PHP - Use date 'yyyy/mm/dd' to show age, POST data

Question

I'm trying to get posted data of someones birthday, to send over to another page and display their age, so far I've got this which works

<?php
$c= date('Y');
$y= date('Y', strtotime('1993-12-29'));
echo $c - $y;
?>

But how would I do this with posted data? I've tried:

$c= date('Y');
$y= date('Y',strtotime($_POST['sBirthday']));
echo $c-$y;

Although this doesn't work, any ideas?

Check the *actual* value with `var_dump($_POST['sBirthday']);`. For php it does not really matter where it comes from. What matters is the value. — zerkms, Feb 10 '15 at 23:28
Show your form... Also you can var_dump($_POST) to see what is sent... — Whirlwind, Feb 10 '15 at 23:28
Problem's most likely in your form. How's it's used, the method, missing name attribute; could be anything. Input in it would need to follow what you have `1993-12-29` or `19951229` — Funk Forty Niner, Feb 10 '15 at 23:34
possible duplicate of [PHP calculate age](http://stackoverflow.com/questions/3776682/php-calculate-age) — kero, Feb 10 '15 at 23:37
`19951229` I meant `19931229` but it's not much different. Give/take a few years ;-) — Funk Forty Niner, Feb 10 '15 at 23:48
Duplicate [http://stackoverflow.com/a/19521323/67332](http://stackoverflow.com/a/19521323/67332). — Glavić, Feb 11 '15 at 05:57
@Glavić I do not think, this is his question, since he assumes that the first code works; even though it includes the same problem, which to solve you imply would be the question... but it clearly is not. — Matteo B., Feb 11 '15 at 09:23

score 1 · Accepted Answer · edited May 23 '17 at 12:05

1

Your posted code will work with the according HTML-Form if the person puts it in correctly. For example this form will work, if the user types 1993-12-29:

<form method="post" action="">
  <input type="date" name="sBirthday" />
  <input type="submit" />
</form>

Your code does not actually calculate the age of a person. Think, somebody is born on 2014-12-31; would you say he is 1 year old (this is what your code would calculate)? In fact he is 0 years old, because he did not live for a full year yet.

To calculate the age of a person try as suggested here:

$date = new DateTime($_POST['sBirthday']);
$now = new DateTime();
$interval = $now->diff($date);
echo $interval->y;

edited May 23 '17 at 12:05

Community

answered Feb 10 '15 at 23:37

Matteo B.

*"For example this form will work, if the user types `1993-12-29`"* - As will `19931229` - It will use the first four integers. – Funk Forty Niner Feb 10 '15 at 23:50
@Fred-ii- Actually, no. The [manual states](http://php.net/manual/en/datetime.construct.php) that the first argument to the constructor of DateTime is 'A date/time string. Valid formats are explained in [Date and Time Formats](http://php.net/manual/en/datetime.formats.date.php).'. The list of viable formats does not contain `YYMMDD` which would be what you suggested (`19931229`). Or did I understand your comment wrong? – Matteo B. Feb 11 '15 at 00:25
My bad, however I was referring to OP's code using `strtotime` and not about yours. – Funk Forty Niner Feb 11 '15 at 01:14

1 Answers1