在codeigniter中检查AJAX调用是否成功

Question

I am using an AJAX call to insert some data into MYSQL

JS code:

$("input.addtruck").click(function (event) {
event.preventDefault();
        var user_id = $("input#user_id").val();
    var numar = $("input#numar").val();
    var serie = $("input#serie").val();
    var marca = $("select#marca").val();
    jQuery.ajax({
        type: "POST",
        url: "<?php echo base_url(); ?>" + "aplicatie/add_truck",
        dataType: 'json',
        data: {user_id: user_id, numar: numar, serie: serie, marca: marca},
    });
    success: function (res) {
        if (res)
        {
            jQuery("div#truck_form").hide();
            jQuery("div#success").show();
        } else {
            jQuery("div#error").show();
        }
    }
});

Method used from controller:

    function add_truck() {
            $data = array(
                'user_id' => $this->input->post('user_id'),
                'marca' => $this->input->post('marca'),
                'serie' => $this->input->post('serie'),
                'numar' => $this->input->post('numar')
            );
//Transfering data to Model
            $this->trucks_model->insert_truck($data);
            $data['confirmare'] = 'Data Inserted Successfully';
        }

And method from models file

function insert_truck($data){
$this->db->insert('trucks', $data);
}

Basicly i need to hide the #truck_form and show #success if the data was inserted, or show #error .

Answer 1

You need to check data is inserted or not in database using affected_rows in model

Model

function insert_truck($data){
$this->db->insert('trucks', $data);
$afftectedRows=$this->db->affected_rows();
if($afftectedRows>0)
{
    return TRUE;
}
else{
    return FALSE;
}

}

YOu need to echo your result in Controller

Controller

function add_truck() {
            $data = array(
                'user_id' => $this->input->post('user_id'),
                'marca' => $this->input->post('marca'),
                'serie' => $this->input->post('serie'),
                'numar' => $this->input->post('numar')
            );
//Transfering data to Model
        $res=$this->trucks_model->insert_truck($data);
        if($res){
        $data['msg'] = 'true';
        }else{
           $data['msg'] = 'false';
        }
         echo json_encode($data);
        }

Ajax

success: function (res) {
        if (res.msg=='true')
        {
            jQuery("div#truck_form").hide();
            jQuery("div#success").show();
        } else {
            jQuery("div#error").show();
        }
    }