mysql join 相关例子

Posted 2022-12-11 信行合一

# 表a数据:
mysql> select * from a;
+----+-----------+------+
| id | name      | age  |
+----+-----------+------+
| 1  | Pirate    | 10   |
| 2  | Monkey    | 20   |
| 3  | Ninja     | 20   |
| 4  | Spaghetti | 20   |
+----+-----------+------+
# 表b数据:
mysql> select * from b;
+----+------------+------+
| id | name       | age  |
+----+------------+------+
| 1  | Rutabaga   | 10   |
| 2  | Pirate     | 20   |
| 3  | Darth Vade | 20   |1
| 4  | Ninja      | 20   |
+----+------------+------+

# inner join: 求交集
select * from a inner join b on a.name = b.name;

+----+--------+------+-------+---------+--------+
| id | name   | age  | id(2) | name(2) | age(2) |
+----+--------+------+-------+---------+--------+
| 1  | Pirate | 10   | 2     | Pirate  | 20     |
| 3  | Ninja  | 20   | 4     | Ninja   | 20     |
+----+--------+------+-------+---------+--------+

# left join,right join 左链左全（左边表记录是全部的）,右链右全（右边表记录是全的）
select * from a left join b on a.name = b.name;
+----+-----------+------+-------+---------+--------+
| id | name      | age  | id(2) | name(2) | age(2) |
+----+-----------+------+-------+---------+--------+
| 1  | Pirate    | 10   | 2     | Pirate  | 20     |
| 3  | Ninja     | 20   | 4     | Ninja   | 20     |
| 2  | Monkey    | 20   | NULL  | NULL    | NULL   |
| 4  | Spaghetti | 20   | NULL  | NULL    | NULL   |
+----+-----------+------+-------+---------+--------+

select * from a right join b on a.name = b.name;
+------+--------+------+-------+------------+--------+
| id   | name   | age  | id(2) | name(2)    | age(2) |
+------+--------+------+-------+------------+--------+
| 1    | Pirate | 10   | 2     | Pirate     | 20     |
| 3    | Ninja  | 20   | 4     | Ninja      | 20     |
| NULL | NULL   | NULL | 1     | Rutabaga   | 10     |
| NULL | NULL   | NULL | 3     | Darth Vade | 20     |
+------+--------+------+-------+------------+--------+

# full outer join: 求并集
# mysql 不支持 full outer join,可以通过两个 join union来实现,没有匹配的记录返回null
SELECT * FROM a left JOIN b ON a.name = b.name
union
SELECT * FROM a right join b ON a.name = b.name;

+------+-----------+------+-------+------------+--------+
| id   | name      | age  | id(2) | name(2)    | age(2) |
+------+-----------+------+-------+------------+--------+
| 1    | Pirate    | 10   | 2     | Pirate     | 20     |
| 3    | Ninja     | 20   | 4     | Ninja      | 20     |
| 2    | Monkey    | 20   | NULL  | NULL       | NULL   |
| 4    | Spaghetti | 20   | NULL  | NULL       | NULL   |
| NULL | NULL      | NULL | 1     | Rutabaga   | 10     |
| NULL | NULL      | NULL | 3     | Darth Vade | 20     |
+------+-----------+------+-------+------------+--------+


# 求没有在对方表中出现的记录
# 求a中有b中没有的记录(b中记录为null): 通过left join来实现(因为a中有,b中没有所以用 a left join b)
select * from a left join b on a.name = b.name
where b.name is null;
+----+-----------+------+-------+---------+--------+
| id | name      | age  | id(2) | name(2) | age(2) |
+----+-----------+------+-------+---------+--------+
| 2  | Monkey    | 20   | NULL  | NULL    | NULL   |
| 4  | Spaghetti | 20   | NULL  | NULL    | NULL   |
+----+-----------+------+-------+---------+--------+

# 求b中有a中没有的记录(a 中记录为null): 通过right join来实现(a中没有,b中有,所以用 a right join b)
select * from a right join b on a.name = b.name
where a.name is null;
+------+------+------+-------+------------+--------+
| id   | name | age  | id(2) | name(2)    | age(2) |
+------+------+------+-------+------------+--------+
| NULL | NULL | NULL | 1     | Rutabaga   | 10     |
| NULL | NULL | NULL | 3     | Darth Vade | 20     |
+------+------+------+-------+------------+--------+

# full outer join: 求差集:
# 通过去除并集来实现,效率比较低
select * from a where a.name not in(
    select a.name from a inner join b on a.name = b.name
    )
union
select * from b where b.name not in(
    select a.name from a inner join b on a.name = b.name
);
+----+------------+------+
| id | name       | age  |
+----+------------+------+
| 2  | Monkey     | 20   |
| 4  | Spaghetti  | 20   |
| 1  | Rutabaga   | 10   |
| 3  | Darth Vade | 20   |
+----+------------+------+
# 通过两个差集合并来实现
select a.* from a left join b on a.name = b.name where b.name is null
union all
select b.* from a right join b b on a.name = b.name where a.name is null;

+----+------------+------+
| id | name       | age  |
+----+------------+------+
| 2  | Monkey     | 20   |
| 4  | Spaghetti  | 20   |
| 1  | Rutabaga   | 10   |
| 3  | Darth Vade | 20   |
+----+------------+------+

# union 与 union all:
# union 查询并集并(去掉重复数据): a 表中所有数据,b表中不在a表中出现的数据
# union all: 查询所有数据,不会去重

# 注意 select * from a union select * from b; 跟 select * from a union all select * from b;结果一样
# 因为去重是针对所查询出来的==所有列==进行的,对于只查询一行的数据来说,它是针对一行去重的,而对于查询多行数据来说,需要对多行进行去重(多行都完全一样才去重)
# 而因为多行来说,a表和b表并没有重复的,所以 union 和 union all的效果是一样的,但是性能上来说,union all会更好,因为它省去了去重的步骤
select a.name from a union select b.name from b;
+------------+
| name       |
+------------+
| Pirate     |
| Monkey     |
| Ninja      |
| Spaghetti  |
| Rutabaga   |
| Darth Vade |
+------------+

select a.name from a union all select b.name from b;
+------------+
| name       |
+------------+
| Pirate     |
| Monkey     |
| Ninja      |
| Spaghetti  |
| Rutabaga   |
| Pirate     |
| Darth Vade |
| Ninja      |
+------------+


# cross join: 笛卡尔积,没有链接条件就会出现笛卡尔积,这种方式很危险,尤其是表比较大的时候
select * from a cross join b on a.name = b.name;
+----+-----------+------+-------+------------+--------+
| id | name      | age  | id(2) | name(2)    | age(2) |
+----+-----------+------+-------+------------+--------+
| 1  | Pirate    | 10   | 1     | Rutabaga   | 10     |
| 2  | Monkey    | 20   | 1     | Rutabaga   | 10     |
| 3  | Ninja     | 20   | 1     | Rutabaga   | 10     |
| 4  | Spaghetti | 20   | 1     | Rutabaga   | 10     |
| 1  | Pirate    | 10   | 2     | Pirate     | 20     |
| 2  | Monkey    | 20   | 2     | Pirate     | 20     |
| 3  | Ninja     | 20   | 2     | Pirate     | 20     |
| 4  | Spaghetti | 20   | 2     | Pirate     | 20     |
| 1  | Pirate    | 10   | 3     | Darth Vade | 20     |
| 2  | Monkey    | 20   | 3     | Darth Vade | 20     |
| 3  | Ninja     | 20   | 3     | Darth Vade | 20     |
| 4  | Spaghetti | 20   | 3     | Darth Vade | 20     |
| 1  | Pirate    | 10   | 4     | Ninja      | 20     |
| 2  | Monkey    | 20   | 4     | Ninja      | 20     |
| 3  | Ninja     | 20   | 4     | Ninja      | 20     |
| 4  | Spaghetti | 20   | 4     | Ninja      | 20     |
+----+-----------+------+-------+------------+--------+


# union 和 left join 区别:
# union 增加行,left join 增加列
select a.name, b.name from a left join b on a.name = b.name;
+-----------+---------+
| name      | name(2) |
+-----------+---------+
| Pirate    | Pirate  |
| Ninja     | Ninja   |
| Monkey    | NULL    |
| Spaghetti | NULL    |
+-----------+---------+

select a.name from a
union
select b.name from b;

+------------+
| name       |
+------------+
| Pirate     |
| Monkey     |
| Ninja      |
| Spaghetti  |
| Rutabaga   |
| Darth Vade |
+------------+