Question

我正在努力使用codeigniter创建它的方式。我正在打开一个编辑页面（订单）。我想更改页面上的数据，这些数据将在离开字段后动态保存。但是，当我没有成功地使用codeigniter的方式或者以“好的方式”使用其他词语时。我不想依赖于uri结构并使用uri段获取数据，就像现在的代码一样。我想改进下面的代码，但我没有成功......

这是打开我的视图的代码

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电话

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();
    }
})

}

我的控制器功能添加商店

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

}

任何帮助将不胜感激!!

Answer 1

您可以使用帖子。修改您的ajax调用并传递如下数据。请参阅此处的文档https://api.jquery.com/jquery.post/

 data: { id: id, shop: shop }

可以访问此数据$this->input->post('id'); $this->input->post('shop');

按照规则codeigniter使用codeiginiter和ajax

1 个答案: