.NET高级工程师面试题之SQL篇
fff8
9年前
原文出处： 岁月如初
1 题目
这确实是一个真实的面试题，琢磨一下吧！知识不用，就会丢掉，我太依赖各种框架和dll了，已经忘记了最基本的东西。有多久没有写过SQL了，我已经不记得了。
已知表信息如下：
Department(depID, depName),depID 系编号，DepName系名
Student(stuID, name, depID) 学生编号，姓名，系编号
Score(stuID, category, score) 学生编码，科目，成绩
找出每一个系的最高分，并且按系编号，学生编号升序排列，要求顺序输出以下信息：
系编号，系名，学生编号，姓名，总分
2 实验
USE [test] GO /****** Object: Table [dbo].[Score] Script Date: 05/11/2015 23:16:23 ******/ SET ANSI_NULLS ON GO SET QUOTED_IDENTIFIER ON GO SET ANSI_PADDING ON GO CREATE TABLE [dbo].[Score]( [stuID] [int] NOT NULL, [category] [varchar](50) NOT NULL, [score] [int] NOT NULL ) ON [PRIMARY] GO SET ANSI_PADDING OFF GO INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (1, N'英语', 80) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (2, N'数学', 80) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (1, N'数学', 70) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (2, N'英语', 89) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (3, N'英语', 81) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (3, N'数学', 71) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (4, N'数学', 91) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (4, N'英语', 61) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (5, N'英语', 91) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (6, N'英语', 89) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (7, N'英语', 77) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (8, N'英语', 97) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (9, N'英语', 57) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (5, N'数学', 87) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (6, N'数学', 89) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (7, N'数学', 80) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (8, N'数学', 81) INSERT [dbo].[Score] ([stuID], [category], [score]) VALUES (9, N'数学', 84) /****** Object: Table [dbo].[Department] Script Date: 05/11/2015 23:16:23 ******/ SET ANSI_NULLS ON GO SET QUOTED_IDENTIFIER ON GO SET ANSI_PADDING ON GO CREATE TABLE [dbo].[Department]( [depID] [int] IDENTITY(1,1) NOT NULL, [depName] [varchar](50) NOT NULL, PRIMARY KEY CLUSTERED ( [depID] ASC )WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY] ) ON [PRIMARY] GO SET ANSI_PADDING OFF GO SET IDENTITY_INSERT [dbo].[Department] ON INSERT [dbo].[Department] ([depID], [depName]) VALUES (1, N'计算机') INSERT [dbo].[Department] ([depID], [depName]) VALUES (2, N'生物') INSERT [dbo].[Department] ([depID], [depName]) VALUES (3, N'数学') SET IDENTITY_INSERT [dbo].[Department] OFF /****** Object: Table [dbo].[Student] Script Date: 05/11/2015 23:16:23 ******/ SET ANSI_NULLS ON GO SET QUOTED_IDENTIFIER ON GO SET ANSI_PADDING ON GO CREATE TABLE [dbo].[Student]( [stuID] [int] IDENTITY(1,1) NOT NULL, [stuName] [varchar](50) NOT NULL, [deptID] [int] NOT NULL, PRIMARY KEY CLUSTERED ( [stuID] ASC )WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY] ) ON [PRIMARY] GO SET ANSI_PADDING OFF GO SET IDENTITY_INSERT [dbo].[Student] ON INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (1, N'计算机张三', 1) INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (2, N'计算机李四', 1) INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (3, N'计算机王五', 1) INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (4, N'生物amy', 2) INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (5, N'生物kity', 2) INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (6, N'生物lucky', 2) INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (7, N'数学_yiming', 3) INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (8, N'数学_haoxue', 3) INSERT [dbo].[Student] ([stuID], [stuName], [deptID]) VALUES (9, N'数学_wuyong', 3) SET IDENTITY_INSERT [dbo].[Student] OFF /****** Object: Default [DF__Departmen__depNa__5441852A] Script Date: 05/11/2015 23:16:23 ******/ ALTER TABLE [dbo].[Department] ADD DEFAULT ('') FOR [depName] GO /****** Object: Default [DF__Score__category__5EBF139D] Script Date: 05/11/2015 23:16:23 ******/ ALTER TABLE [dbo].[Score] ADD DEFAULT ('') FOR [category] GO /****** Object: Default [DF__Score__score__5FB337D6] Script Date: 05/11/2015 23:16:23 ******/ ALTER TABLE [dbo].[Score] ADD DEFAULT ((0)) FOR [score] GO /****** Object: Default [DF__Student__stuName__59063A47] Script Date: 05/11/2015 23:16:23 ******/ ALTER TABLE [dbo].[Student] ADD DEFAULT ('') FOR [stuName] GO /****** Object: ForeignKey [FK__Student__deptID__59FA5E80] Script Date: 05/11/2015 23:16:23 ******/ ALTER TABLE [dbo].[Student] WITH CHECK ADD FOREIGN KEY([deptID]) REFERENCES [dbo].[Department] ([depID]) GO 准备环境 3 结果
面试的时候，没有写出来，当时脑袋昏沉沉的。也确实好久没有写复杂的sql语句了。今天花了2到3个小时，终于试出来了。不知道有没有更好的写法？
-- 每个系里的最高分的学生信息 SELECT Department.depID, Department.depName, Student.stuID, stuName, Dscore.scores FROM Department LEFT JOIN Student on department.depID = student.deptID LEFT JOIN (SELECT Score.stuId, SUM(Score) AS scores FROM Score GROUP by stuID ) AS Dscore on Student.stuID = dScore.stuID where exists ( select * from ( SELECT deptID, MAX(scores) AS topScores FROM Student LEFT JOIN ( SELECT stuID,SUM(score) AS scores FROM Score GROUP BY stuID) AS newScore ON Student.stuID = newScore.stuID group by deptID) AS depScore where Department.depID = depScore.deptID and Dscore.scores=depScore.topScores ) order by Department.depID,Student.stuID;
4 补充
看了那么多的评论，自己写的真的不咋样，可惜今天没有时间细细看了，现在还在公司加班！但百度一下的时间还是有滴，So整理一下相关资料先。
（1）、SQL2005四个排名函数（row_number、rank、dense_rank和ntile）的比较
（2）、关于with as：使用WITH AS提高性能简化嵌套SQL
5 参考SQL
正确的答案的结果是一样的，错误的各有各的不同，正确的答案后的性能也各有各的不同，不过呢，暂时没有水平去分析它，但是有空会把这些全部看一遍.谢谢各位啦！【2015-05-13 23:44】
1、pursuer.chen SELECT B.depID,B.depName,B.stuID ,B.stuName,SUM(A.score )AS SUM_SCORE FROM Score A INNER JOIN (SELECT SA.depID,SA.depName,S.stuID,S.stuName FROM Student S INNER JOIN Score SE ON S.stuID=SE.stuID INNER JOIN ( SELECT D.depID,D.depName ,MAX(SC.score )AS MX_score FROM Student S INNER JOIN Score SC ON S.stuID=SC.stuID INNER JOIN Department D ON S.deptID=D.depID GROUP BY D.depID,D.depName ) SA ON SE.score=SA.MX_score AND S.deptID=SA.depID ) B ON A.stuID=B.stuID GROUP BY B.depID,B.depName,B.stuID ,B.stuName ORDER BY B.depID,B.stuID 结果正确 计算机 2 计算机李四 169 生物 4 生物amy 152 生物 5 生物kity 178 数学 8 数学_haoxue 178 2、Gamain 正确 WITH cte1 as ( SELECT DISTINCT D.depID, D.depName, S.stuID, S.stuName, SUM(Sc.score) OVER (PARTITION BY D.depID,S.stuID) as sumScore FROM Department D LEFT JOIN Student S ON D.depID=S.deptID LEFT JOIN Score Sc ON Sc.stuID=S.stuID ), cte2 as ( SELECT DISTINCT depID, stuID, MAX(sumScore) OVER (PARTITION BY depID) as maxScore FROM cte1 ) SELECT c1.depID, c1.depName, c1.stuID, c1.stuName, c1.sumScore from cte2 c2 INNER JOIN cte1 c1 ON c1.depID=c2.depID AND c1.stuID=c2.stuID and c1.sumScore=c2.maxScore; 3、飞不动 正确 use test; select e.* from ( select c.depID,c.depName,a.stuID,b.stuName,a.total from (select stuID,sum(score) as total from Score group by stuID) a join Student b on b.stuID=a.stuID join Department c on c.depID=b.deptID ) e join (select b.deptID,max(a.total) maxScore from (select stuID,sum(score) as total from Score group by stuID) a join Student b on b.stuID=a.stuID group by b.deptID ) f on e.depID=f.deptID and e.total=f.MaxScore order by e.depID,e.stuID 4、之路 错误 select depID, depName, stuId, stuName, PerTotalScore from ( select stuID, stuName, depID, depName, PerTotalScore, ROW_NUMBER() OVER(partition by depID order by PerTotalScore) as RowId from ( select distinct s.stuID, s.stuName, d.depID, d.depName, SUM(c.score) OVER(partition by d.depID,s.stuID) as PerTotalScore from dbo.student s JOIN dbo.Department d on s.deptID=d.depID JOIN dbo.Score c ON s.StuID=c.StuID ) as T ) as TT WHERE TT.RowId=1 order by depID,stuID 计算机 1 计算机张三 150 生物 4 生物amy 152 数学 9 数学_wuyong 141 5、King兵 正确 WITH a AS (SELECT Department.depID, Department.depName, Student.stuID, stuName, Dscore.scores,ROW_NUMBER() OVER(PARTITION BY Department.depID ORDER BY scores DESC) ROWID FROM Department LEFT JOIN Student on department.depID = student.deptID LEFT JOIN (SELECT Score.stuId, SUM(Score) AS scores FROM Score GROUP by stuID ) AS Dscore on Student.stuID = dScore.stuID), b AS ( SELECT Department.depID, Department.depName, Student.stuID, stuName, Dscore.scores,ROW_NUMBER() OVER(PARTITION BY Department.depID ORDER BY scores DESC) ROWID FROM Department LEFT JOIN Student on department.depID = student.deptID LEFT JOIN (SELECT Score.stuId, SUM(Score) AS scores FROM Score GROUP by stuID ) AS Dscore on Student.stuID = dScore.stuID ) SELECT depID, depName, stuID, stuName, scores,ROWID FROM a WHERE a.scores = (SELECT MAX(scores) FROM b c WHERE a.depid = c.depid) 6、 怪咖Eric 正确 SELECT bb.deptID , cc.depName , bb.stuID , bb.stuName , bb.TotalScore FROM ( SELECT * , RANK() OVER ( PARTITION BY deptID ORDER BY TotalScore DESC ) AS pos FROM ( SELECT SUM(b.score) AS TotalScore , a.stuID , a.stuName , a.deptID FROM Student a JOIN Score b ON a.StuID = b.StuID GROUP BY a.stuID , a.stuName , a.deptID ) aa ) bb JOIN dbo.Department cc ON bb.deptID = cc.depID JOIN dbo.Student dd ON bb.stuID = dd.stuID WHERE pos = '1' ORDER BY bb.deptID , bb.stuID 7、Michael Jiang 手写 改后正确 use test; SELECT D.* FROM ( SELECT de.depID, de.depName, st.stuID, st.stuName, sc.score, RANK() OVER( PARTITION BY st.deptID ORDER BY sc.score DESC ) rowno FROM Student st LEFT JOIN Department de ON de.depID=st.deptID LEFT JOIN ( SELECT sc.stuID, SUM(sc.score) score FROM Score sc GROUP BY sc.stuID ) sc ON sc.stuID=st.stuID ) D WHERE D.rowno = 1 --看错要求，原来只要列出最高分 ORDER BY D.depID, D.rowno 8、正确 Li.zheng use test; select * from ( select (select depName from Department where Department.depID = a.depID) as depName, (select stuName from Student where Student.stuID = a.stuID) as stuName, dense_rank() over(partition by depID order by sumScore desc) as rank, a.sumScore from ( select c.depID,b.stuid,sum(a.score) as sumScore from score as a inner join Student as b on a.stuid = b.stuid inner join Department as c on c.depID = b.deptID group by c.depID,b.stuid ) as a ) as b where b.rank = 1 9、下个路口 错误 漏了并列第一 SELECT * FROM ( SELECT s1.stuID,s1.stuName,s1.deptID,t.totalScore,d.depName, ROW_NUMBER() OVER(PARTITION BY d.depID ORDER BY totalScore DESC) AS Rn FROM Student AS s1 INNER JOIN ( SELECT s.stuID,SUM(s2.score) AS totalScore FROM Student AS s INNER JOIN Department AS d ON d.depID = s.deptID INNER JOIN Score s2 ON s2.stuID = s.stuID GROUP BY s.stuID ) AS t ON t.stuID = s1.stuID INNER JOIN Department AS d ON d.depID = s1.deptID ) result WHERE Rn = 1 ORDER BY result.stuID 9、自由_ 正确 select d.depID,d.depName,s.stuID,s.stuName,t.score from Department d left join (select s.stuID,sum(s.score) as score,st.deptID, rank() over(partition by st.deptID order by sum(s.score) desc) ra from Score s left join Student st on s.stuID = st.stuID group by s.stuID,st.deptID) t on d.depID = t.deptID left join Student s on t.stuID = s.stuID where t.ra = 1 order by d.depID,s. 10、 手写 改了 之后 错误， use test; with Combin AS ( SELECT MAX(score) AS 最高分,deptID AS 系编号,MAX(a.stuID) AS 学生Id FROM Student a LEFT JOIN Score b ON a.stuID=b.stuID GROUP BY a.deptID ) SELECT c.系编号, (SELECT depName FROM Department d WHERE d.depID=c.系编号 ) AS 系名, c.学生Id AS '学生编号', (SELECT stuName FROM Student e WHERE e.stuID=c.学生Id ) AS '姓名', c.最高分 FROM Combin c 计算机 3 计算机王五 89 生物 6 生物lucky 91 数学 9 数学_wuyong 97 11、 舍长 正确 use test; WITH T1 AS ( SELECT A.DEPID,A.DEPNAME,B.STUID,B.STUNAME,SUM(C.SCORE) AS TotalScore FROM Department A INNER JOIN Student B ON A.DEPID = B.DEPTID INNER JOIN Score C ON B.STUID = C.STUID GROUP BY A.DEPID,A.DEPNAME,B.STUID,B.STUNAME ), T2 AS ( SELECT *,RANK() OVER(PARTITION BY DEPID ORDER BY TotalScore DESC) AS RankScore FROM T1 ) SELECT * FROM T2 WHERE RankScore = 1 ORDER BY DEPID,STUID 12、Ender.Lu 正确 with tscore as (select stuID ,sum(score) as score from dbo.Score group by stuID), tinfo as (select Student.deptID ,Department.depName,dbo.Student.stuID,dbo.Student.stuName,tscore.score from dbo.Student inner join [dbo].[Department] on dbo.Department.depID = student.deptID left join tscore on tscore.stuid = Student.stuID), trank as ( select deptID ,depName,stuID,stuName,score ,rank() over(partition by deptID order by score desc) as level from tinfo ) select deptID ,depName,stuID,stuName,score from trank where level = 1 order by deptID ,stuID; 13、McJeremy&Fan 正确 select p.totalscore,p.stuid,p.stuname,p.deptid,x.depname from ( select dense_rank() over(partition by deptid order by totalscore desc) as num, a.totalscore,b.stuid,b.stuname,b.deptid from ( select stuid,sum(score) as totalscore from score group by stuid ) a inner join student b on a.stuid=b.stuid ) as p inner join department x on p.deptid=x.depid where p.num=1 13、清水无大大鱼 正确 with temp as( select a.deptid,a.stuID,a.stuName,b.score from student a,(select stuID,sum(score)as score from score group by stuID)b where a.stuID=b.stuID) select d.depID,d.depName,b.stuID,b.stuName,b.score from Department d,( select * from temp t where t.score=( select max(score) from temp sc where t.deptid=sc.deptid)) b where d.depID=b.deptID order by depID,stuID 14、 BattleHeart 正确 SELECT D.*,DD.depName FROM ( SELECT C.stuID, C.TotleScore, C.stuName, C.deptID, DENSE_RANK() OVER(PARTITION BY C.deptID ORDER BY C.TotleScore DESC ) nubid FROM (SELECT S.stuID, ST.stuName, SUM(S.score) AS TotleScore, ST.deptID FROM dbo.Student AS ST INNER JOIN dbo.Score AS S ON S.stuID = ST.stuID GROUP BY S.stuID,ST.deptID,ST.stuName) AS C) AS D INNER JOIN dbo.Department AS DD ON DD.depID = D.deptID WHERE D.nubid=1