CODEWITHSUNDEEP
phpmysqldate
PHPupil
PHPupil

Reputation: 785

Calculate age based on date of birth

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']}";
}

Upvotes: 76

Views: 209282

Answers (10)

Shailesh Sonare
Shailesh Sonare

Reputation: 3121

$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/>";

Upvotes: 5

Tayyab Hayat
Tayyab Hayat

Reputation: 923

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

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

Upvotes: -1

Shaan Ansari
Shaan Ansari

Reputation: 550

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.

Upvotes: -3

Sushant Samleti
Sushant Samleti

Reputation: 9

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'];

Upvotes: -1

YATHEENDRARAJ
YATHEENDRARAJ

Reputation: 27

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

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

Upvotes: 0

easycodingclub
easycodingclub

Reputation: 331

Very small code to get Age:

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

//output 35

Upvotes: 7

googli.us
googli.us

Reputation: 89

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);
?>

Upvotes: 8

Rutendo
Rutendo

Reputation: 39

 $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;

Upvotes: 0

Subin
Subin

Reputation: 3563

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).

Upvotes: 0

Glavić
Glavić

Reputation: 43552

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()

Upvotes: 220

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