如何在 oracle select 中按等级或第二取第一个值

Question

我使用rank() OVER (PARTITION BY ? ORDER BY ? ASC)函数进行排名。它工作正常。但是如果特定 application_id 的最大 rnk 值为 1，我需要取 1st falue 否则 2nd 一个。

TABLE是这样的

  APPLICATION_ID    |  EMP_CODE   | DATE_ACTION   | ACTION_ID
----------------------------------------------------------------
    24744           |  110147     | 2016-09-13    | -1
    24746           |  162286     | 2016-09-12    |  0 

我的Select如下：

SELECT APPLICATION_ID,EMP_CODE,DATE_ACTION,ACTION_ID,
           rank() OVER (PARTITION BY APPLICATION_ID
                  ORDER BY DATE_ACTION ASC) as rnk
FROM TABLE WHERE ACTION_ID = -1

结果是：

RNK  | APPLICATION_ID    |  EMP_CODE   | DATE_ACTION   | ACTION_ID
------------------------------------------------------------
1    | 53207             | 260829      | 2020-03-03    | -1
2    | 53207             | 161938      | 2020-03-03    | -1
3    | 53207             | 161938      | 2020-03-03    | -1
1    | 51372             | 163668      | 2020-03-04    | -1
1    | 52819             | 260829      | 2020-03-02    | -1
2    | 52819             | 161938      | 2020-03-02    | -1
3    | 52819             | 161938      | 2020-03-02    | -1
4    | 52819             | 161938      | 2020-03-02    | -1

但结果应该是这样的：

RNK  | APPLICATION_ID    |  EMP_CODE   | DATE_ACTION   | ACTION_ID
------------------------------------------------------------
2    | 53207             | 161938      | 2020-03-03    | -1
1    | 51372             | 163668      | 2020-03-04    | -1
2    | 52819             | 161938      | 2020-03-02    | -1

Answer 1

你也可以像这样使用COUNT()解析函数：

SELECT t.APPLICATION_ID, t.EMP_CODE, t.DATE_ACTION, t.ACTION_ID
FROM (
  SELECT APPLICATION_ID,EMP_CODE,DATE_ACTION,ACTION_ID,
    RANK() OVER (PARTITION BY APPLICATION_ID ORDER BY DATE_ACTION DESC) as rnk,
    COUNT(*) OVER (PARTITION BY APPLICATION_ID) counter
  FROM tablename WHERE ACTION_ID = -1
) t
WHERE t.rnk = CASE WHEN t.counter = 1 THEN 1 ELSE 2 END