jQuery中ajax请求后台返回json数据并渲染HTML的方法

<table border="0" class="restaurant_food" cellspacing="0" cellpadding="1">  <input type="text" name="dishes" value="" class="seek_product" placeholder="请输入菜名"/>  <button type='button' class="btn_nor" onclick="seek_product()">搜索</button>  <tr>    <th width="30%">序号</th>    <th width="70%">菜名</th>  </tr>  <tr data-id="">    <td></td>    <td class="tl"> <p></p></td>  </tr></table>

function seek_product(){  var product = $('.seek_product').val();  $.ajax({    type:'get',    url:'/Cash/Index/seek_product',    data:{name:product},    success:function(res){      var data = eval('('+res+')');      var len = data.length;      var cm = "";      if(len > 0){        for(var i = 0; i < len; i++){          cm += '<tr data-id='+data[i]['id']+'>';          cm += '<td>';          cm += i+1;          cm += '</td>';          cm += '<td class="tl">';          cm += '<p>'+data[i]["name"]+'</p>';          cm += '</td>';          cm += '</tr>';          console.log(cm);          $('.restaurant_food').html(cm);        }      }else{        $('.restaurant_food').html('抱歉，没有这道菜！');      }    }  })}

//搜索菜public function seek_product(){  $shop_id = session("cashShopId");  $name = I('get.name');  $map['name'] = array('like','%'.$name.'%');  $map['shop_id'] = $shop_id;  $map['status'] = 1;  $productList = M('product')->field('id,name')->where($map)->select();  echo json_encode($productList);}