微信小程序实现用table显示数据库反馈的多条数据功能示例

<view> <text>我的调查问卷</text> <scroll-view scroll-x="true" style=" white-space: nowrap; display: flex"> <view class="table">  <view class="tr bg-header">   <view class="th">姓名</view>    <view class="th">性别</view>   <view class="th">年龄</view>  </view>  <view class="tr bg-items" wx:for = "{{items}}" wx:key="">    <text class="td">{{item.name}}</text>   <text class="td">{{item.gender}}</text>   <text class="td">{{item.age}}</text>  </view> </view> </scroll-view> <button bindtap="change">查看我的数据</button></view>

Page({ /**  * 页面的初始数据  */ data: {  items:[]//定义变长数组 }, /**  * 生命周期函数--监听页面加载  */ onLoad: function () { }, /**  * 生命周期函数--监听页面显示  */ onShow: function () { }, /**  * 生命周期函数--监听页面隐藏  */ onHide: function () { }, change:function(e){  var that=this;  wx.request({   url: 'https://../data.php',//自己的服务器地址   header: {    "Content-Type": "application/json"   },   method: "POST",   success: function (res) {    for (var i = 0, len = res.data.length; i < len; i++){     that.data.items[i] = res.data[i];    }    that.setData({     items: that.data.items    })   },   fail: function (res) {    wx.showToast({     'title': 'request fail'    })   }  }) }})

<?php$servername = "localhost";$username = "root";$password = "***";//数据库连接密码$dbname = "mydb";// 创建连接$conn = new mysqli($servername, $username, $password, $dbname);// 检测连接if ($conn->connect_error) { die("connect server fail: " . $conn->connect_error);}$query = "select * from table";$result = mysqli_query($conn,$query);$data = array();while ($rows = mysqli_fetch_assoc($result)){$data[] = $rows;}echo json_encode($data);$conn->close();?>