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);
}