php - 我正在尝试使用ajax动态显示数据id

Question

Controller :

因为我使用了两个表类别和第二个子类别 在类别中，我获取类别名称以动态显示 在 Sub_Category 中，我获取了图像、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

<?php 
                        foreach ($res as $ro)
                        {
                    ?>
                    <div data-filter="<?php echo $ro['category_id']; ?>" data_id="<?php echo $ro['category_id']; ?>" name="id" class="cbp-filter-item button_id">
                                <?php echo $ro['category_name']; ?><div class="cbp-filter-counter"></div>
                    </div>
                    <?php 
                        }
                    ?>
                </div>  


            <div id="grid-container" class="cbp" >

                        <?php foreach ($rs as $row)
                              {
                         ?>

                            <div class="cbp-item <?php echo $ro['category_id']; ?>">
                            <div class="cbp-caption">
                                <div class="cbp-caption-defaultWrap">
                            <img src="<?php  echo base_url('image/category/'.$row['image']); ?>" alt="">
                            </div>
                            <div class="cbp-caption-activeWrap">
                            <div class="cbp-l-caption-alignCenter">
                                <div class="cbp-l-caption-body">
                                    <a href="more_get_id/<?php echo $row['sub_category_id'];?>" class="cbp-singlePage cbp-l-caption-buttonLeft">Add to cart</a>
                                    <a href="<?php  echo base_url('image/category/'.$row['image']); ?>" class="cbp-lightbox cbp-l-caption-buttonRight">view larger</a>
                                </div>
                            </div>
                        </div>

                    </div>
                    <div class="cbp-l-grid-projects-title"><?php echo $row['category_name']; ?></div>
                    <div class="cbp-l-grid-projects-desc"><?php echo $row['sub_category_title']; ?></div>
                    <div class="cbp-l-grid-projects-price"><?php echo $row['R_price']; ?></div>

                </div>
                <?php 
                    } 
                ?>

使用 Ajax 获取并传递 ID

<script type="text/javascript">
$(document).ready(function() {
        $(".button_id").click(function(){
         var dataid=$(this).attr('data_id');
         $.ajax({
            url: '<?php echo site_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);
              } 
        });
    });
}); 
 </script>   

Answer 1

我不懂 php，但猜测你的 Controller 函数缺少参数

 public function product_grid()

至

  public function product_grid($dataid)