我有一个 SQL 查询失败(大部分时间),因为连接的行太多。 MySQL 提供的错误是 The SELECT would examine more than MAX_JOIN_SIZE rows;检查您的 WHERE 并使用 SET SQL_BIG_SELECTS=1 或 SET MAX_JOIN_SIZE=# 如果 SELECT 没问题
。我知道我可以通过设置提到的变量 SQL_BIG_SELECTS 和 MAX_JOIN_SIZE 来避免错误,但我觉得这不是正确的方法,并且只会在未来稍微插入这个问题,因为 future 连接计数可能会增加。
事实:我有一个事件计划工具,可以将用户(=工作人员)分配给某些任务。这些表是 users
(userid,username) [ID and name], tasks
(taskid,task,start,end) [ID, task name, start as timestamp, end作为时间戳] 和 userassignment
(id,userid,taskid,deleted) [ID, user assigned to a task, the task, is the assignment still valid..
准确的表定义是这样的:
CREATE TABLE users (
userid INT NOT NULL AUTO_INCREMENT,
username VARCHAR(250),
PRIMARY KEY (userid)
);
CREATE TABLE tasks (
taskid INT NOT NULL AUTO_INCREMENT,
task VARCHAR(250),
start INT,
end INT,
PRIMARY KEY (taskid),
INDEX USING BTREE (start),
INDEX USING BTREE (end)
);
CREATE TABLE userassignment (
id INT NOT NULL AUTO_INCREMENT,
userid INT,
taskid INT,
deleted TINYINT,
PRIMARY KEY (id),
INDEX USING BTREE (userid),
INDEX USING BTREE (userid),
UNIQUE KEY `usertasks` ( `userid` , `taskid` )
);
我需要知道分配了哪些用户以及在事件的哪几天(第 1 天、第 2 天、第 3 天)分配了他们。
我的查询是这样的:
SELECT
u.userid,
u.username,
COUNT(ua.id) AS count_all,
dayone.c AS count_one,
daytwo.c AS count_two,
daythree.c AS count_three
FROM
users AS u
INNER JOIN
userassignment AS ua ON ua.userid = u.userid AND ua.deleted = 0
INNER JOIN
tasks AS t ON ua.taskid = t.taskid
LEFT JOIN (
SELECT
u.userid,
COUNT(ua.id) AS c
FROM
users AS u
INNER JOIN
userassignment AS ua ON
ua.userid = u.userid AND
ua.deleted = 0
INNER JOIN
tasks AS t ON
ua.taskid = t.taskid
WHERE
t.start > UNIX_TIMESTAMP("2014-08-01 00:00:00") AND
t.start < UNIX_TIMESTAMP("2014-08-02 00:00:00")
GROUP BY
u.userid
) AS dayone ON dayone.userid = u.userid
LEFT JOIN (
SELECT
u.userid,
COUNT(ua.id) AS c
FROM
users AS u
INNER JOIN
userassignment AS ua ON
ua.userid = u.userid AND
ua.deleted = 0
INNER JOIN
tasks AS t ON
ua.taskid = t.taskid
WHERE
t.start > UNIX_TIMESTAMP("2014-07-31 00:00:00") AND
t.start < UNIX_TIMESTAMP("2014-08-01 00:00:00")
GROUP BY
u.userid
) AS daytwo ON daytwo.userid = u.userid
LEFT JOIN (
SELECT
u.userid,
COUNT(ua.id) AS c
FROM
users AS u
INNER JOIN
userassignment AS ua ON
ua.userid = u.userid AND
ua.deleted = 0
INNER JOIN
tasks AS t ON
ua.taskid = t.taskid
WHERE
t.start > UNIX_TIMESTAMP("2014-08-02 00:00:00") AND
t.start < UNIX_TIMESTAMP("2014-08-04 00:00:00")
GROUP BY
u.userid
) AS daythree ON daythree.userid = u.userid
WHERE
t.start > UNIX_TIMESTAMP("2014-07-31 00:00:00") AND
t.start < UNIX_TIMESTAMP("2014-08-04 00:00:00")
GROUP BY
u.userid
ORDER BY
username ASC
首先,我选择了在三天中的某一天有任务的所有用户(数据库中的用户比分配给任务的用户多六倍),然后我离开加入了这三天中每一天的分配用户。
那么,有没有办法重建查询以连接更少的行?我只需要知道,哪一天分配给谁,而不是分配的数量。
我已经尝试过 UNION 几个查询,但没有成功。
真实查询(不在 SQL Fiddle 中)的 EXPLAIN 是:
id select_type table type possible_keys key key_len ref rows filtered Extra
1 PRIMARY t range PRIMARY,start start 5 NULL 120 100.00 Using where; Using index; Using temporary; Using filesort
1 PRIMARY ua ref usertasks,userid,taskid taskid 2 db1154575-helfer.t.id 2 100.00 Using where
1 PRIMARY u eq_ref userid userid 2 db1154575-helfer.ua.userid 1 100.00
1 PRIMARY <derived2> ALL NULL NULL NULL NULL 152 100.00
1 PRIMARY <derived3> ALL NULL NULL NULL NULL 94 100.00
1 PRIMARY <derived4> ALL NULL NULL NULL NULL 147 100.00
4 DERIVED t range PRIMARY,start start 5 NULL 53 100.00 Using where; Using index; Using temporary; Using filesort
4 DERIVED ua ref usertasks,userid,taskid taskid 2 db1154575-helfer.t.id 2 100.00 Using where
4 DERIVED u eq_ref userid userid 2 db1154575-helfer.ua.userid 1 100.00 Using index
3 DERIVED t range PRIMARY,start start 5 NULL 21 100.00 Using where; Using index; Using temporary; Using filesort
3 DERIVED ua ref usertasks,userid,taskid taskid 2 db1154575-helfer.t.id 2 100.00 Using where
3 DERIVED u eq_ref userid userid 2 db1154575-helfer.ua.userid 1 100.00 Using index
2 DERIVED t range PRIMARY,start start 5 NULL 44 100.00 Using where; Using index; Using temporary; Using filesort
2 DERIVED ua ref usertasks,userid,taskid taskid 2 db1154575-helfer.t.id 2 100.00 Using where
2 DERIVED u eq_ref userid userid 2 db1154575-helfer.ua.userid 1 100.00 Using index
最佳答案
那么,这真的只是一种冗长的说法吗...
SELECT u.*
, DATE(FROM_UNIXTIME(t.start)) dt
, COUNT(t.taskid) total
FROM users u
LEFT
JOIN userassignment ut
ON ut.userid = u.userid
AND ut.deleted = 0
LEFT
JOIN tasks t
ON t.taskid = ut.taskid
GROUP
BY u.userid
, DATE(FROM_UNIXTIME(t.start))
在上面的示例中,您可以将 COUNT(t.taskid)
更改为 COUNT(CASE WHEN x = 'y' THEN z END) 或 SUM(CASE...
关于mysql - 重建 MySQL 查询以保持低于 MAX_JOIN_SIZE 行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24953577/