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
