Question

我有一个永远运行的MySQL（v5.7.26）查询。这是查询：

SELECT
    ur.user_id      AS user_id,
    sum(r.duration) AS total_time,
    count(user_id)  AS number_of_workouts
FROM user_resource ur
INNER JOIN resource r ON r.id = ur.resource_id
WHERE 
    ur.status = 1 
    AND NOT ur.action_date IS NULL 
    AND ur.user_id IN (
        SELECT user_id
        FROM user_resource ur2
        WHERE ur2.action_date >= now() - INTERVAL 2 DAY
    )
    AND r.type = 'WORKOUT'
    GROUP BY ur.user_id;

通过尝试了解问题出在哪里，我对此做了一些尝试。出于测试目的，我尝试分成两部分。所以：

SELECT user_id
FROM user_resource ur2
WHERE ur2.action_date >= now() - INTERVAL 2 DAY;

（非常快速地）返回用户user_id的列表。当我将返回的结果插入查询的第一部分时，如下所示：

SELECT
    ur.user_id      AS user_id,
    sum(r.duration) AS total_time,
    count(user_id)  AS number_of_workouts
FROM user_resource ur
INNER JOIN resource r ON r.id = ur.resource_id
WHERE 
    ur.status = 1 
    AND NOT ur.action_date IS NULL 
    AND ur.user_id IN (1,1,1,4,4,5,6,7,7,7);
      AND r.type = 'WORKOUT'
GROUP BY ur.user_id

它运行非常快。我的假设是IN（子查询）是瓶颈。

我当时正在考虑提取子查询并获取user_ids，然后将其用作变量，但是我不确定这是否是一种好方法，此外，我还遇到了问题。这是我的尝试：

-- first statement
SET @v1 = (SELECT user_id
FROM user_resource ur2
WHERE ur2.action_date >= now() - INTERVAL 2 DAY)

-- second statement
SELECT
    ur.user_id      AS user_id,
    sum(r.duration) AS total_time,
    count(user_id)  AS prefixes
FROM user_resource ur
INNER JOIN resource r ON r.id = ur.resource_id
WHERE 
    ur.status = 1 
    AND NOT ur.action_date IS NULL 
    AND ur.user_id IN (@v1);
    AND r.type = 'WORKOUT'
GROUP BY ur.user_id

这里的问题是第一条语句返回错误：

子查询返回1行以上。

预期结果是user_id，可以重复。我需要那些重复的计数。

我该如何解决？

Answer 1

尝试用EXISTS代替IN

...
AND EXISTS (SELECT *
                   FROM user_resource ur2
                   WHERE ur2.user_id = ur.user_id
                         AND ur2.action_date >= now() - INTERVAL 2 DAY)
...

以及user_resource (user_id, action_date)，user_resource (status, action_date, user_id)和/或user_resource (type)上的索引。

Answer 2

您可以尝试：

-- first statement
SET @v1 = (SELECT GROUP_CONCAT(user_id)
FROM user_resource ur2
WHERE ur2.action_date >= now() - INTERVAL 2 DAY)

-- second statement
SELECT
ur.user_id      AS user_id,
 sum(r.duration) AS total_time,
 count(user_id)  AS prefixes
FROM user_resource ur
INNER JOIN resource r ON r.id = ur.resource_id
WHERE ur.status = 1 AND NOT ur.action_date IS NULL AND FIND_IN_SET(ur.user_id,@v1)
AND r.type = 'WORKOUT'
GROUP BY ur.user_id

Answer 3

其他联接将比子查询更快：

SELECT
    ur.user_id      AS user_id,
    sum(r.duration) AS total_time,
    count(user_id)  AS number_of_workouts
FROM user_resource ur
INNER JOIN resource r ON r.id = ur.resource_id
INNER JOIN (
    SELECT user_id
    FROM user_resource ur2
    WHERE ur2.action_date >= now() - INTERVAL 2 DAY
) t ON t.user_id = ur.user_id
WHERE 
    ur.status = 1 
    AND NOT ur.action_date IS NULL 
    AND r.type = 'WORKOUT'
    GROUP BY ur.user_id;

MySQL查询运行非常慢

3 个答案: