SQL查询max(), count()

Question

数据库模式看起来像

员工（员工名称，街道，城市）
作品（员工姓名、公司名称、工资）
公司（公司名称，城市）
管理（员工姓名，经理姓名）

需要执行的查询是：
找到拥有最多员工的公司。

我可以通过查询找到最大计数：

SELECT max( cnt ) max_cnt
FROM (

SELECT count( employee_name ) cnt, company_name
FROM works
GROUP BY company_name
)w1;

但是现在我找不到公司的名称。如果有人有任何想法请分享。

Answer 1

要获取包含最大值的整行，可以使用 ORDER BY ... DESC LIMIT 1 而不是 MAX：

SELECT company_name, cnt
FROM (
    SELECT company_name, count(employee_name) AS cnt
    FROM works
    GROUP BY company_name
) w1
ORDER BY cnt DESC
LIMIT 1

Answer 2

怎么样：

SELECT count( employee_name ) cnt, company_name
FROM works
GROUP BY company_name
ORDER BY cnt DESC
LIMIT 1;

编辑：

上面针对 MySQL 进行了更正

Answer 3

SELECT company_name,count(*) as cnt 
FROM works 
GROUP BY company_name 
ORDER BY cnt DESC

Answer 4

select company_name 
from works
group by company_name
having count(distinct employee_name)>=all(select count(distinct employee_name)
from works
group by company_name )

Answer 5

这是工作查询

Select * from(SELECT count(EmpName)cnt, CName FROM works GROUP BY CName Order By cnt desc) where ROWNUM = 1;

Answer 6

这看起来像是一个课程问题。

如果不止一家公司拥有相同的最大员工人数，则使用 LIMIT 进行的查询将不起作用。 “ORDER BY”没有过滤掉无用的信息。因此我们有以下解决方案

SELECT company_name FROM
(SELECT company_name, count(employee_name) cnt
    FROM works
    GROUP BY company_name) 
JOIN 
(SELECT max(cnt) max_cnt
FROM (
    SELECT count(employee_name) cnt
    FROM works
    GROUP BY company_name
)) ON cnt = max_cnt

Answer 7

select company_name from works_for
group by company_name
having count(employee_name) = (select max(count(employee_name))from works_for
group by company_name);

Answer 8

在甲骨文中

    select company_name, count(*) as count 
    from works 
    group by company_name
    having count(*) >= all(select count(*) from works group by company_name)