CodeIgniter + jQuery Ajax runs error but successfully callback is called - php

Question

My Codeigniter: (Do you think there is an error?)
public function KayitOl()
{
$data = array(
'kullaniciadi' => $this->input->post('kullaniciadi'),
'email' => $this->input->post('email'),
'sifre' => $this->input->post('sifre')
);
$kuladi = $this->input->post('kullaniciadi');
$sorgu = $this->db->query("SELECT * FROM uyeler WHERE kullaniciadi='".$kuladi."'");
if ($sorgu->num_rows() > 0)
{
$response_array['status'] = 'error';
echo json_encode($response_array);
}
else
{
$this->db->insert('uyeler',$data);
$response_array['status'] = 'success';
echo json_encode($response_array);
}
}
My jQuery Code: (Do you think there is an error?)
$(".submit").on("click", function(){
var kuladi = $("#kullaniciadi").val();
var email = $("#email").val();
var sifre = $("#sifre").val();
var confirm = $("#sifreonay").val();
var hata = $("#hata").val();
var checkbox = $("#checkbox").is(":checked");
var link = "http://tantunisiparis:8080/main/anasayfa/KayitOl";
var pattern = /^\b[A-Z0-9._%-]+#[A-Z0-9.-]+\.[A-Z]{2,4}\b$/i;
if (!kuladi || !email || !sifre) {
$("p#hata").removeClass("hidden");
$("p#hata").html("Boş bırakılan alanlar var!");
}
else if (!pattern.test(email)) {
$("p#hata").removeClass("hidden");
$("p#hata").html("Lütfen geçerli bir e-mail giriniz!");
}
else if (!checkbox) {
$("p#hata").removeClass("hidden");
$("p#hata").html("Kullanıcı Sözleşmesini Kabul Etmediniz.");
}
else if (sifre != confirm) {
$("p#hata").removeClass("hidden");
$("p#hata").html("Şifreler eşleşmiyor!");
}
else{
$.ajax({
type :"POST",
url :link,
data : $("#kayitform").serialize(),
success: function (data){
console.log(data.status);
alert("Success döndü");
},
error: function (data){
console.log(data.status);
alert("Error döndü");
}
});
}
});
Why I am having a problem like this?
Any answer attempts are appreciated. Any correct answers are doubly appreciated ;)
Thanks!

Answer 1

You need to set HTTP status code. So in case of error call this code in the controller $this->output->set_status_header(500);.
public function KayitOl()
{
$data = array(
'kullaniciadi' => $this->input->post('kullaniciadi'),
'email' => $this->input->post('email'),
'sifre' => $this->input->post('sifre')
);
$kuladi = $this->input->post('kullaniciadi');
$sorgu = $this->db->query("SELECT * FROM uyeler WHERE kullaniciadi='".$kuladi."'");
if ($sorgu->num_rows() > 0)
{
$response_array['status'] = 'error';
$this->output->set_status_header(500); // or any other code
echo json_encode($response_array);
}
else
{
$this->db->insert('uyeler',$data);
$response_array['status'] = 'success';
echo json_encode($response_array);
}
}
You can read more about output class in the docs http://www.codeigniter.com/userguide3/libraries/output.html

Answer 2

$.ajax({
type :"POST",
url :link,
data : $("#kayitform").serialize(),
success: function (data){
if(data.status == 'success'){
console.log(data.status);
alert("Success döndü");
}
if(data.status == 'error'){
console.log(data.status);
alert("Error döndü");
}
}
});
I thing, This code will work for you...