怎么在Mysql中利用join优化sql

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_equipmentunion allselect user_name, 'clothing' as equipment, clothing equip_mame from user1_equipmentunion allselect user_name, 'shoe' as equipment, shoe equip_mame from user1_equipmentorder 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 shoefrom 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_mamefrom 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   |+----+--------------+-----------+

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