[英]Optimize not in sub-query MYSQL
我在MySQL数据库中有一个表,其中> 2000万行,下面的查询在少量行上执行得很好,但是如果有更多行,则需要2-3秒。 如何优化它,使其至少运行<1? 注意-问题出在子查询SELECT read_state FROM消息中...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 | |
+------------+-------------+------+-----+---------+----------------+
扩展说明:
+----+--------------------+----------+-------------+---------------+-----------+---------+--------------------+--------+----------+---------------------------------------------------------------------------+
| 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 |
+----+--------------------+----------+-------------+---------------+-----------+---------+--------------------+--------+----------+---------------------------------------------------------------------------+
进行一些猜测,这样可能会更有效:
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;
这是作为对消息的LEFT JOIN再次进行额外的联接,然后使用case将结果转换为0或1。
当LEFT JOIN可以带回多个匹配行时,DISTINCT应该可以应付(如果不可能的话,您可以消除DISTINCT)
怀疑连接到用户的OR子句不会那么有效。 用2个LEFT OUTER JOIN替换对用户的INNER JOIN可能更好。 像这样的东西:
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;
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