Mysql解决数据库N+1查询问题

[
  {    "id":1,    "name":"test",    "department_id":1,    "department":{      "id":1,      "name":"测试部门"
    }
  }
]

$users = $db->query('SELECT * FROM `user`');foreach($users as &$user) {
  $users['department'] = $db->query('SELECT * FROM `department` WHERE `id` = '.$user['department_id']);
}

$users = $db->query('SELECT * FROM `user` INNER JOIN `department` ON `department`.`id` = `user`.`department_id`');// 手动处理返回结果为需求结构

$users = $db->query('SELECT * FROM `user`');
$departmentIds =[ ];foreach($users as $user) {  if(!in_array($user['department_id'], $departmentIds)) {
    $departmentIds[] = $user['department_id'];
  }
}
$departments = $db->query('SELECT * FROM `department` WHERE id in ('.join(',',$department_id).')');
$map = []; // [部门ID => 部门item]foreach($departments as $department) {
  $map[$department['id']] = $department;
}foreach($users as $user) {
  $user['department'] = $map[$user['department_id']] &＃63;&＃63; null;
 }