将该组中的单元格值与最大值进行比较，出现错误 SELECT 失败 3707

Question

我试图将单元格值与视图中该组中的最大值进行比较，但它抛出了一个错误。

内部查询运行良好，它为我提供了一份患者名单、他们的适应症以及 2017 年和 3 年索赔的数量。在外部查询中，我只需要保持耐心，只需要保持 2017 年索赔 >0 并且在 3 年内拥有最大索赔的适应症。

这里是代码抛出错误的地方

SELECT patient_id, 
       CASE 
         WHEN Count(patient_id) = 1 THEN diagnosis_grouping 
         WHEN claim_2017 > 0 
              AND claims_3yr = Max(claims_3yr) THEN diagnosis_grouping 
         ELSE NULL 
       END AS INDICATION 
FROM   (SELECT patient_id, 
               diagnosis_grouping, 
               Sum (CASE 
                      WHEN LEFT(service_date, 4) = '2017' THEN 1 
                      ELSE 0 
                    END)                       AS CLAIM_2017, 
               Count(DISTINCT claim_id)        AS CLAIMS_3YR, 
               Max(Cast(service_date AS DATE)) AS LATEST_CLAIM 
        FROM   l01_dx_claims A 
               INNER JOIN diagnosis_code_imm B 
                       ON A.diagnosis_code = B.diagnosis_code 
        GROUP  BY 1, 
                  2 
        HAVING Count (DISTINCT claim_id) > 1 
               AND Sum(CASE 
                         WHEN Cast(LEFT(service_date, 4) AS INT) = 2017 THEN 1 
                         ELSE 0 
                       END) > 0 
        ORDER  BY Cast(patient_id AS INT), 
                  diagnosis_grouping) 
GROUP  BY patient_id

Answer 1

问题出现在这里：

WHEN Count(patient_id) = 1 THEN diagnosis_grouping 
     WHEN claim_2017 > 0 
          AND claims_3yr = Max(claims_3yr) THEN diagnosis_grouping 
     ELSE NULL

Count(patent_id) 是聚合，而 claim_2017 不是。

所以你的 group by 语句对于这个查询是低效的。您要么需要将 claim_2017 添加到您的组中，要么找到其他解决方案。

Answer 2

考虑两个聚合级别的多个 CTE：

WITH agg1 AS
       (SELECT patient_id, 
               diagnosis_grouping, 
               Sum (CASE 
                      WHEN LEFT(service_date, 4) = '2017' 
                      THEN 1 
                      ELSE 0 
                    END)                       AS CLAIM_2017, 
               Count(DISTINCT claim_id)        AS CLAIMS_3YR, 
               Max(Cast(service_date AS DATE)) AS LATEST_CLAIM 
        FROM   l01_dx_claims A 
               INNER JOIN diagnosis_code_imm B 
                       ON A.diagnosis_code = B.diagnosis_code 
        GROUP  BY 1, 
                  2 
        HAVING Count (DISTINCT claim_id) > 1 
               AND Sum(CASE 
                         WHEN Cast(LEFT(service_date, 4) AS INT) = 2017
                         THEN 1 
                         ELSE 0 
                       END) > 0
      ),
    agg2 AS 
      (SELECT patient_id,
              COUNT(patient_id) AS patient_count,
              MAX(CLAIMS_3YR) AS max_claims_3yr
       FROM agg1
       GROUP BY patient_id)

SELECT a1.*
FROM agg1 a1
INNER JOIN agg2 a2
  ON a1.patient_id = a2.patient_id
  AND a1.CLAIMS_3YR = a2.max_claims_3yr
ORDER BY CAST(a1.patient_id AS INT),
         a1.diagnosis_grouping

Answer 3

你似乎想比较不同层次的聚合，我是这样理解你的逻辑的：

SELECT patient_id, 
       diagnosis_grouping, 
       Sum (CASE 
              WHEN Left(service_date, 4) = '2017' THEN 1 
              ELSE 0 
            END)                       AS CLAIM_2017, 
       Count(DISTINCT claim_id)        AS CLAIMS_3YR, 
       Max(Cast(service_date AS DATE)) AS LATEST_CLAIM 
FROM   l01_dx_claims A 
       INNER JOIN diagnosis_code_imm B 
               ON A.diagnosis_code = B.diagnosis_code 
GROUP  BY 1, 
          2 
HAVING -- Teradata allows using a Select alias in any place, simplified conditions
       CLAIMS_3YR > 1
   AND CLAIM_2017 > 0
QUALIFY -- using Windowed Aggregates to filter the correct result
        Count(*) -- only a single diagnosis_grouping
        Over (PARTITION BY patient_id) = 1
   OR (      -- there was a claim in 2017
        Max(claim_2017)
        Over (PARTITION BY patient_id) > 0
        AND  -- diagnosis_grouping with the higthest count 
        Row_Number()
        Over(PARTITION BY patient_id
             ORDER BY claims_3yr DESC ) = 1
      )

将该组中的单元格值与最大值进行比较，出现错误 SELECT 失败 3707

Compare Cell value with Max in that group getting error SELECT Failed 3707

sql

max

case

teradata

teradata-sql-assistant