【SQL217】题解 | 获取当前薪水第二多的员工的emp_no以及其对应的薪水salary
获取当前薪水第二多的员工的emp_no以及其对应的薪水salary
https://www.nowcoder.com/practice/c1472daba75d4635b7f8540b837cc719
select e.emp_no, s.salary, e.last_name, e.first_name from employees e left join salaries s on e.emp_no = s.emp_no where s.salary = (select max(salary) from salaries where salary != (select max(salary) from salaries)); # select # e.emp_no, # s.salary, # e.last_name, # e.first_name # from # employees e # left join salaries s on e.emp_no = s.emp_no # where # s.salary = ( # select # max(salary) # from # salaries # where # salary != ( # select # max(salary) # from # salaries # ) # );
这道题是查找薪水排名第二多的员工编号emp_no、薪水salary、last_name以及first_name,不能使用order by完成,以上例子输出为:(温馨提示:sqlite通过的代码不一定能通过mysql,因为SQL语法规定,使用聚合函数时,select子句中一般只能存在以下三种元素:常数、聚合函数,group by 指定的列名。如果使用非group by的列名,sqlite的结果和mysql 可能不一样)
两个关键点:
1.查找排名第二多的薪水数值,思路是先使用工资表salaries里面的最高工资数值max(salary),用where salary != max(salary)去掉最高的工资数值,再取max(salary)得到剩余的里面最高的工资数值,得到整个表里面第二高的工资数值,总共使用了两个子查询,并且是嵌套的
select
max(salary)
from
salaries
where
salary != (
select
max(salary)
from
salaries
)
2.查找薪水排名第二多的员工编号emp_no、薪水salary、last_name以及first_name
select e.emp_no, s.salary, e.last_name, e.first_name from employees e left join salaries s on e.emp_no = s.emp_no where s.salary = (select max(salary) from salaries where salary != (select max(salary) from salaries));
完整代码如下:
select e.emp_no, s.salary, e.last_name, e.first_name from employees e left join salaries s on e.emp_no = s.emp_no where s.salary = (select max(salary) from salaries where salary != (select max(salary) from salaries)); # select # e.emp_no, # s.salary, # e.last_name, # e.first_name # from # employees e # left join salaries s on e.emp_no = s.emp_no # where # s.salary = ( # select # max(salary) # from # salaries # where # salary != ( # select # max(salary) # from # salaries # ) # );
【注意】emp_no 不建议写成 emp_no
因为 两张表都有 emp_no(你 join 就是用它连接的),所以:
SELECT emp_no
会在很多数据库里直接报错:emp_no is ambiguous(列名有歧义)
必须写成:
SELECT e.emp_no
或
SELECT s.emp_no
完结
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