目錄

一、重新格式化部門表

1、題目描述

2、題解

3、源碼

二、第二高的薪資

1、題目描述

2、題解

3、源碼

?三、第n高的薪水

1、題目描述

2、題解

3、源碼

?四、分數(shù)排名

1、題目描述

2、題解

3、源碼

五、連續(xù)出現(xiàn)的數(shù)

1、題目描述

2、題解

?3、源碼

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一、重新格式化部門表

1、題目描述

?

2、題解

3、源碼

# Write your MySQL query statement below select id,sum(case month when 'Jan' then revenue end) as Jan_Revenue,sum(case month when 'Feb' then revenue end) as Feb_Revenue,sum(case month when 'Mar' then revenue end) as Mar_Revenue,sum(case month when 'Apr' then revenue end) as Apr_Revenue,sum(case month when 'May' then revenue end) as May_Revenue,sum(case month when 'Jun' then revenue end) as Jun_Revenue,sum(case month when 'Jul' then revenue end) as Jul_Revenue,sum(case month when 'Aug' then revenue end) as Aug_Revenue,sum(case month when 'Sep' then revenue end) as Sep_Revenue,sum(case month when 'Oct' then revenue end) as Oct_Revenue,sum(case month when 'Nov' then revenue end) as Nov_Revenue,sum(case month when 'Dec' then revenue end) as Dec_Revenue from Department group by id

二、第二高的薪資

1、題目描述

?

2、題解

?

3、源碼

# Write your MySQL query statement below # select ifNull( # (select distinct salary # from Employee # order by Salary Desc # limit 1,1),null # ) as SecondHighestSalary;select max(distinct salary) as SecondHighestSalary from Employee where salary < (select max(distinct salary)from Employee);# select max(distinct 成績) # from 成績表 # where 課程='語文' and # 成績 < (select max(distinct 成績) # from 成績表 # where 課程='語文');

?三、第n高的薪水

1、題目描述

?

2、題解

3、源碼

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT BEGINSET N :=N-1;RETURN (# Write your MySQL query statement below.SELECT salary FROM Employeegroup by salaryORDER by salary DESCLIMIT N,1) ; END

?四、分數(shù)排名

1、題目描述

?

?

2、題解

3、源碼

selectscore,(dense_rank() over (order by Score desc)) AS "rank" fromScores

五、連續(xù)出現(xiàn)的數(shù)

1、題目描述

2、題解

?3、源碼

# Write your MySQL query statement below select distinctNum as ConsecutiveNums from (select Num,Id-cast((row_number() over(partition by Num order by Id asc)) as signed) as ranking from Logs) as t group by Num,ranking having count(*)>=3

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