Question

我一直在“定义不明确的列”。

WITH CountData as (
  SELECT BOARD_RATE_T1,
         BOARD_RATE_T3,
         RENTAL_START_DATE,
         RENTAL_END_DATE,
         (BOARD_RATE_T3 - BOARD_RATE_T1) AS Countt
  FROM PPSS.PA_OA_RATE_JUST
),
GroupCount AS
(
  SELECT BOARD_RATE_T1,
         BOARD_RATE_T3,
         Countt,
         CASE
             WHEN Countt < 500 THEN 'Under 500'
             WHEN Countt BETWEEN 500 AND 5000 THEN '500 - 5000'
             WHEN Countt BETWEEN 5000 AND 10000 THEN '5000 - 10000'
             WHEN Countt > 10000 THEN 'Over 10000'
             ELSE 'Invalid Count'
         END AS Countt
  FROM CountData
)
SELECT COUNT(*) AS CountGrpCount,
      Countt Groups
FROM GroupCount;

Answer 1

最后一个查询引发错误，因为您引用的是Countt，并且在GroupCount中定义了两次。 Oracle不知道您的意思是什么。

一旦克服了模棱两可的列错误，您将无法进行最终查询。 Oracle严格要求对聚合查询中的每个非聚合值都使用GROUP BY，这与MySQL有所不同。有一些例外，例如常量和诸如SYSDATE之类的函数，但这超出了此问题的范围。

如果您要按“低于500”，“ 500-5000”等进行计数，那么以下是我注意到的有关查询的几件事：

您不需要BOARD_RATE_T1，BOARD_RATE_T2，RENTAL_START_DATE和RENTAL_END_DATE列。
可以通过稍微更改逻辑来缩短“费率差异”类别CASE语句，并减少出错的可能性。
在ELSE语句中达到CASE条件的唯一方法是BOARD_RATE_T1和/或BOARD_RATE_T3为空。如果它们都是必需的（NOT NULL）值，则可以省去ELSE。

此查询将给出计数。如果您确实需要其他价值观，并且为了使问题更短而乱砍，我深表歉意：）

WITH CountData AS (
  SELECT BOARD_RATE_T3 - BOARD_RATE_T1 AS RateDiff
  FROM PPSS.PA_OA_RATE_JUST
),
GroupCount AS (
  SELECT
    CASE
      WHEN RateDiff < 500 THEN 'Under 500'
      WHEN RateDiff < 5000 THEN '500 - 5000'
      WHEN RateDiff < 10000 THEN '5000 - 10000'
      ELSE 'Invalid Count'
    END AS RateDiffCategory
  FROM CountData
)
SELECT
  RateDiffCategory,
  COUNT(*)
FROM GroupCount
GROUP BY
  RateDiffCategory
ORDER BY
  RateDiffCategory

不确定为什么说“列定义不明确”

1 个答案: