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();
?>