14

需求

数据表如下:

department表

|id|name|

user表

|id|name|department_id|

需求是得到以下结构的数据:

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

方法一:循环查询

  1. 查询用户列表
  2. 循环用户列表查询对应的部门信息
$users = $db->query('SELECT * FROM `user`');
foreach($users as &$user) {
    $users['department'] = $db->query('SELECT * FROM `department` WHERE `id` = '.$user['department_id']);
}

该方法查询次数为:1+N(1次查询列表,N次查询部门),性能最低,不可取。

方法二:连表

  1. 通过连表查询用户和部门数据
  2. 处理返回数据
$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

该方法其实也有局限性,如果 userdepartment 不在同一个服务器是不可以连表的。

方法三:1+1查询

  1. 该方法先查询1次用户列表
  2. 取出列表中的部门ID组成数组
  3. 查询步骤2中的部门
  4. 合并最终数据

代码大致如下:

$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']] ?? null;
 }

该方法对两个表没有限制,在目前微服务盛行的情况下是比较好的一种做法。


xialeistudio
21.5k 声望5k 粉丝