解决数据库N+1查询问题

需求

数据表如下:

department表

id name
1 测试部门

user表

id name department_id
1 test 1

需求是得到以下结构的数据:

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[
{
"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
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$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
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$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:

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$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。

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