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  • SQL查询练习一(From LeetCode)

     请选用MySQL进行测试.

    1.找出后一天比前一天温度高的日期(E)

    思路:将该表进行自关联,然后选择出日期靠后的一天,最后选择温度高的日期

    1 select Weather.Id
    2 from Weather join Weather w
    3 on datediff(Weather.Date,w.Date) = 1
    4 and Weather.Temperature > w.Temperature
    View Code

    2.找出人口在3百万或者领土面积在25百万的国家(E)

    思路:使用where进行条件过筛选,or进行或判断

    1 select name,population,area
    2 from World
    3 where area > 3000000 or population > 25000000
    View Code
    1 1 select name,population,area
    2 2 from World
    3 3 where area > 3000000 or population > 2500000

    3.判断是否是三角形(E)

    思路,使用case when进行搭配,使用三角形定义进行判断x+y>z,x+z>y,y+z>x

    1 select x,y,z,
    2 case
    3     when x+y>z and y+z>x and x+z>y then 'Yes'
    4     else 'No'
    5 end as triangle 
    6 from triangle 
    View Code

     4.找出薪水第二高的员工

    思路:先找出最多的薪水的员工,在把他的薪水小于最大的工资即可

    1 select Max(Salary) as SecondHighestSalary 
    2 from Employee 
    3 where Salary < (select Max(Salary) from Employee)
    View Code

    5.找出每个学科都有多少名学生(M)

    思路:将两张表进行左连接,一department表作为主表,然后按照dept_name进行分组,最后按照人数进行降序排列

    1 select d.dept_name,count(student_id) as student_number
    2 from department d left join Student s
    3 on d.dept_id = s.dept_id
    4 group by d.dept_name
    5 order by student_number desc,d.dept_name
    View Code

    6.找出每个部门薪水最高的员工(M)

    思路:将两张表进行连接,内层查询根据department表的name进行分组,每组的最大值,既是每个部门的薪水最大值,然后传递给外层的部门id和薪水即可

    1 select d.Name as Department,e.Name as Employee,e.Salary as Salary
    2 from Department d join Employee e
    3 on e.DepartmentId = d.Id
    4 where (e.DepartmentId,e.Salary) in
    5 (
    6     select DepartmentId,max(Salary)
    7     from Employee
    8     group by DepartmentId
    9 )
    View Code

    7.找出至少有5名下属的领导(M)

    思路:使用内层查询查找出有5名下属的ManagerId然后,将外层查询的员工Id=ManagerId就是查询的结果

    1 select e1.Name
    2 from Employee e1
    3 join
    4 (
    5     select ManagerId from Employee
    6     group by ManagerId
    7     having count(*) >= 5
    8 ) as e2
    9 on e1.Id = e2.ManagerId
    View Code

    8.找出得票最多的候选人(M)

     

    思路:先在内层查询中找出最受欢迎的候选人,然后将中间表的候选人Id既是赢家的候选人id,两者相等即可

     1 select c.Name
     2 from Candidate c
     3 join 
     4 (
     5     select CandidateId  from Vote
     6     group by CandidateId  
     7     order by count(*) desc
     8     limit 1
     9 ) as winner
    10 on c.id = winner.CandidateId;
    View Code

    9.根据Score计算等级(M)

    思路:将两张表进行自连接,根据Id进行分组,最后根据Rank进行排序

    1 select s.Score,count(distinct t.Score) as Rank
    2 from Scores s join Scores t
    3 on s.Score <= t.Score
    4 group by s.Id
    5 order by Rank 
    View Code

    10.找出二叉树的节点分布(M)

    思路:使用case when的结构进行循环判断输出

     1 select id,
     2 case
     3     when tree.id = (select atree.id from tree atree where atree.p_id is NULL)
     4         then 'Root'
     5     when tree.id in (select atree.p_id from tree atree)
     6         then 'Inner'
     7     else 
     8         'Leaf'
     9 end as Type
    10 from tree
    11 order by id
    View Code

    11.找出每个部门薪水排前三名的员工(H)

     

    思路:先进行表连接,将内层查询的结果和外部的表的Salary相比较,选择前面3个

     1 select d.Name as Department,e.Name as Employee,e.Salary
     2 from Employee e join Department d
     3 on e.DepartmentId = d.Id
     4 where 3 >
     5 (
     6     select count(distinct e2.Salary)
     7     from Employee e2
     8     where e2.Salary > e.Salary
     9     and e.DepartmentId = e2.DepartmentId
    10 )
    View Code

    12.找出2013-10-01到2013-10-03之间的网约车的取消率(H)

    思路:计算取消率,使用case when语法,找出Trips中Status变量以canceled_开头的比例

    1 select t.Request_at as Day,
    2 round(sum(case when t.Status like 'cancelled_%' then 1 else 0 end)/count(*),2) as "Cancellation Rate"
    3 from Trips t join Users u
    4 on t.Client_Id = u.Users_Id  
    5 where u.Banned = 'No'
    6 and t.Request_at between '2013-10-01' and '2013-10-03'
    7 group by t.Request_at
    View Code

    13.找出每个部门员工薪水的中位数(H)

     

    思路:将此表进行自关联,计算工资的中位数,使用case when计算中间表的中位数

    1 select e.Id,e.Company,e.Salary
    2 from Employee e join Employee aliens
    3 on e.Company = aliens.Company
    4 group by e.Company,e.Salary
    5 having sum(case when e.Salary = aliens.Salary then 1 else 0 end) >=
    6 abs(sum(sign(e.Salary-aliens.Salary)))
    7 order by e.Id
    View Code
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  • 原文地址:https://www.cnblogs.com/luhuajun/p/8471107.html
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