Kueri Anda menghasilkan SQL yang salah
SELECT row_to_json(SELECT ... FROM foo) AS details
FROM (SELECT ... FROM foo) AS details_foo_row_q
Seharusnya
SELECT row_to_json(details_foo_row_q) AS details
FROM (SELECT ... FROM foo) AS details_foo_row_q
Anda perlu menggunakan pilih sebagai literal_column
from sqlalchemy.sql.expression import literal_column
details_foo_q = select([
func.row_to_json(literal_column(details_foo_row_q.name)).label('details')
]).select_from(details_foo_row_q).where(
details_foo_row_q.c.bar_id == Bar.id
).alias('details_foo_q')