如何用PHP将SQL分组查询结果(分类表和详情表)输出为JSON格式？

<?php

// 初始化响应数组
$response = [
    'code' => 0,
    'msg' => 'Ok',
    'data' => null
];

// 暂存查询结果
$result = [];

// 查询分类信息
$queryClass = mysqli_query($conn, "SELECT cid, cname FROM class_wtool WHERE uid = 'common'");
while ($rclass = mysqli_fetch_array($queryClass)) {
    // 创建分类项目
    $item = [];
    
    // 赋值分类名称
    $item['name'] = $rclass['cname'];
    
    // 查询详情信息
    $queryDetail = mysqli_query($conn, "SELECT wid, simpleW, detailW FROM word_wtool WHERE uid = 'common' AND cid = " . $rclass['cid']);
    
    // 遍历详情信息
    while ($rdetail = mysqli_fetch_array($queryDetail)) {
        // 赋值到分类项目的列表中
        $item['list'][] = [
            'wid' => $rdetail['wid'],
            'simpleW' => $rdetail['simpleW'],
            'detailW' => $rdetail['detailW']
        ];
    }
    
    // 添加到结果中
    $result[] = $item;
}

// 赋值响应数组
$response['data'] = $result;

// 设置响应头
header('Content-Type: application/json');

// 输出 JSON 格式的响应
echo json_encode($response, JSON_UNESCAPED_SLASHES | JSON_UNESCAPED_UNICODE);

// 退出脚本
die();
?>