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MySQL-巧用Join来优化SQL

原创 Oracle 作者:云和恩墨 时间:2019-08-21 18:33:40 0 删除 编辑


原文: https://www.enmotech.com/web/detail/1/852/1.html  


导读:本文是读者『小豹子加油』的投稿,通过举出唐僧师徒取经的例子,详述一则使用JOIN来优化SQL的案例。

准备相关表


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            |
+----+-----------+--------------+-----------------+-----------------+

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;

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);

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;

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);

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;

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                       |
+----+-----------+-----------------------------+---------------------------------+

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。


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来自 “ ITPUB博客 ” ,链接:http://blog.itpub.net/29715045/viewspace-2654505/,如需转载,请注明出处,否则将追究法律责任。

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