在你的情况下，有一种方法可以做到这一点。那是一个 union all /窗口函数方法。我想这正是你想要的：

with t as (
      select a.nbr2, b.nbr, b.dt, null as end_date, null as field, b.qty
      from A join
           B
           on A.dt = B.dt
      union all
      select d.nrb2, d.nbr, d.start_date, d.end_date, d.field, null
      from D
    )
select dt, (case when dt < d_end_date then d_field end), sum(qty)
from (select t.*, 
             last_value(field, true) over (partition by nbr, nbr2 order by dt) as d_field,
             last_value(end_date, true) over (partition by nbr, nbr2 order by dt) as d_end_date
      from t
     ) t
group by dt, dt, (case when dt < d_end_date then d_field end);

我不是100%确定这是完全相同的——例如，这假设d中最多有一个匹配的记录，并且没有重叠。但其思想是将这些值交错并使用窗口函数作为 last_value() 与 ignore null s选项来获取正确的值。