Question

只是想知道，我可以这样做来验证用户输入了超过18岁的日期吗？

//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，更有希望。

Answer 1

为什么不呢？对我来说唯一的问题是用户界面 - 如何优雅地向用户发送错误消息。

另一方面，您的功能可能无法正常工作，因为您没有过正常的生日（您使用固定的生日）。你应该将'1988年3月23日'更改为$ 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'); 
    }
}

或者你可以：

// 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;
}

Answer 2

这是我用于多伦多银行系统的简化摘录，考虑到366天的闰年，这种方法总是很完美。

/* $dob is date of birth in format 1980-02-21 or 21 Feb 1980
 * time() is current server unixtime
 * We convert $dob into unixtime, add 18 years, and check it against server's
 * current time to validate age of under 18
 */

if (time() < strtotime('+18 years', strtotime($dob))) {
   echo 'Client is under 18 years of age.';
   exit;
}

Answer 3

我认为最好使用DateTime类。

$bday = new DateTime("22-10-1993");  
$bday->add(new DateInterval("P18Y")); //adds time interval of 18 years to bday  
//compare the added years to the current date  
if($bday < new DateTime()){   
    echo "over 18";  
}else{  
    echo "below 18";
}

DateTime :: diff也可用于将日期与当前日期进行比较。

$today = new DateTime(date("Y-m-d"));
$bday = new DateTime("22-10-1993");
$interval = $today->diff($bday);
if(intval($interval->y) > 18){
    echo "older than 18";
}else{
    echo "younger than 18";
}

N / B：1）对于第二种方法，如果$ bday大于$今年18年或更长时间，它将返回更旧，因此请确保输入的日期小于$今天。 2）DateTime适用于php 5.2.0及以上版本

Answer 4

if( strtotime("1988/03/23") < (time() - (18 * 60 * 60 * 24 * 365))) {
  print "yes";
} else {
  print "no";
}

......但是没有考虑到飞跃年数

Answer 5

HTML输入：

 <input type="date" class="form-control" placeholder="Data of Birth" name="dateOfBirth">

PHP代码：

function validateDateOfBirth($birthDay)
    {
// convert user input date to string and +18 years;
// compare user input date with current date;

        if (time() < strtotime('+18 years', strtotime($birthDay))) {
            return 'Not 18';
        }
        return "user is older than 18 years old";
    }

Answer 6

Php文件

if (isset($_POST['bdate'])){
    $bdate = $_POST['bdate'];
    $age = (date("Y-m-d") - $bdate);



}  //if age if 17 or younger error msg
if ($age < 17) {
    echo "Must 18 or older.";
}
else{ //if age is 120 or greather error msg
    if ($age > 120) {
        echo "Real age please.";
    }
    else{
        echo "$age";
    }
}

HTML code：

<form action="" method="POST"> 

    <p><label>Birth Date &nbsp;&nbsp;: </label>
    <input id="bdate" type="date" name="bdate" required placeholder="" /></p>
    <input class="btn register" type="submit" name="submit" value="Register" />
</form>

如果年龄超过18岁，请确认

6 个答案: