Question

MariaDB的。服务器版本：5.5.5-10.1.18-MariaDB MariaDB服务器

共有3个表格：



1. City: 4,000 rows (has foreign key country_id to Country.id and index)
2. Street: 40,000 rows  (has foreign key city_id to City.id and index)
3. House: 4,000,000,000 rows (has foreign key to Street.id and index)

1. City: 4,000 rows (has foreign key country_id to Country.id and index) 2. Street: 40,000 rows (has foreign key city_id to City.id and index) 3. House: 4,000,000,000 rows (has foreign key to Street.id and index) 以下SQL从未完成：

select count(*) from House , Street  WHERE Street.city_id IN 
(SELECT id FROM City WHERE country_id=177) 
and Street.id=House.street_id;

但是如果在嵌套的SELECT中而不是原始代码：

（SELECT id FROM City WHERE country_id = 177）

我把

SELECT id FROM City WHERE id IN（4617,4618）

运行时＆lt; 0.5秒

以下是慢速查询的解释计划

+------+-------------+-------+--------+----------------------------+------------------------+---------+------------------+------------+-------------+
| id   | select_type | table | type   | possible_keys               | key                   | key_len | ref              | rows       | Extra       |
+------+-------------+-------+--------+-----------------------------+-----------------------+---------+------------------+------------+-------------+
|    1 | PRIMARY     | House | index  | IDX_House_streetIndex       | IDX_House_streetIndex | 11      | NULL             | 4,000,000,000 | Using index |
|    1 | PRIMARY     | Street| eq_ref | PRIMARY,IDX_DF9A1AD51E5D0459| PRIMARY               | 4       | House.street_id  |          1 |             |
|    1 | PRIMARY     | City  | eq_ref | PRIMARY,IDX_784DD132166D1F9C| PRIMARY               | 4       | Street.city_id   |          1 | Using where |
+------+-------------+-------+--------+------------------------------------------+------------------------------------------+---------+———————

Answer 1

试试这个：

CREATE INDEX i_1 ON City(id);
CREATE INDEX i_2 ON City(country_id);

CREATE INDEX i_3 ON Street(id)
CREATE INDEX i_4 ON Street(city_id)

CREATE INDEX i_5 ON House(street_id);

Answer 2

不要使用IN ( SELECT ... );它通常表现不佳。相反，请使用等效的JOIN。

select  count(*)
    FROM  City
    JOIN  Street  ON Street.city_id = City.id
    JOIN  House   ON House.street_id = Street.id
    WHERE  City.country_id=177

然后，您需要这些索引：

City:    INDEX(country_id, id)
Street:  INDEX(city_id, id)
House:   INDEX(street_id)

请，在提出此类问题时，请提供SHOW CREATE TABLE。

4 十亿房屋？我怀疑世界上是否有这么多人！

如果这些建议不充分，则重新考虑表格是否“过度标准化”。

MariaDB中的SQL速度慢

2 个答案: