Calculate age based on date of birth

Question

I have a table of users in sql and they each have birth dates. I want to convert their date of birth to their age (years only), e.g. date: 15.03.1999 age: 14 and 15.03.2014 will change to age: 15

Here I want to show the date of the user:

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']}";
}

possible duplicate of [Calculate Age in MySQL (InnoDb)](http://stackoverflow.com/questions/5773405/calculate-age-in-mysql-innodb) — John Conde, Oct 22 '13 at 14:49
there are plenty of 'calculate age' answers out there. Google is a mighty tool! But by the way: Don't use mysql_* functions. Use PDO or MySQLi — Patrick Manser, Oct 22 '13 at 14:49
Get current date/time and subtract the converted one from database...then convert it back to number of years, use strtotime() and date() with proper formatting. — Bud Damyanov, Oct 22 '13 at 14:50
Is this a MySQL problem, or a PHP problem? Also, be warned that the given query is highly vulnerable for SQL injection. Have a look at prepared statements to avoid getting hacked — Nico Haase, May 04 '23 at 15:39

score 217 · Accepted Answer · edited Jun 20 '20 at 09:12

217

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

^{functions: date_create(), date_diff()}

MySQL ^{>= 5.0.0}

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

demo

^{functions: TIMESTAMPDIFF(), CURDATE()}

edited Jun 20 '20 at 09:12

Community

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answered Oct 22 '13 at 14:53

Glavić

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@PHPupil: `date_diff` is for PHP version >= 5.3.0. – Glavić Oct 22 '13 at 15:07
@PHPupil: upgrade PHP? ;) or use MySQL solution. – Glavić Oct 22 '13 at 15:11
This wouldn't work for dob's that are under 1 years old... showing 0 is not necessarily ideal for age – Ryman Holmes Apr 06 '14 at 15:14
6

@RymanHolmes: if you need to show 1 year old, instead of 0, show it. One simple if statement solves this. Where is the problem? – Glavić Apr 06 '14 at 20:01

score 8 · Answer 2 · answered Oct 11 '15 at 01:25

Got this script from net (thanks to 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);
?>

score 8 · Answer 3 · edited Apr 26 '16 at 01:41

8

Very small code to get Age:

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

//output 35

edited Apr 26 '16 at 01:41

Mogsdad

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answered Feb 11 '16 at 10:26

easycodingclub

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How did this get 11 votes? A person's age depends on the day and month of the year as well. – Nick Bedford Jun 28 '17 at 04:31
2

Your method is incorrect if person have DOB ex. 13-10-1991 and current month is July then the age is 27 which is wrong. The person age is 26 till the current month is equal to or greater then the DOB month. – Shaan Ansari Oct 01 '18 at 06:56

score 5 · Answer 4 · edited May 04 '23 at 15:54

$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 * $minutes_in_hour) + ($diff->days * $hours_in_day * $minutes_in_hour) . " minutest"; echo "<br/>";
echo $seconds   = ($diff->h * $minutes_in_hour * $seconds_in_mins) + ($diff->days * $hours_in_day * $minutes_in_hour * $seconds_in_mins) . " seconds"; echo "<br/>";

Subin · Answer 5 · 2014-06-04T16:10:58.173

For a birthday date with format Date/Month/Year

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;
}

or the same function that accepts day, month, year as parameters :

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;
}

You can use it like this :

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

which will output the correct age (On 4 June, it's 14).

score 0 · Answer 6 · edited Oct 29 '17 at 18:34

0

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

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

edited Oct 29 '17 at 18:34

JH_

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answered Jan 24 '14 at 06:11

YATHI

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score 0 · Answer 7 · answered Jul 16 '15 at 12:54

 $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;

Sushant Samleti · Answer 8 · 2018-11-13T10:10:16.777

-1

I hope you will find this useful.

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

edited Nov 13 '18 at 10:10

answered Nov 13 '18 at 08:38

Sushant Samleti

9
2

Please don't recommend the usage of insecure queries – Nico Haase May 04 '23 at 15:40

score -1 · Answer 9 · edited Sep 02 '19 at 06:55

-1

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

$value['users_dob'] is the database value with format yyyy-mm-dd

edited Sep 02 '19 at 06:55

mnille

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answered Sep 02 '19 at 06:29

Tayyab Hayat

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Shaan Ansari · Answer 10 · 2019-06-28T08:21:18.410

-3

To Calculate age from Date of birth used used query like this.

'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'];
}

Note: don't use reserved keywords column name.

edited Jun 28 '19 at 08:21

answered Oct 01 '18 at 07:43

Shaan Ansari

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Please don't recommend the usage of insecure queries – Nico Haase May 04 '23 at 15:40
Have you seen the question and what creator asking for. I am only trying to help him for getting Date Of Birth. that is why i used creator query not mine. – Shaan Ansari May 06 '23 at 07:34

Calculate age based on date of birth

10 Answers10

PHP ^{>= 5.3.0}

MySQL ^{>= 5.0.0}

Linked

Related

Calculate age based on date of birth

10 Answers10

PHP >= 5.3.0

MySQL >= 5.0.0

Linked

Related

PHP ^{>= 5.3.0}

MySQL ^{>= 5.0.0}