具有多个条件的Oracle SQL连接子查询

Question

编辑：在每个代码段下添加了示例结果，并在最后添加了所需的结果。

我一直在浏览几篇文章，但似乎找不到能够准确应用的解决方案。我有一个表r，其中包含acctnbr，然后是modnbr的数据。对于大多数acctnbrs而言，只有一条记录，而得到的modnbr为1，但是最多可以有5条记录，其中modnbrs为1-5。

表u没有modnbr，因此我想在acctnbr上进行左联接，只返回表u中的值，最大值为{{1} } modnbr中的acctnbr}。

这是我的代码，未尝试连接。我需要在连接中包括另一个限定符，如下所示：

r

不出所料，无论select r.acctnbr, r.modnbr, r.modtypcd, r.moddate, u.userfieldcd, u.value from loanrestructuretracking r LEFT JOIN acctuserfield u on u.acctnbr = r.acctnbr and u.userfieldcd = 'LOMO' order by r.moddate desc 为何，每个u.value都得到相同的r.acctnbr。结果如下：

modnbr

现在，这是我尝试查找最大值+---------+--------+----------+------------+-------------+-------+ | acctnbr | modnbr | modtypcd | moddate | userfieldcd | value | +---------+--------+----------+------------+-------------+-------+ | 1122330 | 1 | LM | 1/1/2018 | LOMO | DLQ | | 1223340 | 1 | LM | 2/5/2018 | LOMO | DLQ | | 2233440 | 1 | LM | 6/9/2017 | LOMO | COT | | 2233440 | 2 | LM | 11/20/2017 | LOMO | COT | | 2233440 | 3 | LM | 3/12/2018 | LOMO | COT | | 3344550 | 1 | LM | 4/27/2018 | LOMO | EXT | | 3344550 | 2 | LM | 6/2/2018 | LOMO | EXT | +---------+--------+----------+------------+-------------+-------+ 的代码，但现在我只看到modnbr的最大值u.value的任何实例（在我的情况下， 5。）

modnbr

这是上面脚本的结果。请注意，记录中只有一个select r.acctnbr, r.modnbr, r.modtypcd, r.moddate, u.userfieldcd, u.value from loanrestructuretracking r LEFT JOIN acctuserfield u on u.acctnbr = r.acctnbr and u.userfieldcd = 'LOMO' and r.modnbr = (select max(modnbr) from loanrestructuretracking) order by r.moddate desc ，最大值为u.value：

modnbr

最后，这是期望的结果-每个+---------+--------+----------+------------+-------------+-------+ | acctnbr | modnbr | modtypcd | moddate | userfieldcd | value | +---------+--------+----------+------------+-------------+-------+ | 1122330 | 1 | LM | 1/1/2018 | LOMO | | | 1223340 | 1 | LM | 2/5/2018 | LOMO | | | 2233440 | 1 | LM | 6/9/2017 | LOMO | | | 2233440 | 2 | LM | 11/20/2017 | LOMO | | | 2233440 | 3 | LM | 3/12/2018 | LOMO | COT | | 3344550 | 1 | LM | 4/27/2018 | LOMO | | | 3344550 | 2 | LM | 6/2/2018 | LOMO | | +---------+--------+----------+------------+-------------+-------+ 的最大值modnbr的u.value，而不是该{{1}的最大值acctnbr的任何记录的null }}：

modnbr

我要怎么做才能使每个acctnbr的最大+---------+--------+----------+------------+-------------+-------+ | acctnbr | modnbr | modtypcd | moddate | userfieldcd | value | +---------+--------+----------+------------+-------------+-------+ | 1122330 | 1 | LM | 1/1/2018 | LOMO | DLQ | | 1223340 | 1 | LM | 2/5/2018 | LOMO | DLQ | | 2233440 | 1 | LM | 6/9/2017 | LOMO | | | 2233440 | 2 | LM | 11/20/2017 | LOMO | | | 2233440 | 3 | LM | 3/12/2018 | LOMO | COT | | 3344550 | 1 | LM | 4/27/2018 | LOMO | | | 3344550 | 2 | LM | 6/2/2018 | LOMO | EXT | +---------+--------+----------+------------+-------------+-------+ 上的联接起作用？ 谢谢！

Answer 1

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没有数据，所以不能百分百确定，但是我认为这将为您提供所需的信息。

max（r.modnbr）超过（由r.acctnbr划分），因为max_modnbr将每个acctnbr的最大modnbr追加到每一行。

然后case语句在不匹配的行上将其替换为null。

可以将整个内容写得更紧凑：

with base_qry as
(
  select
  r.acctnbr,
  r.modnbr,
  r.modtypcd,
  r.moddate,
  u.userfieldcd,
  u.value,
  max(r.modnbr ) over (partition by r.acctnbr) as max_modnbr

  from loanrestructuretracking r 

  LEFT JOIN acctuserfield u on u.acctnbr = r.acctnbr 
  and u.userfieldcd = 'LOMO'
)
select 
  acctnbr,
  modnbr,
  modtypcd,
  moddate,
  userfieldcd,
  case when modnbr = max_modnbr then value else null end as value
from base_query
order by moddate desc

Answer 2

我想你想要

select . . .
from (select r.*, row_number() over (partition by r.acctnbr order by r.modnbr desc) as seqnum
      from loanrestructuretracking r
     ) r left join 
     acctuserfield u
     on u.acctnbr = r.acctnbr and
        u.userfieldcd = 'LOMO'
where seqnum = 1
order by r.moddate desc;