验证年龄是否超过18岁 [英] Validate if age is over 18 years old

问题描述

想知道，我是否可以这样做来验证用户输入的日期超过18?

Just wondering, can I do this to validate that a user has entered a date over 18?

//Validate for users over 18 only
function time($then, $min)
{
    $then = strtotime('March 23, 1988');
    //The age to be over, over +18
    $min = strtotime('+18 years', $then);
    echo $min;
    if (time() < $min) {
        die('Not 18');
    }
}

偶然发现了这个函数date_diff: http://www.php.net/manual/en/function. date-diff.php 看起来，甚至更有希望.

Just stumbled across this function date_diff: http://www.php.net/manual/en/function.date-diff.php Looks, even more promising.

推荐答案

为什么不呢?对我来说，唯一的问题是用户界面-如何将错误消息优雅地发送给用户.

Why not? The only problem to me, is the User Interface - how you send out the error message elegantly to the user.

另一方面，由于您的生日不正确(您使用固定的生日)，因此您的功能可能无法正常使用.您应该将"1988年3月23日"更改为$ then

On another note, your function might not work properly as you did not intake a proper birthday (you are using a fixed birthday). You should change 'March 23, 1988' to $then

//Validate for users over 18 only
function validateAge($then, $min)
{
    // $then will first be a string-date
    $then = strtotime($then);
    //The age to be over, over +18
    $min = strtotime('+18 years', $then);
    echo $min;
    if(time() < $min)  {
        die('Not 18'); 
    }
}

或者您可以:

// validate birthday
function validateAge($birthday, $age = 18)
{
    // $birthday can be UNIX_TIMESTAMP or just a string-date.
    if(is_string($birthday)) {
        $birthday = strtotime($birthday);
    }

    // check
    // 31536000 is the number of seconds in a 365 days year.
    if(time() - $birthday < $age * 31536000)  {
        return false;
    }

    return true;
}

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