我需要帮助才能解决这个特定的sql问题，我无法编写存储过程，因为我需要将其移植到Hive。
有两个tQCles Contr和Lvl，我需要将它们连接起来，并使用前一行的值从连接的tQCle中填充LVL的空值。 我有
中的示例tQClesContr
| 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LVl tQCle填充值为空的ACTY_SEQ_NBR的值。 标准是，从CONTR中找到相应的ACTY_SEQ_NBR（即从连接的tQCle的第4列），并从LVL中找到ACTY_SEQ_NBR，其中ACTY_SEQ_NBR较少 大于或等于CONTR ACTY_SEQ_NBR对同一个id，eff_dt和m_nbr的值。
有关。例如，行＃2的空ACTY_SEQ_NBR。对应的Contr ACTY_SEQ_NBR为2，LVL的小于2的ACTY_SEQ_NBR值为1.
所以我理想的输出应该就像这一行。
| 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，我更改了结果以获得预期值。 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
答案 0 :(得分：1)
感谢更广泛的例子！这可能是单个完全外连接可以做到的，但我认为聚合和过滤会变得混乱。最简单的选择是连接到lvl两次，首先找到＆＃34;之前的act_seq_nbr＆＃34;，然后再次像你一样（但是使用coalesce来使用＆＃34;之前的act_seq_nbr＆＃34;当没有act_seq_nbr是实测值）：
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 |
答案 1 :(得分：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，请帮助处理以下行的有效值：
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
您对ABC614E＆amp; amp; lvl_acty_seq_nbr的期望是什么？ ABC5433。 ABC614E在lvl表中没有acty_Seq_nbr，并且有1＆amp; 4in控制请提供上述ID的预期输出。您想要所有ID中相同ID或最低值的最低值吗？
答案 2 :(得分：0)
我希望我能理解你的问题。试试这个问题：
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;
答案 3 :(得分：0)
我不清楚你想如何处理你提供的两行以外的例子......如果你可以填补其余数据的空白，那会有所帮助。
与此同时，这是一个裂缝：
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