225 lines
5.9 KiB
Plaintext
225 lines
5.9 KiB
Plaintext
[TOC]
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# 查找最晚入职员工的所有信息
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```sql
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select *
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from employees
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order by hire_date desc
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limit 1;
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```
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# 查找入职员工时间排名倒数第三的员工所有信息
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```sql
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select *
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from employees
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order by hire_date desc
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limit 2, 1;
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```
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# 查找各个部门当前领导当前薪水详情以及其对应部门编号dept_no
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```sql
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select s.emp_no, s.salary, s.from_date, s.to_date, d.dept_no
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from salaries as s inner join dept_manager as d
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on d.emp_no = s.emp_no
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and d.to_date = '9999-01-01'
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and s.to_date = '9999-01-01';
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```
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# 查找所有已经分配部门的员工的last_name和first_name
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```sql
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select e.last_name, e.first_name, d.dept_no
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from employees as e inner join dept_emp as d
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on e.emp_no = d.emp_no;
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```
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# 查找所有员工的last_name和first_name以及对应部门编号dept_no
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也包括展示没有分配具体部门的员工
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```sql
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select e.last_name, e.first_name, d.dept_no
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from employees as e left outer join dept_emp as d
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on e.emp_no = d.emp_no;
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```
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# 查找所有员工入职时候的薪水情况
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```sql
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select e.emp_no, s.salary
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from employees as e inner join salaries as s
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on e.emp_no = s.emp_no and e.hire_date = s.from_date
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order by e.emp_no desc;
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```
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# 查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t
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```sql
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select emp_no, count(*) as t
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from salaries
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group by emp_no
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having t > 15;
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```
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# 找出所有员工当前具体的薪水salary情况
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```sql
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select distinct salary
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from salaries
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where to_date = '9999-01-01'
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order by salary desc;
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```
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# 获取所有部门当前manager的当前薪水情况
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```sql
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select d.dept_no, d.emp_no, s.salary
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from salaries as s inner join dept_manager as d
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on d.emp_no = s.emp_no
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and d.to_date = '9999-01-01'
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and s.to_date = '9999-01-01';
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```
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# 获取所有非manager的员工emp_no
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```sql
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select emp_no
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from employees
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where emp_no not in (
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select emp_no
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from dept_manager
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)
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```
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# 获取所有员工当前的manager
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```sql
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select d1.emp_no, d2.emp_no as manager_no
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from dept_emp as d1 inner join dept_manager as d2
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on d1.dept_no = d2.dept_no
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and d1.to_date = '9999-01-01'
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and d2.to_date = '9999-01-01'
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and d1.emp_no <> d2.emp_no
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```
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# 获取所有部门中当前员工薪水最高的相关信息
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```sql
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select d.dept_no, d.emp_no, MAX(s.salary) as salary
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from dept_emp as d inner join salaries as s
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on d.emp_no = s.emp_no
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and d.to_date = '9999-01-01'
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and s.to_date = '9999-01-01'
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group by d.dept_no
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```
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# 从titles表获取按照title进行分组
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```sql
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select title, COUNT(*) as t
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from titles
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group by title
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having t >= 2
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```
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# 从titles表获取按照title进行分组，注意对于重复的emp_no进行忽略。
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```sql
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select title, COUNT(distinct emp_no) as t
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from titles
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group by title
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having t >= 2
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```
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# 查找employees表所有emp_no为奇数
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```sql
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select *
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from employees
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where emp_no % 2 = 1 and last_name != 'Mary'
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order by hire_date desc
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```
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# 统计出当前各个title类型对应的员工当前薪水对应的平均工资
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```sql
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select t.title, AVG(s.salary) as avg
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from titles as t inner join salaries as s
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on t.emp_no = s.emp_no
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and t.to_date = '9999-01-01'
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and s.to_date = '9999-01-01'
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group by t.title
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```
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# 获取当前薪水第二多的员工的emp_no以及其对应的薪水salary
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```sql
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select emp_no, salary
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from salaries
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order by salary desc
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limit 1, 1
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```
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# 查找当前薪水排名第二多的员工编号emp_no
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```sql
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select e.emp_no, MAX(s.salary) as salary, e.last_name, e.first_name
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from employees as e, salaries as s
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where e.emp_no = s.emp_no
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and s.to_date = '9999-01-01'
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and s.salary not in (
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select MAX(salary)
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from salaries
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where s.to_date = '9999-01-01'
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)
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```
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# 查找所有员工的last_name和first_name以及对应的dept_name
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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
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本题思路为运用两次LEFT JOIN连接嵌套
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1、第一次LEFT JOIN连接employees表与dept_emp表，得到所有员工的last_name和first_name以及对应的dept_no，也包括暂时没有分配部门的员工
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2、第二次LEFT JOIN连接上表与departments表，即连接dept_no与dept_name，得到所有员工的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
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```sql
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SELECT em.last_name, em.first_name, dp.dept_name
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FROM (employees AS em LEFT JOIN dept_emp AS de ON em.emp_no = de.emp_no)
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LEFT JOIN departments AS dp ON de.dept_no = dp.dept_no
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```
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# 查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
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```sql
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SELECT (MAX(salary)-MIN(salary)) AS growth
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FROM salaries WHERE emp_no = '10001'
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```
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# 查找所有员工自入职以来的薪水涨幅情况
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```sql
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select a.emp_no, (b.salary - c.salary) as growth
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from
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employees as a
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inner join salaries as b
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on a.emp_no = b.emp_no and b.to_date = '9999-01-01'
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inner join salaries as c
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on a.emp_no = c.emp_no and a.hire_date = c.from_date
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order by growth asc
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```
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# 统计各个部门对应员工涨幅的次数总和
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本题关键是要将 每个部门分组，并分别统计工资记录总数，思路如下：
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1、用INNER JOIN连接dept_emp表和salaries表，并以dept_emp.no分组，统计每个部门所有员工工资的记录总数
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2、再将上表用INNER JOIN连接departments表，限制条件为两表的dept_no相等，找到dept_no与dept_name的对应关系，最后依次输出dept_no、dept_name、sum
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```sql
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SELECT de.dept_no, dp.dept_name, COUNT(s.salary) AS sum
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FROM (dept_emp AS de INNER JOIN salaries AS s ON de.emp_no = s.emp_no)
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INNER JOIN departments AS dp ON de.dept_no = dp.dept_no
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GROUP BY de.dept_no
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``` |