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Calculer l'âge en fonction de la date de naissance

J'ai une table des utilisateurs en SQL et ils ont chacun des dates de naissance. Je souhaite convertir leur date de naissance en âge (années uniquement), par exemple. date: 15.03.1999 age: 14 et 15.03.2014 changera en age: 15

Ici je veux montrer la date de l'utilisateur:

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']}";
}
phpmysqldate
58
PHPupil

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

fonctions: date_create() , date_diff()

MySQL > = 5.0.0

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

demo

fonctions: TIMESTAMPDIFF() , CURDATE()

172
Glavić

Très petit code pour avoir l'âge:

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

//output 35
10
easycodingclub

Vous avez ce script du net (merci à 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);
?>
4
googli.us

Lien de référence 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/>";
2
Shailesh Sonare
 $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;
0
Rutendo

Calculer l'âge à partir de la date de naissance.

$dob = '1991-09-30';
(((int) date("m",strtotime($dob)) >= (int) date('m')) && ((int) date("d",strtotime($dob)) >= (int) date('d'))) 

    ? 
    $age = (date('Y') - date('Y',strtotime($dob))) 
    :  
    $age = (date('Y') - date('Y',strtotime($dob)))-1;

SORTIE: 26

0
Shaan Ansari

déclarer @dateOfBirth dateselect @dateOfBirth = '2000-01-01'

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

0
YATHI

Pour une date d'anniversaire au format Date/Mois/Année 

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;
}

ou la même fonction qui accepte le jour, le mois et l'année en tant que paramètres:

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;
}

Vous pouvez l'utiliser comme ceci:

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

qui produira l’âge correct (le 4 juin, c’est 14).

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

$value['users_dob'] est la valeur de la base de données au format yyyy-mm-dd

0
Tayyab Hayat

J'espère que vous trouverez cela utile.

$query1="SELECT TIMESTAMPDIFF (YEAR, YOUR_DOB_COLUMN, CURDATE()) AS age FROM your_table WHERE id='$user_id'";
$res1=mysql_query($query1);
$row=mysql_fetch_array($res1);
echo $row['age'];
0
Sushant Samleti