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mysql常见面试题

学生成绩表中请找出每门课程都超过80分的人的名字

sql：

select name from test.stu

group by name

having count(score) =sum(case when score>80 then 1 else 0 end )

select name from stu

group by name

having name not in (

select name from stu

where score <80)

select name from test.stu

group by name

having min(score)>=80

查找最晚入职的员工

select * from employees order by hire_date desc limit 1;

入职倒数第三的员工

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

对于可能重复的情况

SELECT * FROM employees WHERE hire_date = (

SELECT DISTINCT hire_date

FROM employees

ORDER BY hire_date DESC – 倒序

LIMIT 1 OFFSET 2 – 去掉排名倒数第一第二的时间，取倒数第三 );

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