MySQL学习笔记（三）刷题篇

持续更新中

• 题目一

获取所有部门中当前(dept_emp.to_date = ‘9999-01-01’)员工当前(salaries.to_date=‘9999-01-01’)薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary
CREATE TABLE dept_emp (
emp_no int(11) NOT NULL,
dept_no char(4) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,dept_no));
CREATE TABLE salaries (
emp_no int(11) NOT NULL,
salary int(11) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,from_date));
如插入：
INSERT INTO dept_emp VALUES(10001,‘d001’,‘1986-06-26’,‘9999-01-01’);
INSERT INTO dept_emp VALUES(10002,‘d001’,‘1996-08-03’,‘9999-01-01’);
INSERT INTO dept_emp VALUES(10003,‘d001’,‘1996-08-03’,‘1997-08-03’);

INSERT INTO salaries VALUES(10001,90000,‘1986-06-26’,‘1987-06-26’);
INSERT INTO salaries VALUES(10001,88958,‘2002-06-22’,‘9999-01-01’);
INSERT INTO salaries VALUES(10002,72527,‘1996-08-03’,‘1997-08-03’);
INSERT INTO salaries VALUES(10002,72527,‘2000-08-02’,‘2001-08-02’);
INSERT INTO salaries VALUES(10002,72527,‘2001-08-02’,‘9999-01-01’);
INSERT INTO salaries VALUES(10003,90000,‘1996-08-03’,‘1997-08-03’);

SELECT
    d.dept_no,
    d.emp_no,
    s.salary
FROM
    (
        SELECT
            d.dept_no,
            MAX(s.salary) as 'maxsalary'
        FROM
            dept_emp d JOIN salaries s
            ON
            d.to_date='9999-01-01' AND s.to_date='9999-01-01' AND d.emp_no=s.emp_no
        GROUP BY
            d.dept_no
    ) AS ss,
    dept_emp d,salaries s
WHERE
    ss.dept_no=d.dept_no 
    AND 
    ss.maxsalary=s.salary
    AND 
    d.emp_no=s.emp_no
    AND
    d.to_date='9999-01-01'
    AND
    s.to_date='9999-01-01'
ORDER BY
    d.dept_no

题目核心：要考虑到薪水最高可能不止一个人，先选出部门-最高薪水表作为筛选条件再进行查询。

• 题目二
  获取当前（to_date=‘9999-01-01’）薪水第二多的员工的emp_no以及其对应的薪水salary
  CREATE TABLE salaries (
  emp_no int(11) NOT NULL,
  salary int(11) NOT NULL,
  from_date date NOT NULL,
  to_date date NOT NULL,
  PRIMARY KEY (emp_no,from_date));

我的第一次答案

SELECT
    emp_no,
    max(salary)
FROM
    salaries
WHERE
    to_date='9999-01-01'
    AND
    salary not in (
    SELECT
        max(salary)
    FROM
        salaries
    WHERE
        to_date='9999-01-01')

未考虑到工资第二高的可能是多个人，这样只选出了第二高工资的编号最前面的员工，之后修改答案如下

SELECT
    emp_no,
    salary
FROM
    salaries
WHERE
    to_date='9999-01-01'
    AND
    salary = (
    SELECT
        DISTINCT salary
    FROM
        salaries
    WHERE
        to_date='9999-01-01'
    ORDER BY
        salary DESC
    LIMIT 1,1)

正确答案,挑选出工资第二高的值再进行筛选

• 题目三
  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
  CREATE TABLE salaries (
  emp_no int(11) NOT NULL,
  salary int(11) NOT NULL,
  from_date date NOT NULL,
  to_date date NOT NULL,
  PRIMARY KEY (emp_no,from_date));

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

考点 自身表的复用