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);}

以上这篇jQuery中ajax请求后台返回json数据并渲染HTML的方法就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持。