oracle如何从重复记录中获取唯一行？答案

【问题标题】：How to get uniq rows from duplicate records in oracle?oracle如何从重复记录中获取唯一行？
【发布时间】：2019-03-20 00:05:54
【问题描述】：

我想从表中为每个 bucket_type 获取最快的 quotaBasedService 行。 packageSpeed 的值在 AdditionalQuota 表中，package_id 和其他字段在 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

这里是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

我的期望就在这里

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

【问题讨论】：

标签： sql oracle subquery inner-join

【解决方案1】：

我认为您只需从group by 和select 中删除id：

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;

如果你想要id，那么你需要一个聚合函数，可能是MIN()。

【讨论】：

【解决方案2】：

我认为你不需要在GROUP BY 子句中使用q.ID：

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;

如果你想要q.ID，然后使用min()/max() 聚合它：

【讨论】：

【解决方案3】：

或者设置 MAX ID 这样，你也会得到 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

【讨论】：

祝福你！