如果条件为“on”的内部联接查询

Question

如何编写查询以获得以下输出（见表 4）

table user : role - 1:manager, 2:employee
+----+------+-----------+--------+
| id | name | role      | salary |
+----+------+-----------+--------+
|  1 | a    | 1         |      0 |
|  2 | b    | 1         |      0 |
|  3 | c    | 2         |     10 |
|  4 | d    | 2         |     20 |
|  5 | e    | 2         |     30 |
|  6 | f    | 2         |     40 |
+----+------+-----------+--------+

table city
+----+--------+------+
| id | name   | type |
+----+--------+------+
|  1 | cityA  |    1 |
|  2 | cityB  |    2 |
+----+--------+------+

table user_city_mapping
+----+-----------+---------+
| id | user_id   | city_id |
+----+-----------+---------+
|  1 | 1         | 1       |
|  2 | 2         | 1       |
|  3 | 2         | 2       |
|  4 | 3         | 1       |
|  5 | 4         | 1       |
|  6 | 5         | 2       |
|  7 | 6         | 2       |
+----+-----------+---------+

output required
+------+-------+
| name | total |
+------+-------+
| a    |    30 |
| b    |   100 |
| c    |    10 |
| d    |    20 |
| e    |    70 |
| f    |    70 |
+------+-------+

• 用户“a”、“b”、“c”、“d”属于类型 1 的“cityA”。

• 用户“b”、“e”、“f”属于类型2的“cityB”。

• 用户“c”和“d”属于经理“a”

• 用户“c”、“d”、“e”、“f”属于经理“b”

所需输出的解释：

1. 用户“a”得到总数 30，即“c”和“d”之和，因为“c”和“d”都属于用户“a”经理
2. 用户“b”得到总数 100，即“c”、“d”、“e”和“f”的总和，因为所有用户都属于用户“b”经理
3. 用户“c”和“d”得到总和10和20，这是他们自己的工资，属于“类型1”的“cityA”
4. 用户“e”和“f”得到总数70和70，即员工工资的总和属于“城市B”，如果属于“类型2”

In short,
if an employee falls under any manager, the manager get the sum of the salary of the employee under him.
if the employee belongs to "type 1" city he gets his salary.
if an employee belongs to the "type 2" city he gets the sum of all employees belongs to that city.

上面提供的细节是需求和输出。我无法获取所需输出的查询。

我尝试过的东西

SELECT b.user_id, sum(salary) 
  FROM user_city_mapping a 
 INNER JOIN user_city_mapping b 
    ON a.city_id = b.city_id 
 INNER JOIN user 
    ON a.user_id = user.id AND role = 2 
 GROUP BY b.user_id

Answer 1

相当混乱，涉及多次解析数据。 最简单的方法是使用基于城市类型的相关子查询按用户计算工资，可以计算类型 2 的总工资。 在子查询 x 中按城市计算工资以及由谁管理该城市，首先计算城市总数，然后由经理汇总 然后外部查询选择 distinct 以消除重复的管理器行。

select distinct uname,
         case when salarybyuser = 0 then salarybycity
                else salarybyuser
         end salary
from
(
select u.id uid,u.name uname,ucm.*,c.id cid,
         case when c.type = 1 then salary
              when c.type = 2 and role <> 1 then
              (select sum(salary)
                from user u
                join user_city_mapping ucm on ucm.user_id = u.id
                join city c on c.id = ucm.city_id
                where c.type = 2 and role <> 1
                )
         else 0
         end as salarybyuser,
         x.salarybycity
from user u
join user_city_mapping ucm on ucm.user_id = u.id
join city c on c.id = ucm.city_id
left join
(
select ucm.user_id uid,u.name, sum(SalaryByCity) salarybycity
from   user_city_mapping ucm
join   user u on u.id = ucm.user_id 
join
(
select  c1.id cid,sum(salary) SalaryByCity
from user u1
join user_city_mapping ucm1 on ucm1.user_id = u1.id
join city c1 on c1.id = ucm1.city_id
group by c1.id
) s on s.cid = ucm.city_id
where u.role = 1
group by u.id,u.name
) x on x.uid = u.id
order by c.id,u.role,u.id
) y
order by uid
;




+-------+--------+
| uname | salary |
+-------+--------+
| a     |     30 |
| b     |    100 |
| c     |     10 |
| d     |     20 |
| e     |     70 |
| f     |     70 |
+-------+--------+
6 rows in set (0.002 sec)

看起来很乱，但这实际上是将 3 个查询压缩为 1 个。