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

比如说：

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

编辑：
针对MySQL更正了上述内容

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

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 )

以下是工作查询

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

这看起来像是一个课程问题。
如果多个公司拥有相同的最大员工数，则使用LIMIT查询将无法工作。“ORDER BY”无法过滤掉无用的信息。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

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);

在Oracle中

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