Question

我想从表中为每个bucket_type获取最快的quotaBasedService行。 packageSpeed值位于带有package_id的AdditionalQuota表中，其他字段位于quotaBased表中。

SELECT
      q.ID,
      q.SUBSCRIBER_IDENTIFIER,
      q.BUCKET_TYPE,
      MAX(CASE WHEN ext.PACKAGE_SPEED = 0 THEN 999999 ELSE ext.PACKAGE_SPEED END) AS PACKAGE_SPEED
    FROM PIMS_QUOTA_BASED_SERVICE q
      INNER JOIN PIMS_ADDITIONAL_QUOTA_PACKAGE ext ON ext.PACKAGE_ID = q.ADDITIONAL_QUOTA_PACKAGE_ID
    WHERE q.PERIOD = '201810' AND q.BUCKET_TYPE IS NOT NULL
    GROUP BY
      q.SUBSCRIBER_IDENTIFIER,
      q.BUCKET_TYPE,
      q.ID
    ORDER BY SUBSCRIBER_IDENTIFIER, BUCKET_TYPE, PACKAGE_SPEED DESC

这是sql结果 quotaBasedId，subs_identifier，bucket_type，速度

108337,subscriber1@domain,Extra,999999
108345,subscriber1@domain,Extra,5120
108334,subscriber1@domain,Extra1,999999
108342,subscriber1@domain,Extra1,5120
108336,subscriber1@domain,Extra2,999999
108344,subscriber1@domain,Extra2,5120
108335,subscriber1@domain,Extra4,999999
108343,subscriber1@domain,Extra4,5120
108253,subscriber2@domain,Extra,999999
108261,subscriber2@domain,Extra,5120
108250,subscriber2@domain,Extra1,999999
108258,subscriber2@domain,Extra1,5120
108252,subscriber2@domain,Extra2,999999
108260,subscriber2@domain,Extra2,5120
108251,subscriber2@domain,Extra4,999999
108259,subscriber2@domain,Extra4,5120

我的期望在这里

108337,subscriber1@domain,Extra,999999
108334,subscriber1@domain,Extra1,999999
108336,subscriber1@domain,Extra2,999999
108335,subscriber1@domain,Extra4,999999
108253,subscriber2@domain,Extra,999999
108250,subscriber2@domain,Extra1,999999
108252,subscriber2@domain,Extra2,999999
108251,subscriber2@domain,Extra4,999999

Answer 1

我认为您只需要从id和group by中删除select：

SELECT q.SUBSCRIBER_IDENTIFIER, q.BUCKET_TYPE,
       MAX(CASE WHEN ext.PACKAGE_SPEED = 0 THEN 999999 ELSE ext.PACKAGE_SPEED END) AS PACKAGE_SPEED
FROM PIMS_QUOTA_BASED_SERVICE q INNER JOIN
     PIMS_ADDITIONAL_QUOTA_PACKAGE ext
     ON ext.PACKAGE_ID = q.ADDITIONAL_QUOTA_PACKAGE_ID
WHERE q.PERIOD = '201810' AND q.BUCKET_TYPE IS NOT NULL
GROUP BY q.SUBSCRIBER_IDENTIFIER, q.BUCKET_TYPE
ORDER BY SUBSCRIBER_IDENTIFIER, BUCKET_TYPE;

如果要使用id，则需要一个聚合函数，可能是MIN()。

Answer 2

我认为您无需在q.ID子句中使用GROUP BY：

SELECT q.SUBSCRIBER_IDENTIFIER, q.BUCKET_TYPE,
       MAX(CASE WHEN ext.PACKAGE_SPEED = 0 THEN 999999 ELSE ext.PACKAGE_SPEED END) AS PACKAGE_SPEED
FROM PIMS_QUOTA_BASED_SERVICE q INNER JOIN
     PIMS_ADDITIONAL_QUOTA_PACKAGE ext
     ON ext.PACKAGE_ID = q.ADDITIONAL_QUOTA_PACKAGE_ID
WHERE q.PERIOD = '201810' AND q.BUCKET_TYPE IS NOT NULL
GROUP BY q.SUBSCRIBER_IDENTIFIER, q.BUCKET_TYPE
ORDER BY SUBSCRIBER_IDENTIFIER, BUCKET_TYPE;

如果要使用q.ID，请使用min()/max()将其聚合：

Answer 3

或设置MAX ID，这样您也将获得ID

SELECT
      MAX(q.ID),
      q.SUBSCRIBER_IDENTIFIER,
      q.BUCKET_TYPE,
      MAX(CASE WHEN ext.PACKAGE_SPEED = 0 THEN 999999 ELSE ext.PACKAGE_SPEED END) AS PACKAGE_SPEED
    FROM PIMS_QUOTA_BASED_SERVICE q
      INNER JOIN PIMS_ADDITIONAL_QUOTA_PACKAGE ext ON ext.PACKAGE_ID = q.ADDITIONAL_QUOTA_PACKAGE_ID
    WHERE q.PERIOD = '201810' AND q.BUCKET_TYPE IS NOT NULL
    GROUP BY
      q.SUBSCRIBER_IDENTIFIER,
      q.BUCKET_TYPE,
    ORDER BY SUBSCRIBER_IDENTIFIER, BUCKET_TYPE, PACKAGE_SPEED DESC

如何从Oracle中的重复记录获取uniq行？

3 个答案: