ajax從javascript請求到codeigniter控制器

Question

我正在從javascript函數中的codeigniter視圖發出ajax請求，但沒有任何反應，並彈出成功警報（確定）

function show_drop_downn(){
    document.getElementById("drop_downn").style.visibility = "visible"; 
    $.ajax({
        type: "POST",
        url: "http://localhost/ok/index.php/search/ajax_delete_ntify",
        success: alert('ok'),
    }); 
}

這是我的控制器，它工作正常，當我在瀏覽器中復制粘貼url（用於ajax請求）時，一切正常，控制器對模型進行了調用，效果很好

function ajax_delete_ntify()
{
    echo "incontroller";
    $email=$this->session->userdata('email_of_user');
    $this->load->model('search_peoplee');
    $data['userid']= $this->search_peoplee->get_userid_from_email($email);
    foreach ($data['userid'] as $row)
    {
        $one=$row->userid;
    }
    $this->search_peoplee->delete_notifications($one);
    return;
}

Answer 1

查看：

function show_drop_downn(){
    document.getElementById("drop_downn").style.visibility = "visible"; 
    $.ajax({
        type: "POST",
        url: "http://localhost/ok/index.php/search/ajax_delete_ntify",
        success:function(
         console.log("success");
        )
    }); 
}

控制器：

function ajax_delete_ntify()
{
    echo "incontroller";
    $email=$this->session->userdata('email_of_user');
    $this->load->model('search_peoplee');
    $data= $this->search_peoplee->get_userid_from_email($email);
    $res=$this->search_peoplee->delete_notifications($data[0]->userid);
    if($res){
      echo json_encode(array('success'=>true));
    }else{
      echo json_encode(array('success'=>false));
    }
}