iOS: ошибка синтаксического анализа JSON при использовании return вместо echo в файле сервиса PHP

У меня есть следующие HTTP запрос с AFNetworking:

NSMutableDictionary *dictionary = [NSMutableDictionary new];
[dictionary setObject:@"login" forKey:@"request"];
[dictionary setObject:@"q" forKey:@"userName"];
[dictionary setObject:@"q" forKey:@"password"];

DDLogDebug(@"LoginView - login - Dictionary: %@", [dictionary description]);

AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
[manager setRequestSerializer:[AFJSONRequestSerializer serializer]];
[manager setResponseSerializer:[AFJSONResponseSerializer serializer]];
//[manager.securityPolicy setAllowInvalidCertificates:true];

AFHTTPRequestOperation *operation =
[manager POST:WEB_SERVICE_URL
parameters:dictionary
success:^(AFHTTPRequestOperation *operation, id responseObject) {

dispatch_async(dispatch_get_main_queue(), ^{
DDLogDebug(@"LoginView - Success Response: %@", responseObject);
NSString *success = [responseObject objectForKey:@"success"];

if ([success isEqualToString:@"1"]) {
DDLogDebug(@"LoginView - Login successful!");
[self performSegueWithIdentifier:kSegueLoginToTimeLine sender:self];
}
else {
DDLogError(@"LoginView - Login failed!");
[HMXCommonMethods showMessage:@"Kullanıcı adı ya da şifre hatalı" withTitle:@"Hata!"];
}
});
}

failure:^(AFHTTPRequestOperation *operation, NSError *error) {

dispatch_async(dispatch_get_main_queue(), ^{
DDLogError(@"LoginView - Error Response: %@", [error localizedDescription]);

NSInteger statusCode = [operation.response statusCode];

switch (statusCode) {
case 500:
[HMXCommonMethods showMessage:@"Sunucu hatası nedeniyle giriş yapılamadı!" withTitle:@"Hata!"];
break;
case 408:
[HMXCommonMethods showMessage:@"İstek zaman aşımına uğradı!" withTitle:@"Hata!"];
break;
case 404:
[HMXCommonMethods showMessage:@"Sunucuya erişilemedi!" withTitle:@"Hata!"];
break;
default:
[HMXCommonMethods showMessage:@"Sunucu hatası nedeniyle giriş yapılamadı!" withTitle:@"Hata!"];
break;
}
});
}];

[operation start];
DDLogError(@"LoginView - login - request started.");

Когда я отвечаю на этот запрос в моем PHP файл со следующим echo код работает нормально:

$response = "{\"success\":\"1\",";
$response = $response . "\"id\":\"" . $id . "\",";
$response = $response . "\"companyid\":\"" . $cid . "\",";
$response = $response . "\"companyname\":\"" . $cname . "\",";
$response = $response . "\"name\":\"" . $name . "\",";
$response = $response . "\"surname\":\"" . $surname . "\",";
$response = $response . "\"email\":\"" . $email2 . "\",";
$response = $response . "\"password\":\"" . $password2 . "\"}";
$stmt2->close();

$log = "Company ID = " . $cid . "\n" .
"Company Name = " . $cname . "\n" .
"Company Address = " . $caddress . "\n" .
"Company Phone = " . $cphone . "\n" .
"Company Web Page = " . $cwebpage . "\n";

$this->logger->debug("GetUser: Company Info:\n" . $log);
$this->logger->debug("GetUser: Response:\n" . $response);

echo $response;

Если я изменю echo с return я получил JSON pars error 3840, Я также пытался

echo json_encode($response)
return json_encode($response)

Но это не сработало, и я получил ту же ошибку. Как я могу использовать return без JSON ошибка?