如何填写postgreSQL中的所有空值直到下一个非空单元格？

Question

我在 postgres 中有一个表“orders”，其中包含“offer_id”和“date”列。我需要编写一个代码，该代码必须使用以前的结果填充“offer_id”列中的所有空案例。我尝试使用“lag”函数，但问题是它只能管理一个 NULL，而不是多个。如何解决这个问题？

图片在这里

我试过这个

with orders as (
                  select 2 as offer_id, '2021-01-01'::date as date union all
                  select 3 as offer_id, '2021-01-02'::date as date union all
                  select null as offer_id, '2021-01-03'::date as date union all
                  select null as offer_id, '2021-01-04'::date as date union all
                  select null as offer_id, '2021-01-05'::date as date union all
                  select 4 as offer_id, '2021-01-07'::date as date union all
                  select 5 as offer_id, '2021-01-08'::date as date union all
                  select null as offer_id, '2021-01-09'::date as date union all
                  select 8 as offer_id, '2021-01-10'::date as date union all
                  select 9 as offer_id, '2021-01-11'::date as date union all
                  select null as offer_id, '2021-01-12'::date as date union all
                  select null as offer_id, '2021-01-13'::date as date union all
                  select 13 as offer_id, '2021-01-14'::date as date union all
                  select 13 as offer_id, '2021-01-15'::date as date union all
                  select null as offer_id, '2021-01-16'::date as date 
                  )
select *, CASE WHEN offer_id IS NULL 
THEN LAG(offer_id) OVER (ORDER BY date) ELSE offer_id END updated_offer_id 
from orders

Answer 1

可以通过形成组的子查询来解决：

SELECT offer_id, date
     , first_value(offer_id) OVER (PARTITION BY grp ORDER BY date) AS updated_offer_id
FROM  (
   SELECT *, count(offer_id) OVER (ORDER BY date) AS grp
   FROM   orders o
   ORDER  BY date
   ) sub;

小提琴

由于

count()

只计算非空值，因此所有包含 null 的行都属于具有前一个非空值的一组。
在外部

SELECT

中，只需为所有人选择第一个组成员的值即可。