Question

这是我目前正在使用的代码，但它不起作用。 Geboortedatum是指荷兰语出生日。

mysql_connect('xxx', 'xxx', 'xxx');

mysql_select_db('xxx'); 

$result = mysql_query("select Geboortedatum from Personen");

while ($row = mysql_fetch_array($result)){

$datum= $row["Geboortedatum"];
}

     //date in mm/dd/yyyy format; or it can be in other formats as well
     $birthDate = $datum;
echo $birthDate;
     //explode the date to get month, day and year
     $birthDate = explode("/", $birthDate);
     //get age from date or birthdate
     $age = (date("md", date("U", mktime(0, 0, 0, $birthDate[0], $birthDate[1], $birthDate[2]))) > date("md") ? ((date("Y")-$birthDate[2])-1):(date("Y")-$birthDate[2]));
     echo "Age is:".$age;
?>

Answer 1

无需PHP计算。 MySQL可以自己完成（在TIMESTAMPDIFF()的帮助下）：

SELECT TIMESTAMPDIFF(YEAR, `Geboortedatum`, NOW()) as `age` FROM `Personen`;

如果您以格式存储日期，其格式与MySQL日期格式不同（即不是YYYY-mm-dd格式），那么您可以尝试使用STR_TO_DATE()函数对其进行格式化。

Answer 2

这样可以正常使用，但您的日期应采用以下格式：mm/dd/yyyy

<?php
         //date in mm/dd/yyyy format; or it can be in other formats as well
         $birthDate = "02/07/1990";
         //explode the date to get month, day and year
         $birthDate = explode("/", $birthDate);
         //get age from date or birthdate
         $age = (date("md", date("U", mktime(0, 0, 0, $birthDate[0], $birthDate[1], $birthDate[2]))) > date("md") ? ((date("Y")-$birthDate[2])-1):(date("Y")-$birthDate[2]));
         echo "Age is:".$age;
    ?>

DEMO

Answer 3

如果您希望按年，月和日分类，这是一种方法：

$secondsInAYear     = 31536000;
$secondsInAMonth    =  2635200; //Using the average (30.5) days in a month.
$secondsInADay      =    86400;


echo $datum;
$birthDate = strtotime($datum);

$ageInSeconds   = time() - $birthDate;

$ageInYears = floor( $ageInSeconds / $secondsInAYear );
$ageRemainder   = ( $ageInSeconds % $secondsInAYear );          // $a % $b  [ Modulus:  Remainder of $a divided by $b ]     

$ageInMonths    = floor( $ageRemainder / $secondsInAMonth );
$monthsRemainder    = ( $ageRemainder % $secondsInAMonth );

$ageInDays      = floor( $monthsRemainder / $secondsInAMonth );    

echo "Age is:" .$ageInYears ." Years, " .$ageInMonths ." Months, and " .$ageInDays ." Days.;

使用我的数据库中的数据计算年龄

3 个答案:

DEMO