Verifying age on PHP

Question

I've looked at many posts on here and I still cant figure this out.

I am trying to verify that someone is older than 13 before they register for my site. This is what i have so far

<?php if (is_string($_POST['birthday']) && $_POST['birthday'] != 'mm/dd/yyyy') { $dateObj = new DateTime($_POST['birthday']); $ageLimit = new DateTime('13 years'); $now = new DateTime(date("Y/m/d")); $checkDate = $now->diff($ageLimit);; if($checkDate > $dateObj) { $errors[]='You must be atleast 13 years old to join.'; } else { $bday = mysqli_real_escape_string($db,$_POST['birthday']); } } else { $errors[]= 'Enter your birthday.'; }

The code will always run to

$bday = mysqli_real_escape_string($db,$_POST['birthday']);}

no matter what is entered in the date field and the outcome is always 1.

Can anyone help me with this? I cant figure this one out on my own.

<b>Birth Date</b><br><input type="date" name="birthday" value=<?php if(isset($_POST['birthday']))echo $_POST['birthday'];?>><br>

Community · Accepted Answer · 2017-05-23 10:24:09Z

1

Comparaison operators work with DateTime, see the answer here. So something like this should work

$dateObj=new DateTime($_POST['birthday']); $ageLimit=new DateTime('-13 years'); if($dateObj > $ageLimit){ //TOO YOUNG }

EDIT per comment

Replace if(isset($_POST['birthday']))echo $_POST['birthday']; with

if(isset($_POST['birthday'])) { echo $_POST['birthday']; } else { echo 'mm/dd/yyyy'; }

Or change

if (is_string($_POST['birthday']) && $_POST['birthday'] != 'mm/dd/yyyy')

To

 if (!empty($_POST['birthday']) && is_string($_POST['birthday']))

edited May 23, 2017 at 10:24

CommunityBot

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answered Feb 21, 2015 at 0:37

Thomas C.

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5 Comments

Cyclonecode Over a year ago

Shouldn't this be if($dateObj < $ageLimit)

Thomas C. Over a year ago

Nope because if the birthday is more recent that 13 years ago it means the user is too young.

Cyclonecode Over a year ago

You're correct, I'm just confused sorry =) Perhaps it's time to go to bed.

Travis Hallet Over a year ago

yea i figured that part but i want to output a message if they have not entered one. i mean the code is good how it is now. i can work with it but i would like it to be more accurate

Travis Hallet Over a year ago

it still goes through and makes it a variable able to be posted. IDK what is wrong if i set it null instead of mm/dd/yyyy and do the check will it work?

John Conde · Accepted Answer · 2015-02-21 00:39:41Z

You have several errors

'13 years' is not a valid value for DateTime()
A date in the 'Y/m/d' format is not a valid format for DateTime()
$checkDate is a DateInterval object and is not comparable to a DateTime object

You can fix this and simplify your code by comparing DateTime objects which are comparable:

$birthday = new DateTime($_POST['birthday']); $ageLimit = new DateTime('-13 years'); if ($birthday < $ageLimit) { // they're old enough } else { // too young }

Demo

cool man thanks. only one thing though. I can not see if they have entered a date or not. i would like to give them a message saying please enter the date. but it appears as a age younger than 13 when no birthday is inputed and the string is mm/dd/yyyy

Bas Slats · Accepted Answer · 2015-02-21 00:40:21Z

It might be easier to use strtotime to calculate the date difference. The higher the number the younger somebody is. So if the persons age is higher than the minimal age number they are not old enough.

if(is_string($_POST['birthday'])&&$_POST['birthday']!='mm/dd/yyyy') { $minAge = strtotime("-13 years"); $dateObject = strtotime($_POST['birthday']); if($dateObject > $minAge) { $errors[]= 'You must be atleast 13 years old to join.'; } } else { $errors[]='Enter your birthday.'; }

cool this is working but i still cant get it to output "Enter your birthday" when the date is not chosen

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