PHP动态创建多维数组

Question

##PHP动态创建多维数组
我需要从数据库中获取所有信息，并且返回成json格式，如下图这样

我已经完成了，但是我需要在data这个里面创建多个数组，这个数组是我数据库里面有多少条信息，就创建多少个数组来完成，我试过了很多方法，依然无法做到，麻烦来个会的人解决一下，感谢🙏！
PHP代码

<?php
$resultArr = array();
$arr = array();
while ($data = mysqli_fetch_assoc($result)) {
    $resultArr = array(array("cid" => $data["id"],"cname" => $data["cname"],"state" => $data["state"]));
    $arr = array("code" => 1,"data" => $resultArr,"msg" => "操作成功");
}
echo json_encode($arr,JSON_UNESCAPED_UNICODE);

Answer 1

<?php
$resultArr = array();
$arr = array();
while ($data = mysqli_fetch_assoc($result)) {
    $resultArr[] = ["cid" => $data["id"],"cname" => $data["cname"],"state" => $data["state"]];
}
$arr = array("code" => 1,"data" => $resultArr,"msg" => "操作成功");
echo json_encode($arr,JSON_UNESCAPED_UNICODE);

Answer 2

<?php

$resultArr = array();
$arr = array();
while ($data = mysqli_fetch_assoc($result)) {
    $resultArr[] = array("cid" => $data["id"], "cname" => $data["cname"], "state" => $data["state"]);
}
$arr = array("code" => 1, "data" => $resultArr, "msg" => "操作成功");
echo json_encode($arr, JSON_UNESCAPED_UNICODE);
?>