我正在尝试动态显示数据aj ajax

控制器:

在我使用两个表类别和secound sub_category在类别我获取类别名称动态显示在Sub_Category我获取图像,价格,标题

public function product_grid() { $id= $this->input->post('dataid'); echo 'Data-Id is form controller: '.$id ; $this->PizzaUp_User_model->getid($id); $data['res'] = $this->PizzaUp_User_model->select('category'); $data['rs'] = $this->PizzaUp_User_model->get_cetegory('sub_category'); $this->load->view('product_grid',$data); } 

查看文件

  <div data-filter="" data_id="" name="id" class="cbp-filter-item button_id"> 
<div class="cbp-item ">
<img src="https://stackoverflow.com/questions/44587319/i-am-trying-to-display-data-dynamically-id-sing-ajax/" alt="">
<a href="https://stackoverflow.com/questions/44587319/i-am-trying-to-display-data-dynamically-id-sing-ajax/more_get_id/" class="cbp-singlePage cbp-l-caption-buttonLeft">Add to cart <a href="https://stackoverflow.com/questions/44587319/i-am-trying-to-display-data-dynamically-id-sing-ajax/" class="cbp-lightbox cbp-l-caption-buttonRight">view larger

使用Ajax获取并传递id

  $(document).ready(function() { $(".button_id").click(function(){ var dataid=$(this).attr('data_id'); $.ajax({ url: '/Home/product_grid', type: "POST", data: { dataid: dataid }, success: function(data) { alert(data); }, error: function(xhr, status, error) { var error=xhr.responseText; alert(error); } }); }); });  

我不知道PHP,但猜你的控制器function缺少参数

  public function product_grid() 

  public function product_grid($dataid) 
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