遵循規則codeigniter使用codeiginiter和ajax

Question

我很努力地使用codeigniter它創建的方式。我正在打開一個編輯頁面（訂單）。我想更改頁面上的數據，這些數據在離開字段後將被動態保存。但是，當我沒有成功的方式代言人的作品或與「好方法」的其他文字。我不想依賴於uri結構，並使用uri段獲取數據，就像現在的代碼一樣。我想下提高代碼但我沒有成功...

這是打開我的看法

public function edit($id = NULL) { 
    // Fetch a page or set a new one 
    $this->load->model('shop_m'); 
    $this->load->model('details_m'); 
    $this->data['subtitle'] = 'incoming'; 
    if ($id) { 
     if($this->transfer_m->get_permission($id) > 0){  
      $this->data['transfer'] = $this->transfer_m->get_rec($id); 
      // $this->data['transfer'] = $this->transfer_m->get($id,true); 
      count($this->data['transfer']) || $this->data['errors'][] = 'page could not be found'; 
      $this->data['details'] = $this->details_m->get_by(array('doc_no' => $id),FALSE); 
     } 
     else{ 
      redirect('admin/Dashboard', 'refresh'); 
     } 
    } else { 
     $this->data['transfer'] = $this->transfer_m->get_new(); 
    } 

    $this->data['shop_from'] = $this->shop_m->get($this->data['transfer']->from_customer,TRUE); 
    $this->data['shop_to'] = $this->shop_m->get($this->data['transfer']->to_customer,TRUE); 

    // Load the view 
    $this->data['subview'] = 'pages/details/transfer_edit'; 
    $this->load->view('pages/main', $this->data); 

} 

我的Ajax調用

function add_shop(shop){ 
var id = $('#doc_no').text(); 

$.ajax({ 
    url: base_url+"/admin/transfer/add_shop/"+id+"/"+shop, 
    async: false, 
    type: "POST", 
    data: "", 
    dataType: "html", 
    success: function(data) { 
     $('#ajax-content-container').hide(); 
    } 
}) 

}

代碼我的控制器功能添加一個店鋪

public function add_shop($id){ 
    $data = array('to_customer' => $this->uri->segment(5)); 
    $this->transfer_m->save($data,$id); 

} 

任何幫助將不勝感激！

Answer 1

您可以使用信息。修改你的ajax調用並傳遞下面的數據。看到這裏的文檔https://api.jquery.com/jquery.post/

data: { id: id, shop: shop } 

這些數據可以被訪問$this->input->post('id'); $this->input->post('shop');