$date2=date_create("2013-02-10");
$diff=date_diff($date1,$date2);

echo $diff->format("Total number of years: %y.");
?>

Answer 7

    <?php

        // ASSUMING YOU HAVE THE DATE IN MYSQL FORMAT: Y-m-d
        $birthDate  = "1980-10-05";
        $today      = date("Y-m-d");

        $dateA      = new DateTime($today);
        $dateB      = new DateTime($birthDate);

        $dateDiff   = $dateA->diff($dateB);
        $age        = "{$dateDiff->y} Years, {$dateDiff->m} Months and {$dateDiff->d} Days";

        var_dump($age);     //<== DUMPS: '35 Years, 8 Months and 3 Days'

使用他们的出生日期获得一个人的年龄

7 个答案: