Question

My new code with the updated changes. Clicking the button after inserting a number has no effect on the form.

Front-End

 $(document).ready(function(){
$("#go").click(function() {
    var id=$("#myid").val();
    $.getJSON('find.php',{num:id},function(obj){
       alert(obj.toSource());
       $("input.fname").val(obj.FirstName); 
       $("input.sname").val(obj.Surname); 
       $("input.age").val(obj.Age); 

     });
 });
});

Form

 ID: <input type="text" name="id" class="myid">
 <input type="button" value="Go" id="go" />

 First Name: <input type="text" name="FirstName" class="fname"><br>
 Surname: <input type="text" name="Surname" class="sname"><br>
 Age: <input type="text" name="Age" class="age"><br>

Back-End

 $id = $_GET['num'];

 $result = array('FirstName' => 1, 'Surname' => 2, 'Age' => 3);

 echo json_encode($result);

Answer 1

obj.length will be undefined and you’re calling the response object obj not data. What you’ll get back from PHP is this;

{
    "FirstName": 1,
    "Surname": 2,
    "Age": 3
}

To cater for this, update your JS to:

// if (obj.length>0){   remove this
   jQuery("input.fname").val(obj.FirstName); 
   jQuery("input.sname").val(obj.Surname); 
   jQuery("input.age").val(obj.Age); 
// }  and this...

Note the case-sensitivity of JavaScript.

Additionally, as pointed out by Nicola Peluchetti, you should be checking for $_GET['num'] rather than $_GET['id'].

You should also have more protection against SQL injection attacks. Escape your input with mysql_real_escape_string at least:

$query = "SELECT * FROM Customers WHERE ID = '" . mysql_real_escape_string($id) . '"';