Question

我尝试创建一个规范化的数据库。因此，我试图加入21个表。我正在使用连接中几乎所有表的索引，如EXPLAIN语句的下面屏幕截图所示。

http://i.imgur.com/V5hQu.png

所有lookup_xxxxx表在2列上都有一个索引（content_id和xxxxxx_id）所有xxxxxx表在1列（xxxxx_id）上都有一个索引

我每张表中只有大约10行。我看到的问题是两个goup_concat将查询延长了200ms。我允许提交字段peripheral和programming language具有多个值。没有它们，查询小于80毫秒。我的问题是我是否应该取消group_concats并为他们做个别查询或重建我的数据库。

lookup_xxxxx表存储每个允许的值，然后其他表（例如peripheral）通过content_id将提交链接到允许的值。所有内容都以提交content_id为参考。 content表包含成员ID，名称等基本信息。

如果我的帖子不够明确，我道歉。

mysql> describe peripheral;
+------------------+----------+------+-----+---------+-------+
| Field            | Type     | Null | Key | Default | Extra |
+------------------+----------+------+-----+---------+-------+
| peripheral_id    | int(2)   | NO   | PRI | NULL    |       |
| peripheral       | char(30) | NO   |     | NULL    |       |
| peripheral_total | int(5)   | NO   |     | NULL    |       |
+------------------+----------+------+-----+---------+-------+

mysql> select * from peripheral;
+---------------+-----------------+------------------+
| peripheral_id | peripheral      | peripheral_total |
+---------------+-----------------+------------------+
|             1 | periph 1        |                0 |
|             2 | periph 2        |                1 |
|             3 | periph 3        |                3 |
+---------------+-----------------+------------------+

mysql> describe lookup_peripheral;
+---------------+---------+------+-----+---------+-------+
| Field         | Type    | Null | Key | Default | Extra |
+---------------+---------+------+-----+---------+-------+
| content_id    | int(10) | NO   | MUL | NULL    |       |
| peripheral_id | int(2)  | NO   |     | NULL    |       |
+---------------+---------+------+-----+---------+-------+  


mysql> mysql> select * from lookup_peripheral;
+------------+---------------+
| content_id | peripheral_id |
+------------+---------------+
|         74 |             2 |
|         74 |             5 |
|         75 |             2 |
|         75 |             5 |
|         76 |             3 |
|         76 |             4 |
+------------+---------------+

SELECT group_concat(DISTINCT peripheral.peripheral_id) as peripheral_id, group_concat(DISTINCT programming_language.programming_language_id) as programming_language_id, c.member_name, c.member_id, c.added_date_time, c.title, c.raw_summary, c.raw_all_content, c.meta_tags, c.main_pic_thumb, application.application_id, architecture.architecture_id, compiler.compiler_id, device_family.device_family_id, difficulty.difficulty_id, ide.ide_id, programmer.programmer_id, table_name.table_name_id, device_name.device_name
FROM (content as c)
INNER JOIN lookup_peripheral ON 76 = lookup_peripheral.content_id
INNER JOIN peripheral ON peripheral.peripheral_id = lookup_peripheral.peripheral_id
INNER JOIN lookup_programming_language ON 76 = lookup_programming_language.content_id
INNER JOIN programming_language ON programming_language.programming_language_id = lookup_programming_language.programming_language_id
.......
LEFT OUTER JOIN device_name ON device_name.content_id = c.content_id
INNER JOIN table_name ON table_name.table_name_id = lookup_table_name.table_name_id
WHERE `c`.`content_id` = '76'

Answer 1

我认为您还应该INDEX_ lookup_peripheral.peripheral_id字段。索引外键使INNER JOIN更快。另外，你真的需要DISTINCT子句，因为你连接ID字段？第二个想法，也许你可以省略GROUP_CONCATs和INNER JOINs（我相信你的一些JOIN条件重叠，如果我弄错了，请纠正我）

... (SELECT GROUP_CONCAT(l.peripheral_id) from lookup_peripheral lp where lp.content_id = 76) as peripheral_id, ...

希望这有帮助。

慢Group_concat查询

1 个答案: