Validate if age is over 18 years old

Question

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 =          


        

Answer 1

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

...not accounting for leaps years however

Answer 2

Here is a simplified extract from what I used for a banking system in Toronto, and this always worked perfectly, taking account of leap years of 366 days.

/* $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

HTML Input:

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

PHP code:

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 4

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

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'); 
    }
}

Or you can:

// 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 5

I think it is best using the DateTime class for this.

$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 can also be used to compare the date with the current date.

$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) for the second method , if $bday is greater than $today by 18 years or more, it will return older , so make sure date entered is less than $today . 2) DateTime works on php 5.2.0 and above

Answer 6

Php file

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>