Development Tip

생년월일을 기준으로 연령 계산

yourdevel 2020. 12. 2. 22:03
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생년월일을 기준으로 연령 계산


SQL에 사용자 테이블이 있으며 각각 생년월일이 있습니다. 나는 그들의 생년월일을 그들의 나이 (연도 만)로 변환하고 싶습니다. 예 : 날짜 : 15.03.1999나이 : 14 그리고 15.03.2014나이로 변경됩니다 :15

여기에 사용자의 날짜를 표시하고 싶습니다.

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']}";
}

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

함수 : date_create(),date_diff()


MySQL > = 5.0.0

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

demo

함수 : TIMESTAMPDIFF(),CURDATE()


Age를 얻기위한 아주 작은 코드 :

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

//output 35

net에서이 스크립트를 받았습니다 (coffeecupweb 덕분에).

<?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);
?>

참조 링크 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/>";

날짜 / 월 / 년 형식의 생일 날짜

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;
}

또는 일, 월, 연도를 매개 변수로 허용하는 동일한 함수 :

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;
}

다음과 같이 사용할 수 있습니다.

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

올바른 연령을 출력합니다 (6 월 4 일, 14 세).


 $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;

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

$value['users_dob'] 형식이있는 데이터베이스 값입니다. yyyy-mm-dd


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

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


이 정보가 유용하기를 바랍니다.

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

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.


There is a simple way to find the date from any birthdate by using substr of PHP

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

Which will just simply give you the output date of that birth date.

In this case, that will be 15.

See substr of PHP for more...

참고URL : https://stackoverflow.com/questions/19521146/calculate-age-based-on-date-of-birth

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