从datebirth sql字段计算年龄？

Question

这是我正在使用的查询，它位于第1页，不在函数内部。

<?php
$sql2= mysql_query("SELECT *
FROM catego WHERE category_id = '$idc'");
$categoryCount = mysql_num_rows($sql2); 
if ($categoryCount>0 )
{
    $row2 = mysql_fetch_array($sql2);     
    $id = $row2["category_id"];


    $birthdate = $row2["birthdate"];
    $address = $row2["address"];
}
?>

Answer 1

SELECT YEAR(SUBDATE(NOW(), INTERVAL DATE_FORMAT("1975,09,02", "%Y-%m") YEAR_MONTH)) fs_age;

Answer 2

一个简单的php函数可能类似于以下内容：

function birth_date ($birth_date){
    list($y,$m,$d) = explode(",",$birth_date);
    $y_diff  = date("Y") - $y;
    $m_diff = date("m") - $m;
    $d_diff   = date("d") - $d;
    if ($m_diff < 0 || $d_diff < 0) { $y_diff--; }
    return $y_diff;
}

即分开你的逗号，计算出差异，调整他们出生日期/月份的哪一方，如果是这样，将年份减少一年，然后返回年份。

然后在你的页面中你可以输入：

<li><span class="label">Age:</spa><?php$age = getage($birthdate); echo $age;?> </li>

修改：

<?php
function birth_date($birth_date){
    list($y,$m,$d) = explode(",",$birth_date);
    $y_diff  = date("Y") - $y;
    $m_diff = date("m") - $m;
    $d_diff   = date("d") - $d;
    if ($m_diff < 0 || $d_diff < 0) { $y_diff--; }
    return $y_diff;
}

$sql2= mysql_query("SELECT * FROM catego WHERE category_id = '$idc'");
$categoryCount = mysql_num_rows($sql2); 
if ($categoryCount>0 )
{
    $row2 = mysql_fetch_array($sql2);     
    $id = $row2["category_id"];
    $birthdate = $row2["birthdate"];
    $address = $row2["address"];

    $age = birth_date($birthdate);
}
?>

Answer 3

在PHP中，有几种方法可以做到这一点。这是一些很好的资源：

http://snippets.dzone.com/posts/show/1310

http://www.dreamincode.net/forums/topic/19953-calculating-age-tutorial/

Answer 4

像这样，用您选择的语言：

function getAge(dateOfBirth)
{
DateTime now    = DateTime.Today;

int years   = now.Year - dateOfBirth.Year;
// subtract another year if we're before the birth day in the current year

if (now.Month < dateOfBirth.Month || (now.Month == dateOfBirth.Month && now.Day < dateOfBirth.Day))
 --years;

    return years;
}