我有三张这样的表
employee_ID(Pk)|部门标识(FK)
department_ID(Pk)| LOCATION_ID(FK)
location_ID(Pk)|城市
我想要的是拥有最少员工的城市的名称。 我尝试过类似下面的sql:
SELECT l.city
FROM employees e, departments d, locations l
WHERE e.department_ID = d.department_ID
AND d.location_ID = l.location_ID
GROUP BY l.city
ORDER BY 2
LIMIT 1

但那不是一个好人。我希望它在子查询和MIN函数可能COUNT函数。我试过但但无法弄明白。 有任何想法吗? 非常感谢!
答案 0 :(得分:0)
你非常接近。试试这个:
select l.city, count(*) as no_of_employees
from locations l
inner join departments d
on d.location_id = l.location_id
inner join employees e
on e.department_id = d.department_id
group by l.city
order by no_of_employees asc
limit 1
示例:
create table locations (location_id int, city varchar(20));
insert into locations values (1, 'LA'), (2, 'NY');
create table departments (department_id int, location_id int);
insert into departments values (1, 1), (2, 1), (3, 2), (4, 2);
create table employees (employee_id int, department_id int);
insert into employees values (1, 1), (2, 1), (3, 1), (4, 3), (5, 4);
Result of the query:
| city | no_of_employees |
|------|-----------------|
| NY | 2 |
SQLFiddle示例:http://sqlfiddle.com/#!9/75aa1/1
根据评论中的要求使用子查询,这里有你怎么做 - 但不要这样做!仅在需要时使用子查询。
select * from (
-- get list of all city and employee count here
select l.city, count(*) as no_of_employees
from locations l
inner join departments d
on d.location_id = l.location_id
inner join employees e
on e.department_id = d.department_id
group by l.city
) subquery1
-- compare the no of employees with min. employees from the same query
where no_of_employees = (
-- find minimum number of employees here
select min(no_of_employees) from (
-- same query as subquery1
select l.city, count(*) as no_of_employees
from locations l
inner join departments d
on d.location_id = l.location_id
inner join employees e
on e.department_id = d.department_id
group by l.city
) subquery2
)
Result:
| city | no_of_employees |
|------|-----------------|
| NY | 2 |
SQLFiddle示例:http://sqlfiddle.com/#!9/75aa1/4