php - 根据出生日期计算年龄

Question

我在 sql 中有一个用户表，他们每个人都有出生日期。我想将他们的出生日期转换为他们的年龄（仅限年），例如日期：15.03.1999年龄：1415.03.2014并将更改为年龄：15

在这里我想显示用户的日期：

if(isset($_GET['id']))
{
    $id = intval($_GET['id']);
    $dnn = mysql_fetch_array($dn);
    $dn = mysql_query('select username, email, skype, avatar, ' .
        'date, signup_date, gender from users where id="'.$id.'"');
    $dnn = mysql_fetch_array($dn);
    echo "{$dnn['date']}";
}

Answer 1

PHP ^{>= 5.3.0}

# object oriented
$from = new DateTime('1970-02-01');
$to   = new DateTime('today');
echo $from->diff($to)->y;

# procedural
echo date_diff(date_create('1970-02-01'), date_create('today'))->y;

demo

^{功能：date_create(),date_diff()}

---

MySQL ^{>= 5.0.0}

SELECT TIMESTAMPDIFF(YEAR, '1970-02-01', CURDATE()) AS age

demo

^{功能：TIMESTAMPDIFF(),CURDATE()}

Answer 2

获取年龄的非常小的代码：

<?php
    $dob='1981-10-07';
    $diff = (date('Y') - date('Y',strtotime($dob)));
    echo $diff;
?>

//output 35

Answer 3

从网上得到这个脚本（感谢coffeecupweb）

<?php
/**
 * Simple PHP age Calculator
 * 
 * Calculate and returns age based on the date provided by the user.
 * @param   date of birth('Format:yyyy-mm-dd').
 * @return  age based on date of birth
 */
function ageCalculator($dob){
    if(!empty($dob)){
        $birthdate = new DateTime($dob);
        $today   = new DateTime('today');
        $age = $birthdate->diff($today)->y;
        return $age;
    }else{
        return 0;
    }
}
$dob = '1992-03-18';
echo ageCalculator($dob);
?>

Answer 4

参考链接http://www.calculator.net/age-calculator.html

$hours_in_day   = 24;
$minutes_in_hour= 60;
$seconds_in_mins= 60;

$birth_date     = new DateTime("1988-07-31T00:00:00");
$current_date   = new DateTime();

$diff           = $birth_date->diff($current_date);

echo $years     = $diff->y . " years " . $diff->m . " months " . $diff->d . " day(s)"; echo "<br/>";
echo $months    = ($diff->y * 12) + $diff->m . " months " . $diff->d . " day(s)"; echo "<br/>";
echo $weeks     = floor($diff->days/7) . " weeks " . $diff->d%7 . " day(s)"; echo "<br/>";
echo $days      = $diff->days . " days"; echo "<br/>";
echo $hours     = $diff->h + ($diff->days * $hours_in_day) . " hours"; echo "<br/>";
echo $mins      = $diff->h + ($diff->days * $hours_in_day * $minutes_in_hour) . " minutest"; echo "<br/>";
echo $seconds   = $diff->h + ($diff->days * $hours_in_day * $minutes_in_hour * $seconds_in_mins) . " seconds"; echo "<br/>";

Answer 5

宣布@dateOfBirth date select @dateOfBirth = '2000-01-01'

SELECT datediff(YEAR,@dateOfBirth,getdate()) as Age

Answer 6

对于格式为Date/Month/Year的生日日期

function age($birthday){
 list($day, $month, $year) = explode("/", $birthday);
 $year_diff  = date("Y") - $year;
 $month_diff = date("m") - $month;
 $day_diff   = date("d") - $day;
 if ($day_diff < 0 && $month_diff==0) $year_diff--;
 if ($day_diff < 0 && $month_diff < 0) $year_diff--;
 return $year_diff;
}

或接受日、月、年作为参数的相同函数：

function age($day, $month, $year){
 $year_diff  = date("Y") - $year;
 $month_diff = date("m") - $month;
 $day_diff   = date("d") - $day;
 if ($day_diff < 0 && $month_diff==0) $year_diff--;
 if ($day_diff < 0 && $month_diff < 0) $year_diff--;
 return $year_diff;
}

你可以像这样使用它：

echo age("20/01/2000");

这将输出正确的年龄（6 月 4 日是 14 岁）。

Answer 7

 $dob = $this->dateOfBirth; //Datetime 
        $currentDate = new \DateTime();
        $dateDiff = $dob->diff($currentDate);
        $years = $dateDiff->y;
        $months = $dateDiff->m;
        $days = $dateDiff->d;
        $age = $years .' Year(s)';

        if($years === 0) {
            $age = $months .' Month(s)';
            if($months === 0) {
                $age = $days .' Day(s)';
            }
        }
        return $age;

Answer 8

我希望你会发现这很有用。

$query1="SELECT TIMESTAMPDIFF (YEAR, YOUR_DOB_COLUMN, CURDATE()) AS age FROM your_table WHERE id='$user_id'";
$res1=mysql_query($query1);
$row=mysql_fetch_array($res1);
echo $row['age'];

Answer 9

$getyear = explode("-", $value['users_dob']);
$dob = date('Y') - $getyear[0];

$value['users_dob']是具有格式的数据库值yyyy-mm-dd

Answer 10

从出生日期计算年龄使用了这样的查询。

'SELECT username, email, skype, avatar, TIMESTAMPDIFF(YEAR, date, CURDATE()) AS age, signup_date, gender FROM users WHERE id="'.$id.'"';


if(isset($_GET['id']))
{
    $id = intval($_GET['id']);

    $dn = mysql_query('select username, email, skype, avatar, TIMESTAMPDIFF(YEAR, date, CURDATE()) AS age, signup_date, gender from users where id="'.$id.'"');

    $dnn = mysql_fetch_array($dn);

    echo $dnn['age'];
}

注意：不要使用保留关键字列名。

Answer 11

有一种简单的方法可以使用 PHP 的 substr 从任何生日中查找日期

$birth_date = '15.03.2014';
$date = substr($birth_date, 0, 2);
echo $date;

这只会简单地为您提供该出生日期的输出日期。

在这种情况下，这将是15。

有关更多信息，请参见PHP 的 substr ...