你在两个不同的层次上聚合。我想不出一个解决方案不需要两个单独的聚合步骤。此外，要聚合物料信息，必须先取消项目id的所有数组，然后才能聚合每个物料id的实际工件值。然后必须将其聚合回json数组。

with pieces as (

  -- the basic aggregation for the "detail pieces"
  select dt.item_id, sum(dt.pieces) as pieces
  from detail_table dt
  where dt.item_id = 2
  group by dt.item_id

), details as (

  -- normalize the material information and aggregate the pieces per material_id
  select dt.item_id, (m.detail -> 'material_id')::int as material_id, sum((m.detail -> 'pieces')::int) as pieces
  from detail_table dt
    cross join lateral jsonb_array_elements(dt.material_detail) as m(detail)
  where dt.item_id in (select item_id from pieces) --<< don't aggregate too much    
  group by dt.item_id, material_id

), material as (

  -- now de-normalize the material aggregation back into a single JSON array
  -- for each item_id
  select item_id, jsonb_agg(to_jsonb(d) - 'item_id') as material_detail
  from details d
  group by item_id

)
-- join both results together
select p.item_id, p.pieces, m.material_detail
from pieces p
  join material m on m.item_id = p.item_id
;

在线示例