自加入sql生成排名

我想根據他獲得的分數來計算每個學生的排名。但是，如果標記處於-ve值，則查詢不會顯示適當的答案。自加入sql生成排名

SELECT a1.ResultByTestId, a1.ObtainedMarks, COUNT(a2.ObtainedMarks) Rank 
FROM ResultByTest a1, ResultByTest a2 
WHERE a1.ObtainedMarks < a2.ObtainedMarks 
     or 
     (a1.ObtainedMarks=a2.ObtainedMarks and a1.StudentId = a2.StudentId) 
GROUP BY a1.ObtainedMarks, a1.ResultByTestId 
ORDER BY a1.ObtainedMarks DESC, Rank DESC

，並回答表是： -

resutlByTestId ObtainedMarks Rank 

11    36    1 
10    22    2 
3     9    3 
9     7    4 
6     5    5 
8     3    6 
7     2    7 
4     0    8 
5     0    8 
1     0    8 
2     0    8 
13    -5    12

有沒有人幫我嗎？

來源

2012-12-05 mitali

你應該在2008年MSSQL使用DENSE_RANK() function：

SELECT ResultByTestId, ObtainedMarks, 
     DENSE_RANK() OVER (order by ObtainedMarks desc) as Rank 
FROM ResultByTest 
ORDER by ObtainedMarks DESC;

也在這裏是你的闕RY與COUNT(DISTINCT <>)：

SELECT a1.ResultByTestId, a1.ObtainedMarks, 
COUNT(DISTINCT a2.ObtainedMarks) Rank 
FROM ResultByTest a1, ResultByTest a2 
WHERE a1.ObtainedMarks < a2.ObtainedMarks 
     or 
     (a1.ObtainedMarks=a2.ObtainedMarks and a1.StudentId = a2.StudentId) 
GROUP BY a1.ObtainedMarks, a1.ResultByTestId 
ORDER BY a1.ObtainedMarks DESC, Rank DESC

來源

2012-12-05 11:24:55 valex

您的幫助，但我想排名9的標記是-5。 – mitali

然後只是改變你的查詢'COUNT（DISTINCT a2.ObtainedMarks）' – valex

非常感謝你。但是你可以告訴我，如果我想通過使用Rank（）函數來計算排名，那你建議我。 – mitali

輸出是正確的，沒有問題，因爲-ve得分爲四名學生得到0和排名第8 所以 8-8 8-9 8-10 8-11

這是它將如何計算這樣下次排名將明顯12

來源

2012-12-05 11:20:53

是的，我知道，這是計算8級4時間，所以接下來的排名是12，這是爲我創造的問題，我想排名後9 8 12.不感謝名單 – mitali

--setup test data 

DECLARE @ResultByTest TABLE (ResultByTestId int, ObtainedMarks int) 

INSERT INTO @ResultByTest VALUES 
(11,    36), 
(10,    22), 
(3,     9), 
(9,     7), 
(6,     5), 
(8,     3), 
(7,     2), 
(4,     0), 
(5,     0), 
(1,     0), 
(2,     0), 
(13,    -5) 

-- this will give you the results you have 

select *, RANK() OVER (ORDER BY ObtainedMarks DESC) AS [Rank] 
from @ResultByTest 

-- this will give you the results I think you want 

select *, DENSE_RANK() OVER (ORDER BY ObtainedMarks DESC) AS [Rank] 
from @ResultByTest

來源

2012-12-05 11:26:34

您可以用Dense_Rank功能

Select 
    StudentID, 
    ObtainedMarks, 
    Dense_Rank() Over (Order By ObtainedMarks Desc) as Rank 
From 
    ResultByTest

http://sqlfiddle.com/#!3/a9e5c/1

，你通常會使用排名之所以這樣做，就是當你問的人誰得到-5你從哪裏來的，他們說「9號」。然後你問，有多少人打你「12」。然後你看起來很困惑。

來源

2012-12-05 11:37:34 Laurence

使用DENSE_RANK來消除間隙。

DECLARE @ObtainedMarks TABLE (
ResultByTestId INT, 
ObtainedMarks INT, 
StudentId INT 

) 

INSERT INTO @ObtainedMarks VALUES 
(1,0,1), 
(2,0,1), 
(3,9,1), 
(4,0,1), 
(5,0,1), 
(6,5,1), 
(7,2,1), 
(8,3,1), 
(9,7,1), 
(10,22,1), 
(11,36,1), 
(13,-5,1) 

SELECT 
    ResultByTestId, 
    ObtainedMarks, 
    DENSE_RANK() OVER (order by ObtainedMarks desc) [Rank] 
FROM @ObtainedMarks

來源

2012-12-05 11:37:54

自加入sql生成排名

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