Solve database N+1 query problem-Mysql Knowledge-php.cn

Solve database N+1 query problem

步履不停

Release： 2019-06-19 09:41:50

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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"
        }
    }
]

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Method 1: Loop query

Query user list
Loop user list to query corresponding department information

$users = $db->query(&#39;SELECT * FROM `user`&#39;);foreach($users as &$user) {
    $users[&#39;department&#39;] = $db->query(&#39;SELECT * FROM `department` WHERE `id` = &#39;.$user[&#39;department_id&#39;]);
}

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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

通过连表查询用户和部门数据
处理返回数据

$users = $db->query(&#39;SELECT * FROM `user` INNER JOIN `department` ON `department`.`id` = `user`.`department_id`&#39;);// 手动处理返回结果为需求结构

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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

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

代码大致如下：

$users = $db->query(&#39;SELECT * FROM `user`&#39;);
$departmentIds =[ ];foreach($users as $user) {    if(!in_array($user[&#39;department_id&#39;], $departmentIds)) {
        $departmentIds[] = $user[&#39;department_id&#39;];
    }
}
$departments = $db->query(&#39;SELECT * FROM `department` WHERE id in (&#39;.join(&#39;,&#39;,$department_id).&#39;)&#39;);
$map = []; // [部门ID => 部门item]foreach($departments as $department) {
    $map[$department[&#39;id&#39;]] = $department;
}foreach($users as $user) {
    $user[&#39;department&#39;] = $map[$user[&#39;department_id&#39;]] ?? null;
 }

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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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