gre*_*ner 8 sql rank sql-server-2008
我有两个表RSLTS和CONTACTS:
RSLTS
QRY_ID | RES_ID | SCORE
-----------------------------
A | 1 | 15
A | 2 | 32
A | 3 | 29
C | 7 | 61
C | 9 | 30
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联系
C_ID | QRY_ID | RES_ID
----------------------------
1 | A | 2
2 | A | 1
3 | C | 9
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我试图创建一份报告,显示每个联系人记录(C_ID)时,RANK()中RES_ID(通过SCORE在)RSLTS其组内(表QRY_ID).使用上面的数据,它看起来像这样:
C_ID | QRY_ID | RES_ID | SCORE | Rank
-----------------------------------------------
1 | A | 2 | 32 | 1
2 | A | 1 | 15 | 3
3 | C | 9 | 30 | 2
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到目前为止,我尝试了这个但是它返回了最后一行的Rank = 1(而第二行的rank = 2也是错误的)
SELECT
C.*
,R.SCORE
,RANK() OVER (PARTITION BY R.QRY_ID ORDER BY R.SCORE DESC)
FROM CONTACTS C LEFT JOIN RSLTS R
ON C.RES_ID = R.RES_ID
AND C.QRY_ID = R.QRY_ID
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更新:SQLFiddle
Joh*_*ret 11
因为排名完全不依赖于联系人
RANKED_RSLTS
QRY_ID | RES_ID | SCORE | RANK
-------------------------------------
A | 1 | 15 | 3
A | 2 | 32 | 1
A | 3 | 29 | 2
C | 7 | 61 | 1
C | 9 | 30 | 2
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因此:
SELECT
C.*
,R.SCORE
,MYRANK
FROM CONTACTS C LEFT JOIN
(SELECT *,
MYRANK = RANK() OVER (PARTITION BY QRY_ID ORDER BY SCORE DESC)
FROM RSLTS) R
ON C.RES_ID = R.RES_ID
AND C.QRY_ID = R.QRY_ID
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