Convert user date of birth html input to age in php

Question

I'm trying to get the date from the date-time-picker name="dob" into the php but it only post 2017 in the database user_age column. However, if I assigned $dob manually to $dob = '10/09/1988' it works.

<div class="form-group date-time-picker label-floating is-empty">
 <label class="control-label">Birthday</label>
 <input name="datetimepicker">
</div>

And here is the php

<?php
 $dob = $_GET['datetimepicker'];
 $dob = explode("/", $dob);
 $agv = (date("md", date("U", mktime(0, 0, 0, $dob[0], $dob[1], $dob[2]))) > 
 date("md") ? ((date("Y") - $dob[2]) -1) : (date("Y") - $dob[2]));

 $age = mysqli_real_escape_string($conn, $agv);
?>

what format is the date being submitted in from the datepicker? dd/mm/yyyy? mm/dd/yyyy? Something else? How are you verifying in the server that the date format was valid and expected? — ADyson, Nov 14 '17 at 13:48
according to the js of the datetimepicker, the format is MM/DD/YYYY — ViajanDee, Nov 14 '17 at 13:50
Possible duplicate of [Calculating age from date of birth in PHP](https://stackoverflow.com/questions/10186811/calculating-age-from-date-of-birth-in-php) — splash58, Nov 14 '17 at 13:53
@jeroen It should pull the value of a datetimepicker inside the div above with format MM/DD/YYYY. — ViajanDee, Nov 14 '17 at 13:55
I don't really care what it should do, I want to know what it contains exactly. — jeroen, Nov 14 '17 at 13:56
@FMeshreky, should do, or actually does do? Big difference. You can test that pretty straightforwardly — ADyson, Nov 14 '17 at 13:56
jcorry's answer is correct assuming you have a valid date in mm/dd/yyyy format (or any other other formats deemed acceptable in the docs: http://php.net/manual/en/datetime.formats.date.php) — ADyson, Nov 14 '17 at 13:58
you have two `name` attributes for you input. The `datetimepicker` one probably should go somewhere else, maybe `type` (as `type="date"`) or `class` — Karsten Koop, Nov 14 '17 at 13:58
@jeroen what I mean is it refers to the with name="dob" value and pulls the data inserted by the user. Sorry if I didn't understand the purpose of your comment. — ViajanDee, Nov 14 '17 at 14:09
You have 1 input field. It has 2 'name' attributes. That's not legal HTML. No wonder PHP has no idea what is in $_GET['dob']. — John Corry, Nov 14 '17 at 14:11
@jcorry I changed it and now it is referring to name="datetimepicker' but I've got the same post to the database as 2017 with my code and 00 with your code below — ViajanDee, Nov 14 '17 at 14:16
@jcorry I've edited that. I have a second doubt at the output of datetimepicker though. would you help me to post as a text it somehow to double check? — ViajanDee, Nov 14 '17 at 14:21
Are you submitting the form with method="GET"? The form data will be in the $_POST array if the method = "POST". — John Corry, Nov 14 '17 at 14:23
@jcorry now it works when I changed both methods to POST. Thank you, very much appreciated. — ViajanDee, Nov 14 '17 at 14:31

score 2 · Accepted Answer · answered Nov 14 '17 at 13:54

2

I think the problem is to get a DOB, subtract that from the date of now and present the difference in years, right?

That seems like a problem for DateTime::diff

$dob = new DateTime($_GET['dob']); // we'll just assume this is a safe date value for now
$now = new DateTime();

$age = $dob->diff($now);

$ageInYears = $age->format('%Y');

answered Nov 14 '17 at 13:54

John Corry

1,567
12
15

1

Thanks, it returns 00 now no matter what the date I pick is. – ViajanDee Nov 14 '17 at 14:07
@FMeshreky I see you updated your question. Try this answer but with `$_GET['datetimepicker']` instead. If you change the field name on the client (i.e. the HTML form), you must change it on the server too. – ADyson Nov 14 '17 at 14:20
@ADyson yes I did changed it in the php as well and made sure that all variables connected but something is wrong with the math or the datetimepicker output – ViajanDee Nov 14 '17 at 14:26

Convert user date of birth html input to age in php

1 Answers1