我在MySQL数据库中有表格，其中> 20 000 000行，下面的查询在少量行上执行很好，但如果有更多行则需要2-3秒。如何优化它以使其运行＆lt;至少1？ 注意 - 问题出在子查询SELECT read_state FROM messages ... 查询：
SELECT sql_no_cache users.id AS uid,
name,
avatar,
avatar_date,
driver,
msg,
DATE,
messages.removed,
from_id = 528798 AS outbox ,
!(0 IN
(SELECT read_state
FROM messages AS msgs FORCE KEY(user_id_2)
WHERE (msgs.from_id = messages.from_id
OR msgs.from_id = messages.user_id)
AND msgs.user_id = 528798
AND removed = 0
)) AS read_state
FROM dialog,
messages,
users
WHERE messages.id = mid
AND ((uid1 = 528798
AND users.id = uid2)
OR (uid2 = 528798
AND users.id = uid1))
ORDER BY DATE DESC;
显示消息中的索引;
+----------+------------+-------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| Table | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | Index_comment |
+----------+------------+-------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| messages | 0 | PRIMARY | 1 | id | A | 27531939 | NULL | NULL | | BTREE | | |
| messages | 1 | to_number | 1 | to_number | A | 22 | NULL | NULL | | BTREE | | |
| messages | 1 | from_id | 1 | from_id | A | 529460 | NULL | NULL | | BTREE | | |
| messages | 1 | from_id | 2 | to_number | A | 529460 | NULL | NULL | | BTREE | | |
| messages | 1 | user_id_2 | 1 | user_id | A | 655522 | NULL | NULL | | BTREE | | |
| messages | 1 | user_id_2 | 2 | read_state | A | 917731 | NULL | NULL | | BTREE | | |
| messages | 1 | user_id_2 | 3 | removed | A | 949377 | NULL | NULL | | BTREE | | |
| messages | 1 | idx_user_id | 1 | user_id | A | 809762 | NULL | NULL | | BTREE | | |
| messages | 1 | idx_from_id | 1 | from_id | A | 302548 | NULL | NULL | | BTREE | | |
+----------+------------+-------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
desc消息;
+------------+-------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------+-------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| from_id | int(11) | NO | MUL | NULL | |
| user_id | int(11) | NO | MUL | NULL | |
| group_id | int(11) | NO | | NULL | |
| to_number | varchar(30) | NO | MUL | NULL | |
| msg | text | NO | | NULL | |
| image | varchar(20) | NO | | NULL | |
| date | bigint(20) | NO | | NULL | |
| read_state | tinyint(1) | NO | | 0 | |
| removed | tinyint(1) | NO | | NULL | |
+------------+-------------+------+-----+---------+----------------+
EXPLAIN EXTENDED：
+----+--------------------+----------+-------------+---------------+-----------+---------+--------------------+--------+----------+---------------------------------------------------------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | filtered | Extra |
+----+--------------------+----------+-------------+---------------+-----------+---------+--------------------+--------+----------+---------------------------------------------------------------------------+
| 1 | PRIMARY | dialog | index_merge | uid1,uid2 | uid1,uid2 | 4,4 | NULL | 1707 | 100.00 | Using sort_union(uid1,uid2); Using where; Using temporary; Using filesort |
| 1 | PRIMARY | users | ALL | PRIMARY | NULL | NULL | NULL | 608993 | 100.00 | Range checked for each record (index map: 0x1) |
| 1 | PRIMARY | messages | eq_ref | PRIMARY | PRIMARY | 4 | numbers.dialog.mid | 1 | 100.00 | |
| 2 | DEPENDENT SUBQUERY | msgs | ref | user_id_2 | user_id_2 | 6 | const,const,const | 2607 | 100.00 | Using where |
+----+--------------------+----------+-------------+---------------+-----------+---------+--------------------+--------+----------+---------------------------------------------------------------------------+
答案 0 :(得分：1)
做出一些猜测，这样的事情可能更有效： -
SELECT DISTINCT users.id AS uid,
name,
avatar,
avatar_date,
driver,
msg,
`DATE`,
messages.removed,
from_id = 528798 AS outbox ,
CASE WHEN msgs.read_state IS NULL THEN 1 ELSE 0 END AS read_state
FROM messages
INNER JOIN dialog ON messages.id = dialog.mid
INNER JOIN users ON (dialog.uid1 = 528798 AND users.id = dialog.uid2) OR (dialog.uid2 = 528798 AND users.id = dialog.uid1)
LEFT OUTER JOIN messages msgs ON msgs.read_state = 0 AND msgs.user_id = 528798 AND removed = 0 AND (msgs.from_id = messages.from_id OR msgs.from_id = messages.user_id)
ORDER BY `DATE` DESC;
这是再次对消息进行LEFT JOIN的额外连接，然后使用case将结果转换为0或1。
当LEFT JOIN可以带回多个匹配的行时，DISTINCT应该应对（如果不可能，那么你可以消除DISTINCT）
怀疑加入用户的OR条款效率不高。用2 LEFT OUTER JOIN替换INNER JOIN可能更好。这样的事情： -
SELECT DISTINCT COALESCE(users1.id, users2.id) AS uid,
COALESCE(users1.name, users2.name),
COALESCE(users1.avatar, users2.avatar),
COALESCE(users1.avatar_date, users2.avatar_date),
COALESCE(users1.driver, users2.driver),
msg,
`DATE`,
messages.removed,
from_id = 528798 AS outbox ,
CASE WHEN msgs.read_state IS NULL THEN 1 ELSE 0 END AS read_state
FROM messages
INNER JOIN dialog ON messages.id = dialog.mid
LEFT OUTER JOIN users users1 ON (dialog.uid1 = 528798 AND users1.id = dialog.uid2)
LEFT OUTER JOIN users users2 ON (dialog.uid2 = 528798 AND users2.id = dialog.uid1)
LEFT OUTER JOIN messages msgs ON msgs.read_state = 0 AND msgs.user_id = 528798 AND removed = 0 AND (msgs.from_id = messages.from_id OR msgs.from_id = messages.user_id)
WHERE users1.id IS NOT NULL
OR users2.id IS NOT NULL
ORDER BY `DATE` DESC;