Solve database N+1 query problem

Requirements

The data table is as follows:

department table

|id|name|

user table

|id|name|department_id|

The requirement is to obtain data with the following structure:

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

Method 1: Loop query

1. Query user list

2. Loop user list to query corresponding department information

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

The number of queries for this method is: 1 N (1 query list , N query departments), the performance is the lowest and not advisable.

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 user 和 department 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。

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