使用我的数据库中的数据计算年龄

问题描述

这是我目前使用的代码，但它不工作。 Geboortedatum是指荷兰语的出生日。

This is the code I'm currently using, but it's not working. Geboortedatum means day of birth in Dutch.

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;
?>

推荐答案

不需要PHP计算。 MySQL可以自己做（借助）：

No need for PHP calculation. MySQL might do it by itself (with help of TIMESTAMPDIFF()):

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

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如果您以格式存储日期，形式MySQL日期格式（即不在 YYYY-mm-dd 格式），那么您可以尝试使用 STR_TO_DATE（） a>函数。

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If you store your date in format, that differs form MySQL Date format (i.e. not in YYYY-mm-dd format), then you may try to format it with STR_TO_DATE() function.

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