SQL错题集

查找最晚入职员工的所有信息

select * from employees where hire_date = (select max(hire_date) from employees)

查找入职员工时间排名倒数第三的员工所有信息

select * from employees order by hire_date desc limit 2, 1;

获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary

select de.dept_no, de.emp_no, s.salary
from dept_emp de, salaries s
where de.emp_no = s.emp_no
and de.to_date = '9999-01-01'
and s.to_date = '9999-01-01'
group by de.dept_no
having max(s.salary)

从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。注意对于重复的emp_no进行忽略。

select t.title, count(distinct emp_no) t
from titles t
group by t.title
having t>=2

查找employees表所有emp_no为奇数，且last_name不为Mary的员工信息，并按照hire_date逆序排列

select *
from employees e
where e.emp_no%2 = 1
and e.last_name != 'Mary'
order by hire_date desc

查找当前薪水(to_date='9999-01-01')排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by

select s.emp_no, max(s.salary) salary, e.last_name, e.first_name
from salaries s, employees e
where s.to_date = '9999-01-01'
and s.emp_no = e.emp_no
and s.salary < (select max(salary) from salaries where s.to_date = '9999-01-01')

查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工

select e.last_name, e.first_name, d.dept_name
from employees e 
left join dept_emp de on e.emp_no = de.emp_no
left join departments d on de.dept_no = d.dept_no

查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_no以及其对应的薪水涨幅growth，并按照growth进行升序

SELECT sCurrent.emp_no, (sCurrent.salary-sStart.salary) AS growth
FROM (SELECT s.emp_no, s.salary FROM salaries s WHERE s.to_date = '9999-01-01') AS sCurrent,
(SELECT s.emp_no, s.salary FROM employees e, salaries s WHERE e.emp_no = s.emp_no AND s.from_date = e.hire_date) AS sStart
WHERE sCurrent.emp_no = sStart.emp_no
ORDER BY growth

对所有员工的当前(to_date='9999-01-01')薪水按照salary进行按照1-N的排名，相同salary并列且按照emp_no升序排列

SELECT s1.emp_no, s1.salary, COUNT(DISTINCT s2.salary) AS rank
FROM salaries AS s1, salaries AS s2
WHERE s1.to_date = '9999-01-01'  AND s2.to_date = '9999-01-01' AND s1.salary <= s2.salary
GROUP BY s1.emp_no
ORDER BY s1.salary DESC, s1.emp_no ASC