PostgreSQL last_value 忽略空值

PostgreSQL last_value ignore nulls

我知道有人问过这个问题,但为什么下面的解决方案不起作用?我想用 idx.

排序的最后一个非空值填充 value

我看到的:

 idx | coalesce 
-----+----------
   1 |        2
   2 |        4
   3 |         
   4 |         
   5 |       10
(5 rows)

我想要的:

 idx | coalesce 
-----+----------
   1 |        2
   2 |        4
   3 |        4 
   4 |        4 
   5 |       10
(5 rows)

代码:

with base as (

    select 1    as idx
         , 2    as value

    union

    select 2    as idx
         , 4    as value

    union

    select 3    as idx
         , null as value

    union

    select 4    as idx
         , null as value

    union

    select 5    as idx
         , 10   as value
)

select idx
     , coalesce(value
              , last_value(value) over (order by case when value is null then -1
                                                 else idx
                                                 end))
from base
order by idx

要了解您的解决方案为何不起作用,如果您按照 window 框架中的顺序进行排序,只需查看输出:

with base as (
    select 1    as idx
         , 2    as value
    union
    select 2    as idx
         , 4    as value
    union
    select 3    as idx
         , null as value
    union
    select 4    as idx
         , null as value
    union
    select 5    as idx
         , 10   as value
)
select idx, value from base
order by case when value is null then -1
                                                 else idx
                                                 end;
 idx | value
-----+-------
   3 |
   4 |
   1 |     2
   2 |     4
   5 |    10

last_value() window 函数将选择当前帧中的最后一个值。在不更改任何框架默认值的情况下,这将是当前行。

你要的是lag(ignore nulls)。这是一种使用两个 window 函数来做你想做的事情的方法。第一个定义 NULL 值的分组,第二个分配值:

select idx, value, coalesce(value, max(value) over (partition by grp))
from (select b.*, count(value) over (order by idx) as grp
      from base b
     ) b
order by idx;

您也可以使用数组在没有子查询的情况下执行此操作。基本上,取最后一个不计算 NULLs:

的元素
select idx, value, 
       (array_remove(array_agg(value) over (order by idx), null))[count(value) over (order by idx)]
from base b
order by idx;

Here 是一个 db<>fiddle.

嗯,这里的 last_value 对我来说没有意义,除非你能指出给我。查看示例,您需要最后一个非值,您可以通过以下方式获取它: 我正在用空值和先前的非空值组成一个组,以便我可以获得第一个非值。

with base as (
select 1    as idx , 2    as value   union
select 2    as idx, -14    as value    union
select 3    as idx , null as value   union
select 4    as idx , null as value   union
select 5    as idx , 1   as value
)
Select idx,value,
first_value(value) Over(partition by rn) as new_val
from(
select idx,value
    ,sum(case when value is not null then 1 end) over (order by idx) as rn
  from   base
) t

这是代码

http://sqlfiddle.com/#!15/fcda4/2