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我有一个查询，它从查询中创建的其他表中提取信息。我的最终 select from 语句的最终输出如下所示：

VISIT_ID | MRN | DAYS SCORE | IP SCORE | ER SCORE | CC SCORE | TOTAL
123456   | 123 | 3          | 3        | 2        | 0        |   8
123456   | 123 | 3          | 3        | 2        | 2        |   10
123456   | 123 | 3          | 3        | 2        | 4        |   12
...

我想做的只是获取带有 MAX(TOTAL) 的行，在这种情况下是带有 MAX(TOTAL) 的行TOTAL = 12

我看过这篇文章，但似乎无法正确理解。我也看过这里。

这是产生结果的查询：

-- @LACE_MSTR TABLE DECLARATION ###################################//
DECLARE @LACE_MSTR TABLE(
MRN VARCHAR(200)
, VISIT_ID VARCHAR(200)
, [LACE DAYS SCORE] INT
, [LACE ACUTE IP SCORE] INT
, [LACE ER SCORE] INT
, [LACE COMORBID SCORE] INT
)
--###################################################################//
INSERT INTO @LACE_MSTR
SELECT
Q1.MRN
, Q1.ENCOUNTER_ID
, Q1.[LACE DAYS SCORE]
, Q1.[ACUTE ADMIT SCORE]
, CASE
    WHEN Q1.VISIT_COUNT IS NULL THEN 0
    WHEN Q1.VISIT_COUNT = 1 THEN 1
    WHEN Q1.VISIT_COUNT = 2 THEN 2
    WHEN Q1.VISIT_COUNT = 3 THEN 3
    WHEN Q1.VISIT_COUNT >= 4 THEN 4
    ELSE 0
  END AS [LACE ER SCORE]
, Q1.[CC LACE SCORE]

FROM
    (
    SELECT
    DISTINCT T1.ENCOUNTER_ID
    , T1.MRN
    , T1.[LACE DAYS SCORE]
    , T1.[ACUTE ADMIT SCORE]
    , CNT.VISIT_COUNT
    , CM.[CC LACE SCORE]

    FROM @T1 T1
    LEFT OUTER JOIN @CNT CNT
    ON T1.MRN = CNT.MRN
    JOIN @CM CM
    ON CM.[MRN CM] = T1.[MRN]
    ) Q1

SELECT DISTINCT VISIT_ID
, MRN
, [LACE DAYS SCORE]
, [LACE ACUTE IP SCORE]
, [LACE ER SCORE]
, [LACE COMORBID SCORE]
, [LACE DAYS SCORE]+[LACE ACUTE IP SCORE]+[LACE ER SCORE]+[LACE COMORBID SCORE] AS [TOTAL LACE]

FROM @LACE_MSTR
Tried this but I did not do it right therefore it does not work
--INNER JOIN
--  (
--  SELECT VISIT_ID
--  , MAX([LACE DAYS SCORE]+[LACE ACUTE IP SCORE]+[LACE ER SCORE]+[LACE COMORBID SCORE]) AS [TOTAL LACE]
--  FROM @LACE_MSTR
--  GROUP BY VISIT_ID
--  ) GROUPEDLACE_MSTR ON @LACE_MSTR.VISIT_ID=GROUPEDLACE_MSTR.VISIT_ID
--  AND @LACE_MSTR.[TOTAL LACE SCORE] = GROUPED@LACE_MSTR.[TOTAL LACE]

GROUP BY VISIT_ID
, MRN
, [LACE DAYS SCORE]
, [LACE ACUTE IP SCORE]
, [LACE ER SCORE]
, [LACE COMORBID SCORE]

请让我知道是否需要澄清。

谢谢，

4

1 回答 1

1

假设分组应基于VISIT_ID和MRN：

;WITH x AS 
(
  SELECT VISIT_ID, MRN, ... etc ..., 
    rn = ROW_NUMBER() OVER (PARTITION BY VISIT_ID, MRN 
      ORDER BY [LACE DAYS SCORE] + [LACE ACUTE IP SCORE]
        + [LACE ER SCORE] + [LACE COMORBID SCORE] DESC)
  FROM @LACE_MSTR
)
SELECT VISIT_ID, MRN, ... etc ... 
FROM x WHERE rn = 1;
于 2013-09-03T19:52:07.423 回答