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Mysql巧用join优化sql的方法详解
站长资源 2024/12/29 佚名
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DDR爱好者之家 Design By 杰米

0. 准备相关表来进行接下来的测试

相关建表语句请看：https://github.com/YangBaohust/my_sql

user1表，取经组
+----+-----------+-----------------+---------------------------------+
| id | user_name | comment   | mobile       |
+----+-----------+-----------------+---------------------------------+
| 1 | 唐僧  | 旃檀功德佛  | 138245623,021-382349   |
| 2 | 孙悟空 | 斗战胜佛  | 159384292,022-483432,+86-392432 |
| 3 | 猪八戒 | 净坛使者  | 183208243,055-8234234   |
| 4 | 沙僧  | 金身罗汉  | 293842295,098-2383429   |
| 5 | NULL  | 白龙马   | 993267899      |
+----+-----------+-----------------+---------------------------------+

user2表，悟空的朋友圈
+----+--------------+-----------+
| id | user_name | comment |
+----+--------------+-----------+
| 1 | 孙悟空  | 美猴王 |
| 2 | 牛魔王  | 牛哥  |
| 3 | 铁扇公主  | 牛夫人 |
| 4 | 菩提老祖  | 葡萄  |
| 5 | NULL   | 晶晶  |
+----+--------------+-----------+

user1_kills表，取经路上杀的妖怪数量
+----+-----------+---------------------+-------+
| id | user_name | timestr    | kills |
+----+-----------+---------------------+-------+
| 1 | 孙悟空 | 2013-01-10 00:00:00 | 10 |
| 2 | 孙悟空 | 2013-02-01 00:00:00 |  2 |
| 3 | 孙悟空 | 2013-02-05 00:00:00 | 12 |
| 4 | 孙悟空 | 2013-02-12 00:00:00 | 22 |
| 5 | 猪八戒 | 2013-01-11 00:00:00 | 20 |
| 6 | 猪八戒 | 2013-02-07 00:00:00 | 17 |
| 7 | 猪八戒 | 2013-02-08 00:00:00 | 35 |
| 8 | 沙僧  | 2013-01-10 00:00:00 |  3 |
| 9 | 沙僧  | 2013-01-22 00:00:00 |  9 |
| 10 | 沙僧  | 2013-02-11 00:00:00 |  5 |
+----+-----------+---------------------+-------+

user1_equipment表，取经组装备
+----+-----------+--------------+-----------------+-----------------+
| id | user_name | arms   | clothing  | shoe   |
+----+-----------+--------------+-----------------+-----------------+
| 1 | 唐僧  | 九环锡杖  | 锦斓袈裟  | 僧鞋   |
| 2 | 孙悟空 | 金箍棒  | 梭子黄金甲  | 藕丝步云履  |
| 3 | 猪八戒 | 九齿钉耙  | 僧衣   | 僧鞋   |
| 4 | 沙僧  | 降妖宝杖  | 僧衣   | 僧鞋   |
+----+-----------+--------------+-----------------+-----------------+

1. 使用left join优化not in子句

例子：找出取经组中不属于悟空朋友圈的人

+----+-----------+-----------------+-----------------------+
| id | user_name | comment   | mobile    |
+----+-----------+-----------------+-----------------------+
| 1 | 唐僧  | 旃檀功德佛  | 138245623,021-382349 |
| 3 | 猪八戒 | 净坛使者  | 183208243,055-8234234 |
| 4 | 沙僧  | 金身罗汉  | 293842295,098-2383429 |
+----+-----------+-----------------+-----------------------+

not in写法：

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

left join写法：

首先看通过user_name进行连接的外连接数据集

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 | 孙悟空 | 斗战胜佛  | 159384292,022-483432,+86-392432 | 1 | 孙悟空 | 美猴王 |
| 1 | 唐僧  | 旃檀功德佛  | 138245623,021-382349   | NULL | NULL  | NULL  |
| 3 | 猪八戒 | 净坛使者  | 183208243,055-8234234   | NULL | NULL  | NULL  |
| 4 | 沙僧  | 金身罗汉  | 293842295,098-2383429   | NULL | NULL  | NULL  |
| 5 | NULL  | 白龙马   | 993267899      | NULL | NULL  | NULL  |
+----+-----------+-----------------+---------------------------------+------+-----------+-----------+

可以看到a表中的所有数据都有显示，b表中的数据只有b.user_name与a.user_name相等才显示，其余都以null值填充，要想找出取经组中不属于悟空朋友圈的人，只需要在b.user_name中加一个过滤条件b.user_name is null即可。

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 | 唐僧  | 旃檀功德佛  | 138245623,021-382349 |
| 3 | 猪八戒 | 净坛使者  | 183208243,055-8234234 |
| 4 | 沙僧  | 金身罗汉  | 293842295,098-2383429 |
| 5 | NULL  | 白龙马   | 993267899    |
+----+-----------+-----------------+-----------------------+

看到这里发现结果集中还多了一个白龙马，继续添加过滤条件a.user_name is not null即可。

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. 使用left join优化标量子查询

例子：查看取经组中的人在悟空朋友圈的昵称

+-----------+-----------------+-----------+
| user_name | comment   | comment2 |
+-----------+-----------------+-----------+
| 唐僧  | 旃檀功德佛  | NULL  |
| 孙悟空 | 斗战胜佛  | 美猴王 |
| 猪八戒 | 净坛使者  | NULL  |
| 沙僧  | 金身罗汉  | NULL  |
| NULL  | 白龙马   | NULL  |
+-----------+-----------------+-----------+

子查询写法：

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

left join写法：

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

3. 使用join优化聚合子查询

例子：查询出取经组中每人打怪最多的日期

+----+-----------+---------------------+-------+
| id | user_name | timestr    | kills |
+----+-----------+---------------------+-------+
| 4 | 孙悟空 | 2013-02-12 00:00:00 | 22 |
| 7 | 猪八戒 | 2013-02-08 00:00:00 | 35 |
| 9 | 沙僧  | 2013-01-22 00:00:00 |  9 |
+----+-----------+---------------------+-------+

聚合子查询写法：

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

join写法：

首先看两表自关联的结果集，为节省篇幅，只取猪八戒的打怪数据来看

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 | 猪八戒 | 2013-01-11 00:00:00 | 20 | 5 | 猪八戒 | 2013-01-11 00:00:00 | 20 |
| 5 | 猪八戒 | 2013-01-11 00:00:00 | 20 | 6 | 猪八戒 | 2013-02-07 00:00:00 | 17 |
| 5 | 猪八戒 | 2013-01-11 00:00:00 | 20 | 7 | 猪八戒 | 2013-02-08 00:00:00 | 35 |
| 6 | 猪八戒 | 2013-02-07 00:00:00 | 17 | 7 | 猪八戒 | 2013-02-08 00:00:00 | 35 |
| 6 | 猪八戒 | 2013-02-07 00:00:00 | 17 | 5 | 猪八戒 | 2013-01-11 00:00:00 | 20 |
| 6 | 猪八戒 | 2013-02-07 00:00:00 | 17 | 6 | 猪八戒 | 2013-02-07 00:00:00 | 17 |
| 7 | 猪八戒 | 2013-02-08 00:00:00 | 35 | 5 | 猪八戒 | 2013-01-11 00:00:00 | 20 |
| 7 | 猪八戒 | 2013-02-08 00:00:00 | 35 | 6 | 猪八戒 | 2013-02-07 00:00:00 | 17 |
| 7 | 猪八戒 | 2013-02-08 00:00:00 | 35 | 7 | 猪八戒 | 2013-02-08 00:00:00 | 35 |
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+

可以看到当两表通过user_name进行自关联，只需要对a表的所有字段进行一个group by，取b表中的max(kills)，只要a.kills=max(b.kills)就满足要求了。sql如下

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. 使用join进行分组选择

例子：对第3个例子进行升级，查询出取经组中每人打怪最多的前两个日期

+----+-----------+---------------------+-------+
| id | user_name | timestr       | kills |
+----+-----------+---------------------+-------+
| 3 | 孙悟空  | 2013-02-05 00:00:00 |  12 |
| 4 | 孙悟空  | 2013-02-12 00:00:00 |  22 |
| 5 | 猪八戒  | 2013-01-11 00:00:00 |  20 |
| 7 | 猪八戒  | 2013-02-08 00:00:00 |  35 |
| 9 | 沙僧   | 2013-01-22 00:00:00 |   9 |
| 10 | 沙僧   | 2013-02-11 00:00:00 |   5 |
+----+-----------+---------------------+-------+

在oracle中，可以通过分析函数来实现

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;

很遗憾，上面sql在mysql中报错ERROR 1064 (42000): You have an error in your SQL syntax; 因为mysql并不支持分析函数。不过可以通过下面的方式去实现。

首先对两表进行自关联，为了节约篇幅，只取出孙悟空的数据

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 | 孙悟空  | 2013-02-12 00:00:00 |  22 | 4 | 孙悟空  | 2013-02-12 00:00:00 |  22 |
| 3 | 孙悟空  | 2013-02-05 00:00:00 |  12 | 3 | 孙悟空  | 2013-02-05 00:00:00 |  12 |
| 3 | 孙悟空  | 2013-02-05 00:00:00 |  12 | 4 | 孙悟空  | 2013-02-12 00:00:00 |  22 |
| 1 | 孙悟空  | 2013-01-10 00:00:00 |  10 | 1 | 孙悟空  | 2013-01-10 00:00:00 |  10 |
| 1 | 孙悟空  | 2013-01-10 00:00:00 |  10 | 3 | 孙悟空  | 2013-02-05 00:00:00 |  12 |
| 1 | 孙悟空  | 2013-01-10 00:00:00 |  10 | 4 | 孙悟空  | 2013-02-12 00:00:00 |  22 |
| 2 | 孙悟空  | 2013-02-01 00:00:00 |   2 | 1 | 孙悟空  | 2013-01-10 00:00:00 |  10 |
| 2 | 孙悟空  | 2013-02-01 00:00:00 |   2 | 3 | 孙悟空  | 2013-02-05 00:00:00 |  12 |
| 2 | 孙悟空  | 2013-02-01 00:00:00 |   2 | 4 | 孙悟空  | 2013-02-12 00:00:00 |  22 |
| 2 | 孙悟空  | 2013-02-01 00:00:00 |   2 | 2 | 孙悟空  | 2013-02-01 00:00:00 |   2 |
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+

从上面的表中我们知道孙悟空打怪前两名的数量是22和12，那么只需要对a表的所有字段进行一个group by，对b表的id做个count，count值小于等于2就满足要求，sql改写如下：

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. 使用笛卡尔积关联实现一列转多行

例子：将取经组中每个电话号码变成一行

原始数据：

+-----------+---------------------------------+
| user_name | mobile             |
+-----------+---------------------------------+
| 唐僧   | 138245623,021-382349      |
| 孙悟空  | 159384292,022-483432,+86-392432 |
| 猪八戒  | 183208243,055-8234234      |
| 沙僧   | 293842295,098-2383429      |
| NULL   | 993267899            |
+-----------+---------------------------------+

想要得到的数据：

+-----------+-------------+
| user_name | mobile   |
+-----------+-------------+
| 唐僧   | 138245623  |
| 唐僧   | 021-382349 |
| 孙悟空  | 159384292  |
| 孙悟空  | 022-483432 |
| 孙悟空  | +86-392432 |
| 猪八戒  | 183208243  |
| 猪八戒  | 055-8234234 |
| 沙僧   | 293842295  |
| 沙僧   | 098-2383429 |
| NULL   | 993267899  |
+-----------+-------------+

可以看到唐僧有两个电话，因此他就需要两行。我们可以先求出每人的电话号码数量，然后与一张序列表进行笛卡儿积关联，为了节约篇幅，只取出唐僧的数据

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 | 唐僧   | 138245623,021-382349      |  2 |
| 2 | 唐僧   | 138245623,021-382349      |  2 |
| 3 | 唐僧   | 138245623,021-382349      |  2 |
| 4 | 唐僧   | 138245623,021-382349      |  2 |
| 5 | 唐僧   | 138245623,021-382349      |  2 |
| 6 | 唐僧   | 138245623,021-382349      |  2 |
| 7 | 唐僧   | 138245623,021-382349      |  2 |
| 8 | 唐僧   | 138245623,021-382349      |  2 |
| 9 | 唐僧   | 138245623,021-382349      |  2 |
| 10 | 唐僧   | 138245623,021-382349      |  2 |
+----+-----------+---------------------------------+------+

a.id对应的就是第几个电话号码，size就是总的电话号码数量，因此可以加上关联条件(a.id <= b.size)，将上面的sql继续调整

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. 使用笛卡尔积关联实现多列转多行

例子：将取经组中每件装备变成一行

原始数据：

+----+-----------+--------------+-----------------+-----------------+
| id | user_name | arms     | clothing    | shoe      |
+----+-----------+--------------+-----------------+-----------------+
| 1 | 唐僧   | 九环锡杖   | 锦斓袈裟    | 僧鞋      |
| 2 | 孙悟空  | 金箍棒    | 梭子黄金甲   | 藕丝步云履   |
| 3 | 猪八戒  | 九齿钉耙   | 僧衣      | 僧鞋      |
| 4 | 沙僧   | 降妖宝杖   | 僧衣      | 僧鞋      |
+----+-----------+--------------+-----------------+-----------------+

想要得到的数据：

+-----------+-----------+-----------------+
| user_name | equipment | equip_mame   |
+-----------+-----------+-----------------+
| 唐僧   | arms   | 九环锡杖    |
| 唐僧   | clothing | 锦斓袈裟    |
| 唐僧   | shoe   | 僧鞋      |
| 孙悟空  | arms   | 金箍棒     |
| 孙悟空  | clothing | 梭子黄金甲   |
| 孙悟空  | shoe   | 藕丝步云履   |
| 沙僧   | arms   | 降妖宝杖    |
| 沙僧   | clothing | 僧衣      |
| 沙僧   | shoe   | 僧鞋      |
| 猪八戒  | arms   | 九齿钉耙    |
| 猪八戒  | clothing | 僧衣      |
| 猪八戒  | shoe   | 僧鞋      |
+-----------+-----------+-----------------+

union的写法：

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;

join的写法：

首先看笛卡尔数据集的效果，以唐僧为例

select a.*, b.* from user1_equipment a cross join tb_sequence b where b.id <= 3;
+----+-----------+--------------+-----------------+-----------------+----+
| id | user_name | arms     | clothing    | shoe      | id |
+----+-----------+--------------+-----------------+-----------------+----+
| 1 | 唐僧   | 九环锡杖   | 锦斓袈裟    | 僧鞋      | 1 |
| 1 | 唐僧   | 九环锡杖   | 锦斓袈裟    | 僧鞋      | 2 |
| 1 | 唐僧   | 九环锡杖   | 锦斓袈裟    | 僧鞋      | 3 |
+----+-----------+--------------+-----------------+-----------------+----+

使用case对上面的结果进行处理

select user_name, 
case when b.id = 1 then 'arms' 
when b.id = 2 then 'clothing'
when b.id = 3 then 'shoe' end 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      |
+-----------+-----------+--------------+-----------------+-----------------+
| 唐僧   | arms   | 九环锡杖   | NULL      | NULL      |
| 唐僧   | clothing | NULL     | 锦斓袈裟    | NULL      |
| 唐僧   | shoe   | NULL     | NULL      | 僧鞋      |
+-----------+-----------+--------------+-----------------+-----------------+

使用coalesce函数将多列数据进行合并

select user_name, 
case when b.id = 1 then 'arms' 
when b.id = 2 then 'clothing'
when b.id = 3 then 'shoe' end 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. 使用join更新过滤条件中包含自身的表

例子：把同时存在于取经组和悟空朋友圈中的人，在取经组中把comment字段更新为"此人在悟空的朋友圈"

我们很自然地想到先查出user1和user2中user_name都存在的人，然后更新user1表，sql如下

update user1 set comment = '此人在悟空的朋友圈' where user_name in (select a.user_name from user1 a join user2 b on (a.user_name = b.user_name));

很遗憾，上面sql在mysql中报错：ERROR 1093 (HY000): You can't specify target table 'user1' for update in FROM clause，提示不能更新目标表在from子句的表。

那有没有其它办法呢？我们可以将in的写法转换成join的方式

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 | 孙悟空 | 斗战胜佛 | 159384292,022-483432,+86-392432 | 孙悟空 |
+----+-----------+--------------+---------------------------------+-----------+

然后对join之后的视图进行更新即可

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 = '此人在悟空的朋友圈';

再查看user1，可以看到user1已修改成功

select * from user1;
+----+-----------+-----------------------------+---------------------------------+
| id | user_name | comment           | mobile             |
+----+-----------+-----------------------------+---------------------------------+
| 1 | 唐僧   | 旃檀功德佛         | 138245623,021-382349      |
| 2 | 孙悟空  | 此人在悟空的朋友圈     | 159384292,022-483432,+86-392432 |
| 3 | 猪八戒  | 净坛使者          | 183208243,055-8234234      |
| 4 | 沙僧   | 金身罗汉          | 293842295,098-2383429      |
| 5 | NULL   | 白龙马           | 993267899            |
+----+-----------+-----------------------------+---------------------------------+

8. 使用join删除重复数据

首先向user2表中插入两条数据

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');

例子：将user2表中的重复数据删除，只保留id号大的

+----+--------------+-----------+
| id | user_name  | comment  |
+----+--------------+-----------+
| 1 | 孙悟空    | 美猴王  |
| 2 | 牛魔王    | 牛哥   |
| 3 | 铁扇公主   | 牛夫人  |
| 4 | 菩提老祖   | 葡萄   |
| 5 | NULL     | 晶晶   |
| 6 | 孙悟空    | 美猴王  |
| 7 | 牛魔王    | 牛哥   |
+----+--------------+-----------+

首先查看重复记录

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 | 孙悟空  | 美猴王  | 孙悟空  | 美猴王  |  6 |
| 6 | 孙悟空  | 美猴王  | 孙悟空  | 美猴王  |  6 |
| 2 | 牛魔王  | 牛哥   | 牛魔王  | 牛哥   |  7 |
| 7 | 牛魔王  | 牛哥   | 牛魔王  | 牛哥   |  7 |
+----+-----------+-----------+-----------+-----------+------+

接着只需要删除(a.id < b.id)的数据即可

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;

查看user2，可以看到重复数据已经被删掉了

select * from user2;
+----+--------------+-----------+
| id | user_name  | comment  |
+----+--------------+-----------+
| 3 | 铁扇公主   | 牛夫人  |
| 4 | 菩提老祖   | 葡萄   |
| 5 | NULL     | 晶晶   |
| 6 | 孙悟空    | 美猴王  |
| 7 | 牛魔王    | 牛哥   |
+----+--------------+-----------+

总结：

给大家就介绍到这里，大家有兴趣可以多造点数据，然后比较不同的sql写法在执行时间上的区别。本文例子取自于慕课网《sql开发技巧》。

好了，以上就是这篇文章的全部内容了，希望本文的内容对大家的学习或者工作具有一定的参考学习价值，谢谢大家对的支持。

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