mysqljoin好用吗（读得懂唐僧师徒取经）

原文：https://www.enmotech.com/web/detail/1/852/1.html （复制链接至浏览器，即可查看），下面我们就来说一说关于mysqljoin好用吗?我们一起去了解并探讨一下这个问题吧!

mysqljoin好用吗

原文：https://www.enmotech.com/web/detail/1/852/1.html （复制链接至浏览器，即可查看）

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 | ---- ----------- ----------------------------- ---------------------------------

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。

出处：https://www.cnblogs.com/ddzj01/p/11346954.html

作者：小豹子加油

,