PHP和SQL数据库：如何实现基于分类的JSON分组输出？

<?php// 初始化响应数组$response = [    'code' => 0,    'msg' => 'ok',    'data' => null,];// 用于存储分组结果$result = [];// 查询分类表$queryclass = mysqli_query($conn, "select cid, cname from class_wtool where uid = 'common'");while ($rclass = mysqli_fetch_array($queryclass)) {    // 创建一个空数组来存储特定分类的详情    $item = [];    // 设置分类名称    $item['name'] = $rclass['cname'];    // 根据分类查询详情表    $querydetail = mysqli_query($conn, "select wid, simplew, detailw from word_wtool where uid = 'common' and cid = " . $rclass['cid']);    // 遍历详情表记录并填充数组    while ($rdetail = mysqli_fetch_array($querydetail)) {        // 将详情记录添加到列表数组中        $item['list'][] = [            'wid' => $rdetail['wid'],            'simplew' => $rdetail['simplew'],            'detailw' => $rdetail['detailw']        ];    }    // 将分类和详情数组添加到结果数组中    $result[] = $item;}// 将分组结果分配给响应数组$response['data'] = $result;// 设置响应头，使其以 json 格式输出header('content-type: application/json');// 将响应数组转换为 json 字符串并输出echo json_encode($response, json_unescaped_slashes | json_unescaped_unicode);// 退出脚本die();

{  "code": 0,  "msg": "ok",  "data": [    {      "name": "分类 1",      "list": [        {          "wid": 1,          "simpleW": "简介1",          "detailW": "详情1"        },        {          "wid": 2,          "simpleW": "简介2",          "detailW": "详情2"        }      ]    },    {      "name": "分类 2",      "list": [] // 如果该分类没有详情记录，则此列表为空    }  ]}