在PHP中将用户输入日期html输入转换为年龄

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);
?>

Answer 1

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');