php - 无法在 Controller 中制作数据的 html 并显示在下拉列表选择上

Question

我想在从下拉列表中选择餐厅区域后显示餐厅。但我的代码没有显示餐厅名称和该餐厅的菜单按钮，请告诉我哪里出错了

这是我的查看代码

    <div id="restaurant">


    </div>

这是 Controller 代码

     public function get_rests()
    {

        $cit_id = $this->input->post('cit_id');
        $area = $this->input->post('areaID');
        $where = array('city_id'=>$cit_id,'city_area_id'=>$area);


        $html = '

        <div class="container" id="">


            <table align="centre" class="table table-condensed table-striped table-hover no-margin"style="width:70%" id="">

                <thread>

                    <tr style="width: 56%;">
                        <th>
                            No.

                        </th>
                        <th style="">
                            Restaurant Names
                        </th>
                    </tr>

                </thread>
                <tbody>

                <?php $i=1;foreach($result as $row){
                    ?>

                    <tr id="<?php echo $row->restaurant_id; ?>" class="res_id">
                        <th style="">
                            <?php echo  $i++; ?>

                        </th>
                        <th style="">
                            <?php echo $row->restaurant_name; ?>




                        </th>

                        <th style="width: 1%" >
                            <a href="<?php echo base_url();?>index.php/BulkRecipe_Controller/bulk_recipe/<?php echo $row->restaurant_id;?>"  class="btn btn-warning" <i class="glyphicon-edit"></i>See Menu</a>

                        </th>
                    </tr>
                <?php } ?>

                </tbody>
            </table>
        </div>

        ';
        echo $html;

这是我的模型代码 View

function select_record($table, $where = NULL){

    $this->db->select();
    if($where)
        $this->db->where($where);
    $this->db->from($table);
    $query = $this->db->get();
    //  echo $this->db->last_query();
    return $query->result();
}

脚本代码

 function get_rests(){

        var city_id = $('#city_id').val();
        var area_id = $("#area_id").val();
        $.ajax({
            type: "POST",
            url: "<?=base_url();?>index.php/Bulk_Controller/get_rests",
            data: {cit_id: city_id,areaID: area_id},
            dataType: "text",
            cache:false,
            success:
                function(data){
                    // alert(data);
                    $('#restaurant').html(data);
                }
        });
    }

Answer 1

您没有调用模型函数来从数据库表中获取数据。

public function get_rests()
{

    $cit_id = $this->input->post('cit_id');
    $area = $this->input->post('areaID');
    $where = array('city_id'=>$cit_id,'city_area_id'=>$area);

    $result= $this->your_model->select_record('table_name',$where);
    $html = '

    <div class="container" id="">


        <table align="centre" class="table table-condensed table-striped table-hover no-margin"style="width:70%" id="">

            <thread>

                <tr style="width: 56%;">
                    <th>
                        No.

                    </th>
                    <th style="">
                        Restaurant Names
                    </th>
                </tr>

            </thread>
            <tbody>

            <?php $i=1;foreach($result as $row){
                ?>

                <tr id="<?php echo $row->restaurant_id; ?>" class="res_id">
                    <th style="">
                        <?php echo  $i++; ?>

                    </th>
                    <th style="">
                        <?php echo $row->restaurant_name; ?>




                    </th>

                    <th style="width: 1%" >
                        <a href="<?php echo base_url();?>index.php/BulkRecipe_Controller/bulk_recipe/<?php echo $row->restaurant_id;?>"  class="btn btn-warning" <i class="glyphicon-edit"></i>See Menu</a>

                    </th>
                </tr>
            <?php } ?>

            </tbody>
        </table>
    </div>

    ';
    echo $html;