解决了微信小程序自定义表格，并显示的多条数据的问题。
index.wxml
<view> <text>我的调查问卷</text> <scroll-view scroll-x="true"> <view> <view> <view>姓名</view> <view>性别</view> <view>年龄</view> </view> <view wx:for = "{{items}}" wx:key=""> <text>{{item.name}}</text> <text>{{item.gender}}</text> <text>{{item.age}}</text> </view> </view> </scroll-view> <button bindtap="change">查看我的数据</button> </view>
index.js
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' }) } }) } })
data.php
<?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(); ?>
文章有借鉴，并非全部自创