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（仅限年份），例如日期：15.03.1999年龄：14岁，15.03.2014将更改为年龄：15

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 ^{＆gt; = 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

^{函数：date_create()，date_diff()}

---

MySQL ^{＆gt; = 5.0.0}

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

demo

^{函数：TIMESTAMPDIFF()，CURDATE()}

Answer 2

获得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

从网上获取此脚本（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

格式为日期/月/年

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");

将输出正确的年龄（6月4日，它是14）。

Answer 6

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

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

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

从出生日期算起年龄。

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

输出：26

Answer 9

希望您会发现它有用。

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

Answer 10

#include <iostream>
#include <algorithm>
int main()
{
    const int length = 5;
    int array[length] = {35, 67, 75, 60, 11};

    std::sort(std::begin(array), std::end(array));
    for(int i = 0; i < length; i++)
    {
      std::cout << array[i] << ' ';
    }

    return 0;
}

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

的数据库值

Answer 11

有一种简单的方法可以通过使用PHP的substr来查找任何生日的日期

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

这只是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 。

有关详情，请参阅substr of PHP