加入后另一个表的第二个最低值

Question

我需要帮助来解决这个特定的 sql 问题，我无法编写存储过程，因为我需要将它移植到 Hive。

有两个 tQCles Contr 和 Lvl，我需要将它们左连接并使用前一行的值填充连接的 tQCle 中 LVL 的空值。 我在

中有示例 tQCles

Contr
|      id |     EFF_DT | M_NBR | ACTY_SEQ_NBR | L_CD |
|---------|------------|-------|--------------|------|
| QQFAE46 | 2000-12-24 |    11 |            1 |  POT |
| QQFAE46 | 2000-12-24 |    11 |            2 |  POT |
| QQFAE46 | 2000-12-24 |    11 |            3 |  POT |
| QCC5433 | 2013-04-21 |    00 |            1 |  MIC |
| QCC5433 | 2013-04-21 |    00 |            2 |  MIC |
| QCC614E | 2015-07-18 |    00 |            1 |  MIC |
| QCC614E | 2015-07-18 |    00 |            4 |  MIC |
| QC56DDF | 1999-10-01 |    14 |            2 |  POT |
| QC56DDF | 1999-10-01 |    14 |            3 |  POT |
| QC56DDF | 1999-10-01 |    14 |            4 |  POT |
| ACB3DC2 | 1999-10-01 |    14 |            1 |  POT |

LVL
|      id |     EFF_DT | M_NBR | ACTY_SEQ_NBR | OCCR |
|---------|------------|-------|--------------|------|
| QQFAE46 | 2000-12-24 |    11 |            1 |  100 |
| QQFAE46 | 2000-12-24 |    11 |            3 |  100 |
| QCC5433 | 2013-04-21 |    00 |            2 |  200 |
| QCC614E | 2015-07-18 |    00 |            3 |  200 |
| QC56DDF | 1999-10-01 |    14 |            1 |    0 |

LEFT JOIn of Contr and Lvl

|      id |     EFF_DT | M_NBR | ACTY_SEQ_NBR | L_CD |      id |     EFF_DT |  M_NBR | ACTY_SEQ_NBR |   OCCR |
|---------|------------|-------|--------------|------|---------|------------|--------|--------------|--------|
| QQFAE46 | 2000-12-24 |    11 |            1 |  POT | QQFAE46 | 2000-12-24 |     11 |            1 |    100 |
| QQFAE46 | 2000-12-24 |    11 |            2 |  POT |  (null) |     (null) | (null) |       (null) | (null) |
| QQFAE46 | 2000-12-24 |    11 |            3 |  POT | QQFAE46 | 2000-12-24 |     11 |            3 |    100 |
| QCC5433 | 2013-04-21 |    00 |            1 |  MIC |  (null) |     (null) | (null) |       (null) | (null) |
| QCC5433 | 2013-04-21 |    00 |            2 |  MIC | QCC5433 | 2013-04-21 |     00 |            2 |    200 |
| QCC614E | 2015-07-18 |    00 |            1 |  MIC |  (null) |     (null) | (null) |       (null) | (null) |
| QCC614E | 2015-07-18 |    00 |            4 |  MIC |  (null) |     (null) | (null) |       (null) | (null) |
| QC56DDF | 1999-10-01 |    14 |            2 |  POT |  (null) |     (null) | (null) |       (null) | (null) |
| QC56DDF | 1999-10-01 |    14 |            3 |  POT |  (null) |     (null) | (null) |       (null) | (null) |
| QC56DDF | 1999-10-01 |    14 |            4 |  POT |  (null) |     (null) | (null) |       (null) | (null) |
| ACB3DC2 | 1999-10-01 |    14 |            1 |  POT |  (null) |     (null) | (null) |       (null) | (null) |

现在我需要用值填充来自 LVl tQCle 的空 ACTY_SEQ_NBR 的值。 标准是，从 CONTR（即从加入的 tQCle 的第 4 列）中找到相应的 ACTY_SEQ_NBR，并从 LVL 中找到 ACTY_SEQ_NBR，其中 ACTY_SEQ_NBR 较小 对于相同的 id、eff_dt 和 m_nbr，大于或等于来自 CONTR ACTY_SEQ_NBR 的值。

为了。例如第2行的ACTY_SEQ_NBR为空，对应的Contr ACTY_SEQ_NBR为2，LVL小于2的ACTY_SEQ_NBR值为1。

所以我的理想输出应该是这一行。

|      id |     EFF_DT | M_NBR | ACTY_SEQ_NBR | L_CD |      id |     EFF_DT |  M_NBR | ACTY_SEQ_NBR |   OCCR |
|---------|------------|-------|--------------|------|---------|------------|--------|--------------|--------|
| QQFAE46 | 2000-12-24 |    11 |            1 |  POT | QQFAE46 | 2000-12-24 |     11 |            1 |    100 |
| QQFAE46 | 2000-12-24 |    11 |            2 |  POT |  (null) |     (null) | (null) |            1 | (null) |

I tried a lag query but its not giving correct output for all values.

我执行了 Amniders 查询并更改了结果以获取预期值。 这是我的期望值

|      ID |     EFF_DT | M_NBR | ACTY_SEQ_NBR | L_CD |  LVL_ID | LVL_EFF_DT | LVL_M_NBR | LVL_ACTY_SEQ_NBR |   OCCR | CALC_LVL_ACTY_SEQ_NBR |
|---------|------------|-------|--------------|------|---------|------------|-----------|------------------|--------|-----------------------|
| QQFAE46 | 2000-12-24 |    11 |            1 |  POT | QQFAE46 | 2000-12-24 |        11 |                1 |    100 |                     1 |
| QQFAE46 | 2000-12-24 |    11 |            2 |  POT |  (null) |     (null) |    (null) |           (null) | (null) |                     1 |
| QQFAE46 | 2000-12-24 |    11 |            3 |  POT | QQFAE46 | 2000-12-24 |        11 |                3 |    100 |                     3 |
| QC56DDF | 1999-10-01 |    14 |            2 |  POT |  (null) |     (null) |    (null) |           (null) | (null) |                     1 |
| QC56DDF | 1999-10-01 |    14 |            3 |  POT |  (null) |     (null) |    (null) |           (null) | (null) |                     1 |
| QC56DDF | 1999-10-01 |    14 |            4 |  POT |  (null) |     (null) |    (null) |           (null) | (null) |                     1 |
| QCC5433 | 2013-04-21 |    00 |            1 |  MIC |  (null) |     (null) |    (null) |           (null) | (null) |                   -99 |
| QCC5433 | 2013-04-21 |    00 |            2 |  MIC | QCC5433 | 2013-04-21 |        00 |                2 |    200 |                     2 |
| QCC614E | 2015-07-18 |    00 |            1 |  MIC |  (null) |     (null) |    (null) |           (null) | (null) |                   -99 |
| QCC614E | 2015-07-18 |    00 |            4 |  MIC |  (null) |     (null) |    (null) |           (null) | (null) |                     3 |
| ACB3DC2 | 1999-10-01 |    14 |            1 |  POT |  (null) |     (null) |    (null) |           (null) | (null) |                   -99 |

感谢任何帮助

Answer 1

试试这个：

SELECT
     ID, EFF_DT, M_NBR, ACTY_SEQ_NBR, L_CD, LVL_ID, LVL_EFF_DT, LVL_M_NBR, LVL_ACTY_SEQ_NBR, OCCR
   ,COALESCE(CASE WHEN LVL_ACTY_SEQ_NBR IS NULL THEN COALESCE(LAG(ACTY_SEQ_NBR) OVER (PARTITION BY ID, EFF_DT, M_NBR ORDER BY ACTY_SEQ_NBR),ACTY_SEQ_NBR) ELSE LVL_ACTY_SEQ_NBR END,'NA') LVL_NMBR
    FROM(
    SELECT A.ID, A.EFF_DT, A.M_NBR, A.ACTY_SEQ_NBR, A.L_CD
    , B.ID LVL_ID, B.EFF_DT LVL_EFF_DT, B.M_NBR LVL_M_NBR, B.ACTY_SEQ_NBR LVL_ACTY_SEQ_NBR , B.OCCR
    FROM EDWT.CONTR A
    LEFT JOIN EDWT.LVL B
    ON A.ID = B.ID AND A.ACTY_SEQ_NBR = B.ACTY_SEQ_NBR and a.eff_dt=b.eff_dt and a.m_nbr=b.m_nbr ) A;

另外，请提供以下行的有效值：

ID EFF_DT M_NBR ACTY_SEQ_NBR L_CD LVL_ID LVL_EFF_DT LVL_M_NBR LVL_ACTY_SEQ_NBR OCCR LVL_NMBR

ABC5433 2013-04-21 00 1 IMC 1

对于 ABC614E 和 ABC5433，您期望 lvl_acty_seq_nbr 的数据是什么。 ABC614E 在 lvl 表中没有 acty_Seq_nbr 并且有 1 和 4in 控制。请提供您对上述 id 的预期输出。您想要相同 id 中的最低值还是所有 id 中最低的值？

Answer 2

感谢您提供更广泛的示例！这可能通过一个完整的外部连接来实现，但我认为聚合和过滤会变得混乱。最简单的赌注是左加入到 lvl 两次，第一次找到“previous act_seq_nbr”，然后再次像以前一样（但在没有找到 act_seq_nbr 时使用 coalesce 使用“previous act_seq_nbr”）：

SELECT  c.id,c.eff_dt,c.m_nbr,c.acty_seq_nbr, 
        l.id,l.eff_dt,l.m_nbr,
        coalesce(l.acty_seq_nbr, prev_acty_seq_nbr, -99) l_acty_seq_nbr
from 
    (       
    select c.id,c.eff_dt,c.m_nbr,c.acty_seq_nbr, 
           MAX(L.acty_seq_nbr) prev_acty_seq_nbr
    from   contr c 
    left join lvl l 
        on 
            c.id=l.id 
            and c.eff_dt=l.eff_dt 
            and c.m_nbr=l.m_nbr 
            and c.acty_seq_nbr>l.acty_seq_nbr
    GROUP BY   
        c.id,c.eff_dt,c.m_nbr,c.acty_seq_nbr
    ) c
left join lvl l 
    on 
        c.id=l.id 
        and c.eff_dt=l.eff_dt 
        and c.m_nbr=l.m_nbr     
        and c.acty_seq_nbr=l.acty_seq_nbr;

小提琴： http://www.sqlfiddle.com/#!6/1270f/74/0

结果：

|      id |     eff_dt | m_nbr | acty_seq_nbr |      id |     eff_dt |  m_nbr | l_acty_seq_nbr |
|---------|------------|-------|--------------|---------|------------|--------|----------------|
| AAFAE46 | 2000-12-24 |    11 |            1 | AAFAE46 | 2000-12-24 |     11 |              1 |
| AAFAE46 | 2000-12-24 |    11 |            2 |  (null) |     (null) | (null) |              1 |
| AAFAE46 | 2000-12-24 |    11 |            3 | AAFAE46 | 2000-12-24 |     11 |              3 |
| AB56DDF | 1999-10-01 |    14 |            2 |  (null) |     (null) | (null) |              1 |
| AB56DDF | 1999-10-01 |    14 |            3 |  (null) |     (null) | (null) |              1 |
| AB56DDF | 1999-10-01 |    14 |            4 |  (null) |     (null) | (null) |              1 |
| ABC5433 | 2013-04-21 |    00 |            1 |  (null) |     (null) | (null) |            -99 |
| ABC5433 | 2013-04-21 |    00 |            2 | ABC5433 | 2013-04-21 |     00 |              2 |
| ABC614E | 2015-07-18 |    00 |            1 |  (null) |     (null) | (null) |            -99 |
| ABC614E | 2015-07-18 |    00 |            4 |  (null) |     (null) | (null) |              3 |
| ACB3DC2 | 1999-10-01 |    14 |            1 |  (null) |     (null) | (null) |            -99 |            

Answer 3

我不清楚您希望如何处理您提供的两行之外的示例...如果您可以填写其余数据的空白，那将有所帮助。

与此同时，这里有一个破解：

with foo as (
  select
    c.id as cid, c.eff_dt as c_eff_dt,
    c.m_nbr as c_m_nbr, c.acty_seq_nbr as c_acty_seq_nbr,
    l.id as lid, l.eff_dt as l_eff_dt, l.m_nbr as l_m_nbr,
    l.acty_seq_nbr as l_acty_seq_nbr,
    sum (case when l.id is null then 0 else 1 end)
      over (partition by c.id order by c.acty_seq_nbr) as idx
  from
    contr c
    left join lvl l on
      c.id=l.id and 
      c.eff_dt=l.eff_dt and 
      c.m_nbr=l.m_nbr and 
      c.acty_seq_nbr=l.acty_seq_nbr
)
select
  cid, c_eff_dt, c_m_nbr, c_acty_seq_nbr,
  lid, l_eff_dt, l_m_nbr, l_acty_seq_nbr,
  min (c_acty_seq_nbr) over (partition by cid, idx) as acty_seq_nbr
from foo

Answer 4

我希望我能够理解您的问题。试试这个查询：

 SELECT
 ID, EFF_DT, M_NBR, ACTY_SEQ_NBR, L_CD, LVL_ID, LVL_EFF_DT, LVL_M_NBR, LVL_ACTY_SEQ_NBR, OCCR
 --,LAG(ACTY_SEQ_NMBR) OVER (PARTITION BY ID, EFF_DT, M_NBR ORDER BY ACTY_SEQ_NMBR)
 ,COALESCE(CASE WHEN LVL_ACTY_SEQ_NBR IS NULL THEN LAG(ACTY_SEQ_NBR) OVER (PARTITION BY ID, EFF_DT, M_NBR ORDER BY ACTY_SEQ_NBR) ELSE LVL_ACTY_SEQ_NBR END,'NA') CALC_LVL_ACTY_SEQ_NBR    FROM(
SELECT A.ID, A.EFF_DT, A.M_NBR, A.ACTY_SEQ_NBR, A.L_CD
, B.ID LVL_ID, B.EFF_DT LVL_EFF_DT, B.M_NBR LVL_M_NBR, B.ACTY_SEQ_NBR LVL_ACTY_SEQ_NBR , B.OCCR
FROM EDWT.CONTR A
LEFT JOIN EDWT.LVL B
ON A.ID = B.ID AND A.ACTY_SEQ_NBR = B.ACTY_SEQ_NBR and a.eff_dt=b.eff_dt and a.m_nbr=b.m_nbr ) A;