Question

我想根据最新的“ APP_START_DT”返回每人1行。

我想确保它查看MAX（APP_START_DT），然后跟踪该人在该行上的所有日期，而忽略该行上的其他行和日期。

This is what I have tried:
select distinct ap1.ID,
MAX(ap1.APP_START_DT) over (PARTITION by ap1.ID) APP_START_DT,
ap1.APP_SUBMIT_DT
from table1 ap1
where ap1.ID in ('444','555');

Getting 2 rows for both of these ids.



My Table:
ID   NAME    Acct_CR_DT    App_Ap    APP_START_DT    APP_SUBMIT_DT
444  ABC1    9/5/2018      My univ   9/5/2018        9/14/2018
444  ABC1    9/5/2018      {null}    {null}          {null}
555  ABC2    1/9/2019      {null}    {null}          {null}
555  ABC2    1/9/2019      Univ4     1/9/2019        1/9/2019




Desired Results:   
ID     NAME   Acct_CR_DT  App_Ap   APP_START_DT  APP_SUBMIT_DT App_stat
444  ABC1     9/5/2018    My univ  9/5/2018      9/14/2018     Submt
555  ABC2     1/9/2019    Univ4    1/9/2019      1/9/2019      Submt

Answer 1

按ap1.ID分组可以让您在获得MAX APP_START_DT的同时，每个ID获得一行

SELECT distinct ap1.ID, MAX(ap1.APP_START_DT)
FROM table1 ap1
WHERE ap1.ID in ('444','555')
GROUP BY ap1.ID ;

Answer 2

您可以应用null处理功能以获得所需的结果。

select distinct ap1.ID,
MAX(ap1.APP_START_DT) over (PARTITION by ap1.ID) APP_START_DT,
ap1.APP_SUBMIT_DT
from (select COALESCE(APP_START_DT, 'lowest_date') as APP_START_DT, ID, APP_SUBMIT_ID from table1 where ID in ('444','555')) ap1;

Answer 3

关于FIRST_VALUE / LAST_VALUE（以及根据用户ID划分窗口？）：

SELECT DISTINCT
    ID, NAME,
    LAST_VALUE(Acct_CR_DT)    OVER(PARTITION BY ID ORDER BY APP_START_DT NULLS FIRST RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING),
    LAST_VALUE(App_Ap)        OVER(PARTITION BY ID ORDER BY APP_START_DT NULLS FIRST RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING),
    LAST_VALUE(APP_START_DT)  OVER(PARTITION BY ID ORDER BY APP_START_DT NULLS FIRST RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING),
    LAST_VALUE(APP_SUBMIT_DT) OVER(PARTITION BY ID ORDER BY APP_START_DT NULLS FIRST RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)
  FROM table1 ap1

对于每一行，我们为每个用户ID /名称对找到每一列的“最后一个值”。我们需要DISTINCT来删除重复项，因为此“最后一个值”是为每一行（不仅是每个组中的最后一个）计算的。

如果可行，如果可以保证，将只有APP_START_DT = MAX（APP_START_DT）OVER（PARTITION BY ID）的行，或者如果存在更多此类行，则它们至少与所选列保持一致。如果您不能保证，我会考虑在ORDER BY子句中添加更多列（该行实际上应该是第一行）。

Answer 4

select * 
FROM table1
where (id, app_start_dt) IN (
    SELECT ap1.ID, MAX(ap1.APP_START_DT)
    FROM table1 ap1
    WHERE ap1.ID in ('444','555')
    GROUP BY ap1.ID 
    )   ;

哈维尔很近。如果您使用他的查询作为子选择，我相信它会起作用。很高兴知道数据是什么样子，以及在任何给定日期是否存在重复项，因为这样做会破坏这种方法。

Answer 5

SELECT *
FROM
(
    SELECT *
    FROM mytable
    WHERE id IN(444, 555)
) T1
INNER JOIN
(
    SELECT id, 
           MAX(APP_START_DT) APP_START_DT
    FROM mytable
    WHERE id IN(444, 555)
    GROUP BY ID
) T2 ON T1.id = T2.ID
        AND T1.APP_START_DT = T2.APP_START_DT;

每人返回1行

5 个答案: