我有两张表EMP(id,name,DEPT_id)
和DEPT(id ,name)
。我需要找到最多员工工作的部门。请帮忙。
答案 0 :(得分:1)
尝试此查询。
SELECT a.name,Max(a.NumEmp) AS maxEmpCount FROM ( SELECT d.name,COUNT(*) AS NumEmp FROM EMP e INNER JOIN DEPT d ON e.DEPT_id = d.id GROUP BY e.DEPT_id ) AS a GROUP BY a.name
答案 1 :(得分:0)
您可以尝试此查询。
Select Id, Name from Dept
Where Id = (Select Top(1) DeptId from Emp
Group By DeptId
order by Count(DeptId) desc)
答案 2 :(得分:0)
比其他两个解决方案稍微冗长一点,但它会完成工作......随时调整以方便您。
select countbydept.*
from
(
-- from EMP table, let's count number of records per dept
-- and then sort it by count (highest to lowest)
-- and take just the first value. We just care about the highest
-- count
select dept_id, count(*) as counter
from emp
group by dept_id
order by counter desc
limit 1
) as maxcount
inner join
(
-- let's repeat the exercise, but this time let's join
-- EMP and DEPT tables to get a full list of dept and
-- employe count
select
dept.id,
dept.`name`,
count(*) as numberofemployees
from dept
inner join emp on emp.dept_id = dept.id
group by dept.id, dept.`name`
) countbydept
-- combine the two queries's results by matching the employee count
on countbydept.numberofemployees = maxcount.counter
答案 3 :(得分:0)
您可以创建视图来查找它。
JOINS
答案 4 :(得分:0)
现在,给出了这两个表的EMP(id,name,DEPT_id)和DEPT(id,name)。现在,我在表格中插入一些条目:
SELECT COUNT(*) AS NO_OF_EMPLOYEES,
DEPARTMENT.DEPT_NAME
FROM EMP, DEPARTMENT
WHERE EMP.DEPT_ID=DEPARTMENT.DEPT_ID
GROUP BY EMP.DEPT_ID
ORDER BY NO_OF_EMPLOYEES;
此查询生成以下内容:
NO_OF_EMPLOYEES DEPT_NAME
3 Research
3 Finance
4 Sales
4 Product
现在,提供正确结果的查询:
SELECT COUNT(*) AS MAX_NO_OF_EMPLOYEES,
DEPARTMENT.DEPT_NAME
FROM EMP, DEPARTMENT
WHERE EMP.DEPT_ID=DEPARTMENT.DEPT_ID
GROUP BY EMP.DEPT_ID
HAVING MAX_NO_OF_EMPLOYEES=(
SELECT COUNT(*) AS NO_OF_EMPLOYEES
FROM EMP
GROUP BY DEPT_ID
ORDER BY NO_OF_EMPLOYEES DESC
LIMIT 1
);
它将生成:
MAX_NO_OF_EMPLOYEES DEPT_NAME
4 Sales
4 Product
答案 5 :(得分:0)
这个问题可以通过多种方式解决
使用子查询
SELECT name FROM dept WHERE id IN (SELECT dept_id FROM emp HAVING COUNT(dept_id) IN (SELECT MAX(COUNT(dept_id)) FROM emp) GROUP BY dept_id)
使用加入
SELECT name FROM emp e INNER JOIN dept d ON e. dept_id = d. id HAVING COUNT(e.dept_id) IN (SELECT MAX(COUNT(dept_id)) from emp) group by dept_id)
答案 6 :(得分:0)
这将给出员工人数最多的部门的部门名称。
Select DEPT_NAME from department where DEPT_ID = (select DEPT_ID from (Select DEPT_ID, count(DEPT_ID) from Employee group by DEPT_ID order by count(DEPT_ID) desc) where rownum = 1);
答案 7 :(得分:0)
您可以使用with
这样的语句来解决此问题:
with deps as
(select dep.department_name as dep_name, count(emp.employee_id) as cnt
from departments dep
inner join employees emp
on emp.department_id = dep.department_id
group by dep.department_name)
select deps.dep_name,cnt from deps
where cnt=(select max(cnt) from deps)
OR
select dep.department_name as dep_name, count(emp.employee_id) as cnt
from departments dep
inner join employees emp
on emp.department_id = dep.department_id
group by dep.department_name
having count(emp.employee_id) >= all (select count(emp.employee_id) as cnt
from departments dep
inner join employees emp
on emp.department_id =
dep.department_id
group by dep.department_name)
OR
with s1 as
(select dep.department_name as dep_name,
count(emp.employee_id) over(partition by dep.department_name) as cnt
from departments dep
inner join employees emp
on emp.department_id = dep.department_id
order by cnt desc),
s2 as
(select s1.dep_name,
s1.cnt,
row_number() over(order by cnt desc) as row_num
from s1)
select dep_name from s2 where row_num = 1
这些解决方案适用于我们没有top(1)
或limit 1
答案 8 :(得分:0)
select Top 1 d.DNAME,count(e.ename) as counts from emp e,dept d where d.DEPTNO=e.DEPTNO
group by d.DNAME
order by counts desc
或
select d.DNAME,count(e.ename) as counts from emp e,dept d where d.DEPTNO=e.DEPTNO
group by d.DNAME
having count(e.ename) = (select max(micount) from (select count(deptno) micount from emp group by DEPTNO) a)
答案 9 :(得分:0)
如果您只有emp表,则下面的查询将为您提供结果-
select a.* from (select deptno, dense_rank() over(order by count(*) desc ) as rank from dbo.emp group by deptno) a where a.rank =1
答案 10 :(得分:-1)
SELECT department_id, count(employee_id) as 'No_of_Emp'
FROM employees
GROUP BY department_id
ORDER BY No_of_Emp DESC