Question

我正在尝试形成一个转换，我需要连接每个10行值。

第1步：表输入（来自Postgres DB的查询：select id from tablename）

上述查询的结果示例：

                  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

这里我必须将每10行ID连成一行。例如。单行1-10个行ID，另一行11-20行ID，依此类推。

预期产出：

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

我知道Group By或Memory Group将会连续行，但在这种情况下我可以使用它，如果是这样，我该如何使用它。

请帮我解决这个问题。提前谢谢！

Answer 1

如果您没有合适的字段来分组您的ID，请自己创建一个。

在这种情况下，我会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

这应该给你10行rnum 1,10行rnum 2等。将这个字段配置为Group By字段，你就完成了。

Answer 2

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

Answer 3

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。

Concat每行n行

3 个答案: