在 Redshift 中从 bigquery 实现以下内容

Question

尝试使用

listagg()

但结果如下：

1 a,b,c
1 a,b,c
1 a,b,c
2 MMM

需要与图像bigquery类似的实现：

with data 1 as (
select
  1 sk,
  'a' st,
  timestamp('1900-05-08 04:00:00') dt
union all
select
  1 sk,
  'b',
  timestamp('1901-05-08 04:00:00')
union all
select
  1 sk,
  'c',
  timestamp('1902-05-08 04:00:00')
union all
select
  2 sk,
  'MMM',
  timestamp('1902-05-08 04:00:00')
)

select
  sk,
    string_agg(st, ',') over (partition by sk order by dt)
from data1

输出：

sk  f0_
 1  a
 1  a,b
 1  a,b,c
 2  MMM

通过连接值与运行总计不同的地方查找上述内容

Answer 1

正如您可能已经读过的那样，Redshift 的 listagg() 窗口函数不支持框架子句。不过，我们可以通过绕道至 SUPER 数据类型来模拟这一点。

下面的代码创建测试数据，执行 listagg()，将这些字符串转换为超级数组，并使用 row_number() 窗口捕获分区内字符串的顺序。 Redshift 有一个 subarray() 函数，允许仅选择超级数组的部分。删除括号和引号即可产生所需的结果。

with data as (
select
  1 sk,
  'a' st,
  '1900-05-08 04:00:00'::timestamp dt
union all
select
  1 sk,
  'b',
  '1901-05-08 04:00:00'::timestamp
union all
select
  1 sk,
  'c',
  '1902-05-08 04:00:00'::timestamp
union all
select
  2 sk,
  'MMM',
  '1902-05-08 04:00:00'::timestamp
),
arrayed as (
select sk, 
  json_parse('["'||listagg(st, '","') within group (order by dt) over (PARTITION by sk)||'"]') as arr,
  row_number() over (partition by sk order by dt)::int as rn
from data
)
select sk, translate(json_serialize(subarray(arr,0,rn)),'[]"','') as foo
from arrayed
order by sk, rn;