用以前的记录填充 NULLS - Netezza SQL

Question

我在 Aginity Workbench 上使用 Netezza SQL 并且有以下数据：

id           DATE1              DATE2
1            2013-07-27         NULL
2            NULL               NULL
3            NULL               2013-08-02
4            2013-09-10         2013-09-23
5            2013-12-11         NULL
6            NULL               2013-12-19

我需要用 DATE1 字段中已填充的先前值填充 DATE1 中的所有 NULL 值。对于 DATE2，我需要执行相同的操作，但顺序相反。所以我想要的输出如下：

id           DATE1              DATE2
1            2013-07-27         2013-08-02
2            2013-07-27         2013-08-02
3            2013-07-27         2013-08-02
4            2013-09-10         2013-09-23
5            2013-12-11         2013-12-19
6            2013-12-11         2013-12-19

我只有数据的读取权限。所以创建表或视图是不可能的

Answer 1

我认为 Netezza 支持 max() 和 min() 的 order by 子句。所以，你可以这样做：

select max(date1) over (order by date1) as date1,
       min(date2) over (order by date2 desc) as date2
 . . .

编辑：

在 Netezza 中，您可以使用 last_value() 和 first_value():

执行此操作

select last_value(date1 ignore nulls) over (order by id rows between unbounded preceding and 1 preceding) as date1,
       first_value(date1 ignore nulls) over (order by id rows between 1 following and unbounded following) as date2

Netezza 似乎不支持 LAG() 上的 IGNORE NULLs，但支持这些功能。

Answer 2

我只在 Oracle 中测试过，希望它能在 Netezza 中运行：

Fiddle: http://www.sqlfiddle.com/#!4/7533f/1/0

select id,
       coalesce(date1, t1_date1, t2_date1) as date1,
       coalesce(date2, t3_date2, t4_date2) as date2
  from (select t.*,
               t1.date1 as t1_date1,
               t2.date1 as t2_date1,
               t3.date2 as t3_date2,
               t4.date2 as t4_date2,
               row_number() over(partition by t.id order by t.id) as rn
          from tbl t
          left join tbl t1
            on t1.id < t.id
           and t1.date1 is not null
          left join tbl t2
            on t2.id > t.id
           and t2.date1 is not null
          left join tbl t3
            on t3.id < t.id
           and t3.date2 is not null
          left join tbl t4
            on t4.id > t.id
           and t4.date2 is not null
         order by t.id) x
 where rn = 1

Answer 3

这是一种使用自联接使用最近的 min/max 非空日期填充 NULL 日期的方法。此查询应该适用于大多数数据库

select t1.id, max(t2.date1), min(t3.date2)
from tbl t1
join tbl t2 on t1.id >= t2.id
join tbl t3 on t1.id <= t3.id
group by t1.id

http://www.sqlfiddle.com/#!4/acc997/2

Answer 4

this怎么样？

select
  id
  ,last_value(date1 ignore nulls) over (
    order by id
    rows between unbounded preceding and current row
  ) date1
  ,first_value(date2 ignore nulls) over (
    order by id
    rows between current row and unbounded following
  ) date2

您也可以手动计算，而不是依赖窗口函数。

with chain as (
  select 
    this.*,
    prev.date1 prev_date1,
    case when prev.date1 is not null then abs(this.id - prev.id) else null end prev_distance,
    next.date2 next_date2,
    case when next.date2 is not null then abs(this.id - next.id) else null end next_distance
  from 
    Table1 this 
    left outer join Table1 prev on this.id >= prev.id
    left outer join Table1 next on this.id <= next.id
), min_distance as (
  select
    id,
    min(prev_distance) min_prev_distance,
    min(next_distance) min_next_distance
  from
    chain
  group by
    id
)
select
  chain.id,
  chain.prev_date1,
  chain.next_date2
from
  chain
  join min_distance on 
    min_distance.id = chain.id
    and chain.prev_distance = min_distance.min_prev_distance
    and chain.next_distance = min_distance.min_next_distance
order by chain.id

如果您无法通过减法计算 ID 之间的距离，只需将排序方案替换为 row_number() 调用即可。

用以前的记录填充 NULLS - Netezza SQL

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