mysql 7 +加入一个表,如何加快速度?
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【中文标题】mysql 7 +加入一个表,如何加快速度?【英文标题】:mysql 7 + joins on a table, how to speed up? 【发布时间】:2017-11-19 05:11:33 【问题描述】:我有 7 张桌子。表 1 是“父表”和has_many
与表 2-7 的关系。
我想做一个查询来获取表 1 的行,其中包含表 2-7 中的 id 列表。我已经尝试了以下查询,但是对于大型数据库,我的查询需要 15 秒。我想知道如何使这个查询更快?
编辑几点说明: -> 删除 Distinct 只节省 4 秒,我仍然得到 10-11 秒的查询
-> 从查询中删除 1 个连接表(不管哪个)将时间从 15 秒减少到 2-3 秒。删除 2 个连接表(同样无关紧要)将查询减少到 1/2 秒。
SELECT
table1.table1_id as table1Id,
GROUP_CONCAT(DISTINCT table2.table2_id) AS table2Ids,
GROUP_CONCAT(DISTINCT table3.table3_id) AS table3Ids,
GROUP_CONCAT(DISTINCT table4.table4_id) AS table4Ids,
GROUP_CONCAT(DISTINCT table5.table5_id) AS table5Ids,
GROUP_CONCAT(DISTINCT table6.table6_id) AS table6Ids,
GROUP_CONCAT(DISTINCT table7.table7_id) AS table7Ids
FROM table1
LEFT JOIN table2 ON table1.table1_id = table2.table1_id
LEFT JOIN table3 ON table1.table1_id = table3.table1_id
LEFT JOIN table4 ON table1.table1_id = table4.table1_id
LEFT JOIN table5 ON table1.table1_id = table5.table1_id
LEFT JOIN table6 ON table1.table1_id = table6.table1_id
LEFT JOIN table7 ON table1.table1_id = table7.table1_id
WHERE table1.archived = false
GROUP BY table1.table1_id LIMIT 1000
我的解释查询:
+----+-------------+------------------+------------+-------+----------------------------------------------------------------------------------------------------+---------------------------+---------+--------------------------+------+----------+-------------+
| id | select_type | table | partitions | type | possible_keys | key | key_len | ref | rows | filtered | Extra |
+----+-------------+------------------+------------+-------+----------------------------------------------------------------------------------------------------+---------------------------+---------+--------------------------+------+----------+-------------+
| 1 | SIMPLE | table1 | NULL | index | PRIMARY,unique_name_in_table8, table8_idx,table9_idx,table10_idx | PRIMARY | 4 | NULL | 1 | 10.00 | Using where |
| 1 | SIMPLE | table2 | NULL | ref | PRIMARY | PRIMARY | 4 | db.table1.table1_id | 20 | 100.00 | Using index |
| 1 | SIMPLE | table3 | NULL | ref | table3_to_table1_id_idx | table3_to_table1_id_idx | 4 | db.table1.table1_id | 824 | 100.00 | Using index |
| 1 | SIMPLE | table4 | NULL | ref | table4_word_unique,table4_to_table1_id_idx | table4_to_table1_id_idx | 4 | db.table1.table1_id | 4 | 100.00 | Using index |
| 1 | SIMPLE | table5 | NULL | ref | table5_to_table1_id_idx | table5_to_table1_id_idx | 4 | db.table1.table1_id | 26 | 100.00 | Using index |
| 1 | SIMPLE | table6 | NULL | ref | table6_to_table1_id_idx | table6_to_table1_id_idx | 4 | db.table1.table1_id | 3 | 100.00 | Using index |
| 1 | SIMPLE | table7 | NULL | ref | table7_to_table1_id_idx | table7_to_table1_id_idx | 4 | db.table1.table1_id | 483 | 100.00 | Using index |
+----+-------------+------------------+------------+-------+----------------------------------------------------------------------------------------------------+---------------------------+---------+--------------------------+------+----------+-------------+
【问题讨论】:
是否可以消除distinct
? AFAIK,它会给后端带来负担。
@PrabhatG 我添加了一条评论。消除 distinct 只节省 4 秒,我宁愿没有重复。有没有更有效的方法来运行这个查询?
你能加入这个Stack Overflow Chat (chat.***.com/rooms/146828/…)
也许将 table#_ID 添加到索引中,这样 table1_ID、table#_ID 就是一个复合索引。
【参考方案1】:
我试图重写查询。试着让我知道。我没有测试过。
SELECT
t1.table1Id,
GROUP_CONCAT(DISTINCT table2.table2_id) AS table2Ids,
GROUP_CONCAT(DISTINCT table3.table3_id) AS table3Ids,
GROUP_CONCAT(DISTINCT table4.table4_id) AS table4Ids,
GROUP_CONCAT(DISTINCT table5.table5_id) AS table5Ids,
GROUP_CONCAT(DISTINCT table6.table6_id) AS table6Ids,
GROUP_CONCAT(DISTINCT table7.table7_id) AS table7Ids
FROM (
SELECT table1_id as table1Id
FROM table1
WHERE table1.archived = false
GROUP BY table1.table1_id LIMIT 1000
) AS t1
INNER JOIN table2 ON t1.table1Id = table2.table1_id
INNER JOIN table3 ON t1.table1Id = table3.table1_id
INNER JOIN table4 ON t1.table1Id = table4.table1_id
INNER JOIN table5 ON t1.table1Id = table5.table1_id
INNER JOIN table6 ON t1.table1Id = table6.table1_id
INNER JOIN table7 ON t1.table1Id = table7.table1_id
【讨论】:
【参考方案2】:解释说table2
在table1Id
上没有索引。
试试:
create index table2_to_table1_id_idx on table2(table1Id);
这可能会充分提高性能,但要最大限度地提高此特定查询的性能,请创建覆盖索引:
create index table2_table1_id_idx on table2(table1Id, table2Id);
create index table3_table1_id_idx on table3(table1Id, table3Id);
create index table4_table1_id_idx on table4(table1Id, table4Id);
create index table5_table1_id_idx on table5(table1Id, table5Id);
create index table6_table1_id_idx on table6(table1Id, table6Id);
create index table7_table1_id_idx on table7(table1Id, table7Id);
拥有这些索引意味着您可以完全避免访问基表。
【讨论】:
学习者的好奇心。你能解释一下你的答案吗?您如何通过explain
确定索引是否不存在。覆盖索引如何避免完全命中基表?以上是关于mysql 7 +加入一个表,如何加快速度?的主要内容,如果未能解决你的问题,请参考以下文章