Populate input fields with values when option is selected in php mysql - php

Question

I have a form that i would like to submit details when an option is selected from the database.
The mysql database table :-
USERS contains fields like [email],[age],[name].
I want to be able to populate the other input fields values when one field is selected from the menu.
<form>
User
<select name="user" id="user">
<option>-- Select User --</option>
<option value="Mark">Mark</option>
<option value="Paul">Paul</option>
<option value="Hannah">Hannah</option>
</select>
<p>
Age
<input type="text" name="age" id="age">
</p>
<p>
Email
<input type="text" name="email" id="email">
</p>
</form>
How do i acheive this using jquery or javascript.

Answer 1

Please Try this,
on your HTML page:
write this to your html page,
<script>
$(document).ready(function(){
$('#user').on('change',function(){
var user = $(this).val();
$.ajax({
url : "getUser.php",
dataType: 'json',
type: 'POST',
async : false,
data : { user : user},
success : function(data) {
userData = json.parse(data);
$('#age').val(userData.age);
$('#email').val(userData.email);
}
});
});
});
</script>
in getUser.php:
getUser.php
<?php
$link = mysqli_connect("localhost", "user", "pass","mydb");
$user = $_REQUEST['user'];
$sql = mysqli_query($link, "SELECT age,email FROM userstable WHERE name = '".$user."' ");
$row = mysqli_fetch_array($sql);
json_encode($row);die;