Print exception in Ajax thrown from PHP - php

Question

I am trying to make a POST request to a Url using Ajax. My JS/JQuery code looks like this
var params = {
'username' : 'eddard.stark#got.com',
'name' : 'Eddard Stark'
};
$.post("/user/add", params, function(data) {
// no errors
var user = eval('(' + data + ')');
$("#spanId").html("User Id " + user.id);
// Here - http://stackoverflow.com/questions/16472116/ajax-error-cannot-catch-thrown-exception-from-php
// they are saying that it can be handled here. But How?
}).fail(function(err, status) {
// error 4xx : client side errors (e.g. controller/action does not exist)
// error 5xx : server side errors (like db failure)
$("#spanId").html("Error " + err);
}).always(function() {
$(elm).hide();
$('#spanId').show();
});
PHP code for /user/add action is
function add()
{
$username = null;
$name = null;
if (!empty($_POST)) {
if (isset($_POST['username']) { $username = $_POST['username']; }
if (isset($_POST['name']) { $name = $_POST['name']; }
}
if (is_null($username) || is_null($name)) {
throw new Exception('Invalid request');
}
$user = $this->User->search($username);
if (isset($user)) {
thorw new Exception('User not available');
}
// ... more code ...
}
How can I print those exceptions in Ajax?
Edit:
There is another way to handle this. Set below header before throwing an exception
header("HTTP/1.1 400 User not available");
// throw exception
Then fail handler from my Ajax code can print it like
$("#spanId").html("Error : " + err.statusText)
But I don't want to do this. I want to print it in success handler itself.

Answer 1

here a little demonstration of my comment.
in your ajax
.done(function (response) {
if( (response.exception) === "noExp" )
{
alert("success!!!");
//no exception
}
else
{
//handle it
}
}
in php
function add()
{
$username = null;
$name = null;
$exception = "noExp";
if (!empty($_POST)) {
if (isset($_POST['username']) { $username = $_POST['username']; }
if (isset($_POST['name']) { $name = $_POST['name']; }
}
if (is_null($username) || is_null($name)) {
//throw new Exception('Invalid request');
$exception = "invalid request";
}
$user = $this->User->search($username);
if (isset($user)) {
//thorw new Exception('User not available');
$exception = "User not available";
}
// ... more code ...
$result = array(
"exception" => $exception,
//other returns
)
echo json_encode($result);
}