layui的页码显示不出来，有“异常，错误提示：parsererror”这样的报错

Question

<script>         
                        layui.use('table', function(){
                          var table = layui.table;
                          
                          table.render({
                             elem: '#test'
                            ,url:'/demo/select.php'
                            ,method:'post'
                            ,toolbar: '#toolbarDemo' //开启头部工具栏，并为其绑定左侧模板
                            ,response:{
                             code:'code',
                             data:'data'
                             }
                            ,limit: 10
                            ,limits: [10, 20, 30]
                            ,defaultToolbar: ['filter', 'exports', 'print', { //自定义头部工具栏右侧图标。如无需自定义，去除该参数即可
                            title: '提示'
                            ,layEvent: 'LAYTABLE_TIPS'
                            ,icon: 'layui-icon-tips',
                            }]
                            ,title: '用户数据表'
                            ,cols: [[
                              {type: 'checkbox', fixed: 'left'}
                              ,{field:'id', title:'ID', width:80, fixed: 'left', unresize: true, sort: true}
                              ,{field:'username', title:'用户名', width:120, edit: 'text'}
                              ,{field:'age', title:'年龄', width:120, edit: 'text'}
                              ,{field:'sex', title:'性别', width:120, edit: 'text'}
                              ,{field:'phonenumber', title:'电话号码', width:120, edit: 'text'}
                            
                            ]]
                            ,page: true
                            ,parseData: function (res) {
                            return data = {
                            "code": res.code, //解析接口状态
                            "msg": res.message, //解析提示文本
                            "count":res.count, //解析数据长度 //这里是总数量 
                            "data": res.data //解析数据列表 这里的data即是渲染到表格中的数据, 因为我传过来的json数据是list作为key值的,所以这里是个list 根据你自己的json数据进行更改
                      };
  
                    
                  }
                          });

<?php
$servername = "localhost";
$usern = "admin";
$password = "123456";
$dbname = "admin";
// 创建连接
header("Content-type:text/html;charset=utf-8");
$link = mysqli_connect($servername,$usern,$password,$dbname);
// Check connection

$sql = "SELECT * FROM test3";
$result = mysqli_query($link, $sql);
class User 
{
public $id;
public $yonghu;
public $age;
public $sex;
public $phonenumber;

}



if($result){
    //echo "查询成功";
    while ($row = mysqli_fetch_array($result))
    {
    $user = new User();
    $user->id = $row["id"];
    $user->username = $row["username"];
    $user->age = $row["age"];
    $user->sex = $row["sex"];
    $user->phonenumber = $row["phonenumber"];
    $data[]=$user;
    }
    $sql_count = "SELECT COUNT(*) FROM test3";
    $a = mysqli_query( $link, $sql_count );
    $b = mysqli_fetch_assoc( $a );

    $json = json_encode($data);//把数据转换为JSON数据.
   
    echo '{"code":0,"count":'.$b.',"data":'.$json.'}';

    }else{
    echo "查询失败";
    }



?>  

Answer 1

这里，改成：

// $json = json_encode($data);//把数据转换为JSON数据.
   $res = [
  'code' => 0,
'message' => '请求成功',
'count' => $b,
'data' => $data
];
    echo json_encode($res);