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怎么在 Mysql 中利用 join 优化 sql？针对这个问题，这篇文章详细介绍了相对应的分析和解答，希望可以帮助更多想解决这个问题的小伙伴找到更简单易行的方法。
0. 准备相关表来进行接下来的测试
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 +-----------+-----------+--------------+-----------------+-----------------+
| 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 |
+----+--------------+-----------+关于怎么在 Mysql 中利用 join 优化 sql 问题的解答就分享到这里了，希望以上内容可以对大家有一定的帮助，如果你还有很多疑惑没有解开，可以关注丸趣 TV 行业资讯频道了解更多相关知识。
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