0

I am trying to form a transformation where I need to concatenate each 10 row value.

1st step: Table input (query from Postgres DB: select id from tablename)

sample result from the above query:

                  id
 00000191-555c-11e4-922d-29fb57a42e4c
 00000192-555c-11e4-922d-29fb57a42e4c
 00000193-555c-11e4-922d-29fb57a42e4c
 00000194-555c-11e4-922d-29fb57a42e4c
 00000195-555c-11e4-922d-29fb57a42e4c
 00000196-555c-11e4-922d-29fb57a42e4c
 00000197-555c-11e4-922d-29fb57a42e4c
 00000198-555c-11e4-922d-29fb57a42e4c
 00000199-555c-11e4-922d-29fb57a42e4c
 0000019a-555c-11e4-922d-29fb57a42e4c
 000001a3-3cf2-11e4-b398-e52ee0ec6a4c
 000002ad-3768-4242-88cf-96f27d0263af
 000003ea-26e3-11e4-ace7-15c7d609fa6e
 00000684-73fb-4d65-a502-87c4eb6607c1
 0000087a-f587-44fa-8e88-7bcae5bcb22c
 00000889-39c5-11e4-bd0e-c3f9d65ac856
 0000094c-be98-4456-8b49-6357a36581aa
 00000987-2f19-4574-ab85-6744a65ee4e3
 00000cd0-4097-11e4-a4e6-af71a3d902c0
 00000e1e-3b55-11e4-9897-d958d55e6784

here I have to concat each 10 rows ids into a single row. Eg. 1-10 row ids in a single row, 11-20 rows ids in another row and so on.

Expected Output:

ids

00000191-555c-11e4-922d-29fb57a42e4c,00000192-555c-11e4-922d-29fb57a42e4c,00000193-555c-11e4-922d-29fb57a42e4c,00000194-555c-11e4-922d-29fb57a42e4c,00000195-555c-11e4-922d-29fb57a42e4c,00000196-555c-11e4-922d-29fb57a42e4c,00000197-555c-11e4-922d-29fb57a42e4c,00000198-555c-11e4-922d-29fb57a42e4c,00000199-555c-11e4-922d-29fb57a42e4c,0000019a-555c-11e4-922d-29fb57a42e4c
000001a3-3cf2-11e4-b398-e52ee0ec6a4c,000002ad-3768-4242-88cf-96f27d0263af,000003ea-26e3-11e4-ace7-15c7d609fa6e,00000684-73fb-4d65-a502-87c4eb6607c1,0000087a-f587-44fa-8e88-7bcae5bcb22c,00000889-39c5-11e4-bd0e-c3f9d65ac856,0000094c-be98-4456-8b49-6357a36581aa,00000987-2f19-4574-ab85-6744a65ee4e3,00000cd0-4097-11e4-a4e6-af71a3d902c0,00000e1e-3b55-11e4-9897-d958d55e6784

I know Group By or Memory Group by will concat rows but in this scenario can I use it if so, how can I use it.

Please help me with this. Thanks in advance!

4

3 回答 3

1

如果您没有合适的字段来对您的 id 进行分组，请自己创建一个。

在这种情况下，我将在查询中添加行号并将它们除以 10 以获得一个体面且易于配置的组。

select row_number()/10 + 1 OVER (ORDER BY id) as rnum, id from tablename ORDER BY rnum

这应该为您提供 10 行 rnum 1、10 行 rnum 2 等。将此字段配置为 Group By 字段，您就完成了。

于 2017-03-06T13:17:43.323 回答
0

我认为解决方案是：

select string_agg(id, ',')
from (select t.*, row_number() over (order by id) - 1 as seqnum
      from t
     ) t
group by floor(seqnum / 10);

虽然这使用string_agg()，但我可能会使用数组作为结果。

于 2017-03-07T02:22:50.703 回答
0

像这样？

t=# \x
Expanded display is on.
t=# with a as 
(
  select ntile(2) over (order by id),id from tablename
) 
select 
  string_agg(id,',') 
from a 
group by ntile;
-[ RECORD 1 ]-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
string_agg | 00000191-555c-11e4-922d-29fb57a42e4c, 00000192-555c-11e4-922d-29fb57a42e4c, 00000193-555c-11e4-922d-29fb57a42e4c, 00000194-555c-11e4-922d-29fb57a42e4c, 00000195-555c-11e4-922d-29fb57a42e4c, 00000196-555c-11e4-922d-29fb57a42e4c, 00000197-555c-11e4-922d-29fb57a42e4c, 00000198-555c-11e4-922d-29fb57a42e4c, 00000199-555c-11e4-922d-29fb57a42e4c, 0000019a-555c-11e4-922d-29fb57a42e4c
-[ RECORD 2 ]-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
string_agg |  000001a3-3cf2-11e4-b398-e52ee0ec6a4c, 000002ad-3768-4242-88cf-96f27d0263af, 000003ea-26e3-11e4-ace7-15c7d609fa6e, 00000684-73fb-4d65-a502-87c4eb6607c1, 0000087a-f587-44fa-8e88-7bcae5bcb22c, 00000889-39c5-11e4-bd0e-c3f9d65ac856, 0000094c-be98-4456-8b49-6357a36581aa, 00000987-2f19-4574-ab85-6744a65ee4e3, 00000cd0-4097-11e4-a4e6-af71a3d902c0, 00000e1e-3b55-11e4-9897-d958d55e6784
于 2017-03-06T13:19:36.940 回答