jQuery中ajax请求后台返回json数据并渲染HTML的方法

html实例

<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>

jquery实例

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！');
      }
    }
  })
}

php实例

//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);
}