Detailed explanation of how to use join to optimize SQL in MySQL

Detailed explanation of how to use join to optimize SQL in MySQL

0. Prepare relevant tables for the following tests

For related table creation statements, please see: https://github.com/YangBaohust/my_sql

User1 table, learning group +----+-----------+-----------------+---------------------------------+
| id | user_name | comment | mobile |
+----+-----------+-----------------+---------------------------------+
| 1 | Tang Monk | Sandalwood Merit Buddha | 138245623,021-382349 |
| 2 | Sun Wukong | Fighting Buddha | 159384292,022-483432,+86-392432 |
| 3 | Zhu Bajie | Altar Cleanser | 183208243,055-8234234 |
| 4 | Sha Seng | Golden Body Arhat | 293842295,098-2383429 |
| 5 | NULL | White Dragon Horse | 993267899 |
+----+-----------+-----------------+---------------------------------+

User2 table, Wukong's circle of friends +----+--------------+-----------+
| id | user_name | comment |
+----+--------------+-----------+
| 1 | Sun Wukong | The Monkey King |
| 2 | Bull Demon King | Brother Niu |
| 3 | Princess Iron Fan | Mrs. Niu |
| 4 | Bodhi Patriarch | Grapes |
| 5 | NULL | Jingjing|
+----+--------------+-----------+

user1_kills table, the number of monsters killed on the journey to the West +----+-----------+---------------------+-------+
| id | user_name | timestr | kills |
+----+-----------+---------------------+-------+
| 1 | Sun Wukong | 2013-01-10 00:00:00 | 10 |
| 2 | Sun Wukong | 2013-02-01 00:00:00 | 2 |
| 3 | Sun Wukong | 2013-02-05 00:00:00 | 12 |
| 4 | Sun Wukong | 2013-02-12 00:00:00 | 22 |
| 5 | Zhu Bajie | 2013-01-11 00:00:00 | 20 |
| 6 | Zhu Bajie | 2013-02-07 00:00:00 | 17 |
| 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 |
| 8 | Sha Seng | 2013-01-10 00:00:00 | 3 |
| 9 | Sha Seng | 2013-01-22 00:00:00 | 9 |
| 10 | Sha Seng | 2013-02-11 00:00:00 | 5 |
+----+-----------+---------------------+-------+

user1_equipment table, quest group equipment +----+-----------+--------------+-----------------+-----------------+
| id | user_name | arms | clothing | shoe |
+----+-----------+--------------+-----------------+-----------------+
| 1 | Tang Monk| Nine-ring staff| Colorful cassock| Monk shoes|
| 2 | Sun Wukong| Golden Cudgel| Golden Armor| Lotus Silk Cloud-Walking Shoes|
| 3 | Zhu Bajie| Nine-toothed rake| Monk's robe| Monk's shoes|
| 4 | Sha Monk| Demon-Subduing Staff| Monk's Robe| Monk's Shoes|
+----+-----------+--------------+-----------------+-----------------+

1. Use left join to optimize the not in clause

Example: Find out the people in the pilgrimage group who are not in Wukong’s circle of friends

+----+-----------+-----------------+-----------------------+
| id | user_name | comment | mobile |
+----+-----------+-----------------+-----------------------+
| 1 | Tang Monk | Sandalwood Merit Buddha | 138245623,021-382349 |
| 3 | Zhu Bajie | Altar Cleanser | 183208243,055-8234234 |
| 4 | Sha Seng | Golden Body Arhat | 293842295,098-2383429 |
+----+-----------+-----------------+-----------------------+

Not in writing:

select * from user1 a where a.user_name not in (select user_name from user2 where user_name is not null);

Left join writing method:

First, look at the outer join data set connected by user_name

select a.*, b.* from user1 a left join user2 b on (a.user_name = b.user_name);
+----+-----------+-----------------+---------------------------------+------+-----------+-----------+
| id | user_name | comment | mobile | id | user_name | comment |
+----+-----------+-----------------+---------------------------------+------+-----------+-----------+
| 2 | Sun Wukong | Fighting Buddha | 159384292,022-483432,+86-392432 | 1 | Sun Wukong | The Monkey King |
| 1 | Tang Monk | Sandalwood Merit Buddha | 138245623,021-382349 | NULL | NULL | NULL |
| 3 | Zhu Bajie | Altar Cleanser | 183208243,055-8234234 | NULL | NULL | NULL |
| 4 | Sha Seng | Golden Body Arhat | 293842295,098-2383429 | NULL | NULL | NULL |
| 5 | NULL | White Dragon Horse | 993267899 | NULL | NULL | NULL |
+----+-----------+-----------------+---------------------------------+------+-----------+-----------+

It can be seen that all the data in table a are displayed, and the data in table b are only displayed if b.user_name is equal to a.user_name, and the rest are filled with null values. To find out the people in the journey to the West who are not in Wukong's circle of friends, you only need to add a filter condition b.user_name is null in b.user_name.

select a.* from user1 a left join user2 b on (a.user_name = b.user_name) where b.user_name is null;
+----+-----------+-----------------+-----------------------+
| id | user_name | comment | mobile |
+----+-----------+-----------------+-----------------------+
| 1 | Tang Monk | Sandalwood Merit Buddha | 138245623,021-382349 |
| 3 | Zhu Bajie | Altar Cleanser | 183208243,055-8234234 |
| 4 | Sha Seng | Golden Body Arhat | 293842295,098-2383429 |
| 5 | NULL | White Dragon Horse | 993267899 |
+----+-----------+-----------------+-----------------------+

Here we find that there is one more white dragon horse in the result set, so we can continue to add the filter condition a.user_name is not null.

select a.* from user1 a left join user2 b on (a.user_name = b.user_name) where b.user_name is null and a.user_name is not null;

2. Use left join to optimize scalar subqueries

Example: Check the nicknames of people in the Journey to the West group in Wukong’s circle of friends

+-----------+-----------------+-----------+
| user_name | comment | comment2 |
+-----------+-----------------+-----------+
| Tang Monk | Sandalwood Merit Buddha | NULL |
| Sun Wukong| Fighting Buddha| The Monkey King|
| Zhu Bajie| Altar Cleanser| NULL |
| Sha Monk | Golden Body Arhat | NULL |
| NULL | White Dragon Horse | NULL |
+-----------+-----------------+-----------+

Subquery writing:

select a.user_name, a.comment, (select comment from user2 b where b.user_name = a.user_name) comment2 from user1 a;

Left join writing method:

select a.user_name, a.comment, b.comment comment2 from user1 a left join user2 b on (a.user_name = b.user_name);

3. Use join to optimize aggregate subqueries

Example: Find the date when each member of the pilgrimage group killed the most monsters

+----+-----------+---------------------+-------+
| id | user_name | timestr | kills |
+----+-----------+---------------------+-------+
| 4 | Sun Wukong | 2013-02-12 00:00:00 | 22 |
| 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 |
| 9 | Sha Seng | 2013-01-22 00:00:00 | 9 |
+----+-----------+---------------------+-------+

Aggregate subquery writing method:

select * from user1_kills a where a.kills = (select max(b.kills) from user1_kills b where b.user_name = a.user_name);

Join writing method:

First, let's look at the result set of the two tables' self-association. To save space, we'll only take Zhu Bajie's monster-killing data.

select a.*, b.* from user1_kills a join user1_kills b on (a.user_name = b.user_name) order by 1;
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+
| id | user_name | timestr | kills | id | user_name | timestr | kills |
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+
| 5 | Zhu Bajie | 2013-01-11 00:00:00 | 20 | 5 | Zhu Bajie | 2013-01-11 00:00:00 | 20 |
| 5 | Zhu Bajie | 2013-01-11 00:00:00 | 20 | 6 | Zhu Bajie | 2013-02-07 00:00:00 | 17 |
| 5 | Zhu Bajie | 2013-01-11 00:00:00 | 20 | 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 |
| 6 | Zhu Bajie | 2013-02-07 00:00:00 | 17 | 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 |
| 6 | Zhu Bajie | 2013-02-07 00:00:00 | 17 | 5 | Zhu Bajie | 2013-01-11 00:00:00 | 20 |
| 6 | Zhu Bajie | 2013-02-07 00:00:00 | 17 | 6 | Zhu Bajie | 2013-02-07 00:00:00 | 17 |
| 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 | 5 | Zhu Bajie | 2013-01-11 00:00:00 | 20 |
| 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 | 6 | Zhu Bajie | 2013-02-07 00:00:00 | 17 |
| 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 | 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 |
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+

It can be seen that when the two tables are self-associated through user_name, we only need to perform a group by on all fields in table a and take the max(kills) in table b. As long as a.kills=max(b.kills), the requirement is met. The sql is as follows

select a.* from user1_kills a join user1_kills b on (a.user_name = b.user_name) group by a.id, a.user_name, a.timestr, a.kills having a.kills = max(b.kills);

4. Use join for group selection

Example: Upgrade the third example to find out the first two dates when each person in the pilgrimage group killed the most monsters

+----+-----------+---------------------+-------+
| id | user_name | timestr | kills |
+----+-----------+---------------------+-------+
| 3 | Sun Wukong | 2013-02-05 00:00:00 | 12 |
| 4 | Sun Wukong | 2013-02-12 00:00:00 | 22 |
| 5 | Zhu Bajie | 2013-01-11 00:00:00 | 20 |
| 7 | Zhu Bajie | 2013-02-08 00:00:00 | 35 |
| 9 | Sha Seng | 2013-01-22 00:00:00 | 9 |
| 10 | Sha Seng | 2013-02-11 00:00:00 | 5 |
+----+-----------+---------------------+-------+

In Oracle, this can be achieved through analytical functions

select b.* from (select a.*, row_number() over(partition by user_name order by kills desc) cnt from user1_kills a) b where b.cnt <= 2;

Unfortunately, the above SQL will report an error in MySQL: ERROR 1064 (42000): You have an error in your SQL syntax; because MySQL does not support analytical functions. However, it can be achieved in the following way.

First, the two tables are self-associated. In order to save space, only the data of Sun Wukong is taken out.

select a.*, b.* from user1_kills a join user1_kills b on (a.user_name=b.user_name and a.kills<=b.kills) order by a.user_name, a.kills desc;
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+
| id | user_name | timestr | kills | id | user_name | timestr | kills |
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+
| 4 | Sun Wukong | 2013-02-12 00:00:00 | 22 | 4 | Sun Wukong | 2013-02-12 00:00:00 | 22 |
| 3 | Sun Wukong | 2013-02-05 00:00:00 | 12 | 3 | Sun Wukong | 2013-02-05 00:00:00 | 12 |
| 3 | Sun Wukong | 2013-02-05 00:00:00 | 12 | 4 | Sun Wukong | 2013-02-12 00:00:00 | 22 |
| 1 | Sun Wukong | 2013-01-10 00:00:00 | 10 | 1 | Sun Wukong | 2013-01-10 00:00:00 | 10 |
| 1 | Sun Wukong | 2013-01-10 00:00:00 | 10 | 3 | Sun Wukong | 2013-02-05 00:00:00 | 12 |
| 1 | Sun Wukong | 2013-01-10 00:00:00 | 10 | 4 | Sun Wukong | 2013-02-12 00:00:00 | 22 |
| 2 | Sun Wukong | 2013-02-01 00:00:00 | 2 | 1 | Sun Wukong | 2013-01-10 00:00:00 | 10 |
| 2 | Sun Wukong | 2013-02-01 00:00:00 | 2 | 3 | Sun Wukong | 2013-02-05 00:00:00 | 12 |
| 2 | Sun Wukong | 2013-02-01 00:00:00 | 2 | 4 | Sun Wukong | 2013-02-12 00:00:00 | 22 |
| 2 | Sun Wukong | 2013-02-01 00:00:00 | 2 | 2 | Sun Wukong | 2013-02-01 00:00:00 | 2 |
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+

From the table above, we know that the number of monsters killed by Sun Wukong is 22 and 12. Then we only need to perform a group by on all fields of table a and count the id of table b. If the count value is less than or equal to 2, the requirement is met. The SQL is rewritten as follows:

select a.* from user1_kills a join user1_kills b on (a.user_name=b.user_name and a.kills<=b.kills) group by a.id, a.user_name, a.timestr, a.kills having count(b.id) <= 2;

5. Use Cartesian product join to convert one column to multiple rows

Example: Convert each phone number in the query group into a line

Original data:

+-----------+---------------------------------+
| user_name | mobile |
+-----------+---------------------------------+
| Tang Monk | 138245623,021-382349 |
| Sun Wukong | 159384292,022-483432,+86-392432 |
| Zhu Bajie | 183208243,055-8234234 |
| Sha Seng | 293842295,098-2383429 |
| NULL | 993267899 |
+-----------+---------------------------------+

The data you want to get:

+-----------+-------------+
| user_name | mobile |
+-----------+-------------+
| Tang Monk | 138245623 |
| Tang Monk | 021-382349 |
| Sun Wukong | 159384292 |
| Sun Wukong | 022-483432 |
| Sun Wukong | +86-392432 |
| Zhu Bajie | 183208243 |
| Zhu Bajie | 055-8234234 |
| Sha Seng | 293842295 |
| Sha Seng | 098-2383429 |
| NULL | 993267899 |
+-----------+-------------+

You can see that Tang Seng has two phones, so he needs two lines. We can first find the number of phone numbers for each person, and then perform a Cartesian product with a sequence table. In order to save space, only Tang Seng’s data is taken out.

select a.id, b.* from tb_sequence a cross join (select user_name, mobile, length(mobile)-length(replace(mobile, ',', ''))+1 size from user1) b order by 2,1;
+----+-----------+---------------------------------+------+
| id | user_name | mobile | size |
+----+-----------+---------------------------------+------+
| 1 | Tang Monk | 138245623,021-382349 | 2 |
| 2 | Tang Monk | 138245623,021-382349 | 2 |
| 3 | Tang Monk | 138245623,021-382349 | 2 |
| 4 | Tang Monk | 138245623,021-382349 | 2 |
| 5 | Tang Monk | 138245623,021-382349 | 2 |
| 6 | Tang Monk | 138245623,021-382349 | 2 |
| 7 | Tang Monk | 138245623,021-382349 | 2 |
| 8 | Tang Monk | 138245623,021-382349 | 2 |
| 9 | Tang Monk | 138245623,021-382349 | 2 |
| 10 | Tang Monk | 138245623,021-382349 | 2 |
+----+-----------+---------------------------------+------+

a.id corresponds to the number of phone numbers, and size is the total number of phone numbers. Therefore, we can add a correlation condition (a.id <= b.size) and adjust the above SQL statement.

select b.user_name, replace(substring(substring_index(b.mobile, ',', a.id), char_length(substring_index(mobile, ',', a.id-1)) + 1), ',', '') as mobile from tb_sequence a cross join (select user_name, concat(mobile, ',') as mobile, length(mobile)-length(replace(mobile, ',', ''))+1 size from user1) b on (a.id <= b.size);

6. Use Cartesian product association to convert multiple columns to multiple rows

Example: Turn each piece of equipment in the quest group into a row

Original data:

+----+-----------+--------------+-----------------+-----------------+
| id | user_name | arms | clothing | shoe |
+----+-----------+--------------+-----------------+-----------------+
| 1 | Tang Monk| Nine-ring staff| Colorful cassock| Monk shoes|
| 2 | Sun Wukong| Golden Cudgel| Golden Armor| Lotus Silk Cloud-Walking Shoes|
| 3 | Zhu Bajie| Nine-toothed rake| Monk's robe| Monk's shoes|
| 4 | Sha Monk| Demon-Subduing Staff| Monk's Robe| Monk's Shoes|
+----+-----------+--------------+-----------------+-----------------+

The data you want to get:

+-----------+----------+-----------------+
| user_name | equipment | equip_mame |
+-----------+----------+-----------------+
| Tang Monk | arms | Nine-ring staff |
| Tang Monk | clothing | Colorful cassock |
| Tang Monk | shoe | monk shoes |
| Sun Wukong| arms | Golden Cudgel|
| Sun Wukong| clothing | Shuttle Golden Armor|
| Sun Wukong | shoe | Lotus silk cloud shoes |
| Sha Seng| arms | Demon-subduing Staff|
| Sha Seng| clothing | monk's clothes|
| Sha Seng| shoe | monk shoes|
| Zhu Bajie| arms | Nine-toothed rake|
| Zhu Bajie| clothing | monk's clothing|
| Zhu Bajie| shoe | monk shoes|
+-----------+----------+-----------------+

How to write union:

select user_name, 'arms' as equipment, arms equip_mame from user1_equipment
union all
select user_name, 'clothing' as equipment, clothing equip_mame from user1_equipment
union all
select user_name, 'shoe' as equipment, shoe equip_mame from user1_equipment
order by 1, 2;

How to write join:

First, let’s look at the effect of the Descartes dataset, taking Tang Seng as an example

select a.*, b.* from user1_equipment a cross join tb_sequence b where b.id <= 3;
+----+-----------+--------------+-----------------+-----------------+----+
| id | user_name | arms | clothing | shoe | id |
+----+-----------+--------------+-----------------+-----------------+----+
| 1 | Tang Monk| Nine-ring staff| Colorful cassock| Monk shoes| 1 |
| 1 | Tang Monk | Nine-ring staff | Colorful cassock | Monk shoes | 2 |
| 1 | Tang Monk | Nine-ring staff | Colorful cassock | Monk shoes | 3 |
+----+-----------+--------------+-----------------+-----------------+----+

Use case to process the above results

select user_name, 
case when b.id = 1 then 'arms' 
when b.id = 2 then 'clothing'
when b.id = 3 then 'shoe' ends as equipment,
case when b.id = 1 then arms end arms,
case when b.id = 2 then clothing end clothing,
case when b.id = 3 then shoe end shoe
from user1_equipment a cross join tb_sequence b where b.id <=3;
+-----------+----------+--------------+-----------------+-----------------+
| user_name | equipment | arms | clothing | shoe |
+-----------+----------+--------------+-----------------+-----------------+
| Tang Monk | arms | Nine-ring staff | NULL | NULL |
| Tang Monk | clothing | NULL | Colorful cassock | NULL |
| Tang Monk | shoe | NULL | NULL | monk shoes |
+-----------+----------+--------------+-----------------+-----------------+

Use the coalesce function to merge multiple columns of data

select user_name, 
case when b.id = 1 then 'arms' 
when b.id = 2 then 'clothing'
when b.id = 3 then 'shoe' ends as equipment,
coalesce(case when b.id = 1 then arms end,
case when b.id = 2 then clothing end,
case when b.id = 3 then shoe end) equip_mame
from user1_equipment a cross join tb_sequence b where b.id <=3 order by 1, 2;

7. Use join to update the table that contains itself in the filter condition

Example: For people who are in both the Journey to the West group and Wukong's friend circle, update the comment field in the Journey to the West group to "This person is in Wukong's friend circle"

It is natural for us to first find out the people whose user_name exists in both user1 and user2, and then update the user1 table. The sql is as follows

update user1 set comment = 'This person is in Wukong's circle of friends' where user_name in (select a.user_name from user1 a join user2 b on (a.user_name = b.user_name));

Unfortunately, the above SQL reports an error in MySQL: ERROR 1093 (HY000): You can't specify target table 'user1' for update in FROM clause, which indicates that the target table in the from clause cannot be updated.

Is there any other way? We can convert the in method into a join method.

select c.*, d.* from user1 c join (select a.user_name from user1 a join user2 b on (a.user_name = b.user_name)) d on (c.user_name = d.user_name);
+----+-----------+--------------+---------------------------------+-----------+
| id | user_name | comment | mobile | user_name |
+----+-----------+--------------+---------------------------------+-----------+
| 2 | Sun Wukong | Fighting Buddha | 159384292,022-483432,+86-392432 | Sun Wukong |
+----+-----------+--------------+---------------------------------+-----------+

Then update the view after join

update user1 c join (select a.user_name from user1 a join user2 b on (a.user_name = b.user_name)) d on (c.user_name = d.user_name) set c.comment = 'This person is in Wukong's circle of friends';

Check user1 again, you can see that user1 has been modified successfully

select * from user1;
+----+-----------+-----------------------------+---------------------------------+
| id | user_name | comment | mobile |
+----+-----------+-----------------------------+---------------------------------+
| 1 | Tang Monk | Sandalwood Merit Buddha | 138245623,021-382349 |
| 2 | Sun Wukong | This person is in Wukong's circle of friends | 159384292,022-483432,+86-392432 |
| 3 | Zhu Bajie | Altar Cleanser | 183208243,055-8234234 |
| 4 | Sha Seng | Golden Body Arhat | 293842295,098-2383429 |
| 5 | NULL | White Dragon Horse | 993267899 |
+----+-----------+-----------------------------+---------------------------------+

8. Use join to remove duplicate data

First, insert two data into the user2 table

insert into user2(user_name, comment) values ​​('孙悟空', '美猴王');
insert into user2(user_name, comment) values ​​('牛魔王', '牛哥');

Example: Delete duplicate data in the user2 table and keep only the data with large ID numbers

+----+--------------+-----------+
| id | user_name | comment |
+----+--------------+-----------+
| 1 | Sun Wukong | The Monkey King |
| 2 | Bull Demon King | Brother Niu |
| 3 | Princess Iron Fan | Mrs. Niu |
| 4 | Bodhi Patriarch | Grapes |
| 5 | NULL | Jingjing|
| 6 | Sun Wukong | The Monkey King |
| 7 | Bull Demon King | Brother Niu |
+----+--------------+-----------+

Check duplicate records first

select a.*, b.* from user2 a join (select user_name, comment, max(id) id from user2 group by user_name, comment having count(*) > 1) b on (a.user_name=b.user_name and a.comment=b.comment) order by 2;
+----+-----------+-----------+-----------+-----------+------+
| id | user_name | comment | user_name | comment | id |
+----+-----------+-----------+-----------+-----------+------+
| 1 | Sun Wukong| The Monkey King| Sun Wukong| The Monkey King| 6 |
| 6 | Sun Wukong| The Monkey King| Sun Wukong| The Monkey King| 6 |
| 2 | Bull Demon King| Brother Niu| Bull Demon King| Brother Niu| 7 |
| 7 | Bull Demon King| Brother Niu| Bull Demon King| Brother Niu| 7 |
+----+-----------+-----------+-----------+-----------+------+

Then just delete the data where (a.id < b.id)

delete a from user2 a join (select user_name, comment, max(id) id from user2 group by user_name, comment having count(*) > 1) b on (a.user_name=b.user_name and a.comment=b.comment) where a.id < b.id;

Check user2 and you can see that the duplicate data has been deleted.

select * from user2;
+----+--------------+-----------+
| id | user_name | comment |
+----+--------------+-----------+
| 3 | Princess Iron Fan | Mrs. Niu |
| 4 | Bodhi Patriarch | Grapes |
| 5 | NULL | Jingjing|
| 6 | Sun Wukong | The Monkey King |
| 7 | Bull Demon King | Brother Niu |
+----+--------------+-----------+

Summarize:

That’s all I have to say. If you are interested, you can create more data and compare the execution time of different SQL statements. The examples in this article are taken from the course "SQL Development Skills" on MOOC.com.

Well, that’s all for this article. I hope the content of this article will be of certain reference value to your study or work. Thank you for your support of 123WORDPRESS.COM.

You may also be interested in:
  • MySQL joint table query basic operation left-join common pitfalls
  • Analysis of the difference between the usage of left join setting conditions in on and where in mysql
  • Summary of various common join table query examples in MySQL
  • MySQL 8.0.18 Hash Join does not support left/right join left and right join issues
  • Mysql 8.0.18 hash join test (recommended)
  • MySQL 8.0 New Features: Hash Join
  • MySQL 8.0.18 stable version released! Hash Join is here as expected
  • Query process and optimization method of (JOIN/ORDER BY) statement in MySQL
  • In-depth understanding of MySQL self-connection and join association
  • Mysql join table and id auto-increment example analysis

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