您可以动态创建视图。 in this answer. 描述了比您更简单的案例的想法和解决方案@ 请在继续之前阅读它。
我们将使用以下查询来创建视图：
with all_locations(location) as (
select distinct location_a
from locations
union
select distinct location_b
from locations
)
select location_a as location, json_object_agg(location_b, count order by location_b) as data
from (
select a.location as location_a, b.location as location_b, count(l.*)
from all_locations a
cross join all_locations b
left join locations l on location_a = a.location and location_b = b.location
group by 1, 2
) s
group by 1
order by 1;
结果：
location | data
----------+----------------------------------------------------------------------------
Atlanta | { "Atlanta" : 0, "London" : 1, "New York" : 0, "Sydney" : 0, "Tokyo" : 0 }
London | { "Atlanta" : 0, "London" : 0, "New York" : 0, "Sydney" : 0, "Tokyo" : 0 }
New York | { "Atlanta" : 0, "London" : 0, "New York" : 0, "Sydney" : 0, "Tokyo" : 1 }
Sydney | { "Atlanta" : 0, "London" : 1, "New York" : 0, "Sydney" : 0, "Tokyo" : 0 }
Tokyo | { "Atlanta" : 0, "London" : 0, "New York" : 0, "Sydney" : 2, "Tokyo" : 0 }
(5 rows)
城市列表将在函数内部使用两次，因此存储在数组cities中。请注意，您可以将函数中的第一个查询替换为更简单的查询（它只是不同城市的有序列表）。
create or replace function create_locations_view()
returns void language plpgsql as $$
declare
cities text[];
list text;
begin
-- fill array with all cities in alphabetical order
select array_agg(location_a order by location_a)
from (
select distinct location_a
from locations
union
select distinct location_b
from locations
) s
into cities;
-- construct list of columns to use in select list
select string_agg(format($s$data->>'%1$s' "%1$s"$s$, city), ', ')
from unnest(cities) city
into list;
-- create view from select based on the above list
execute format($ex$
drop view if exists locations_view;
create view locations_view as
select location, %1$s
from (
select location_a as location, json_object_agg(location_b, count order by location_b) as data
from (
select a.location as location_a, b.location as location_b, count(l.*)
from unnest(%2$L::text[]) a(location)
cross join unnest(%2$L::text[]) b(location)
left join locations l on location_a = a.location and location_b = b.location
group by 1, 2
) s
group by 1
) s
order by 1
$ex$, list, cities);
end $$;
使用该函数并从创建的视图中选择数据：
select create_locations_view();
select * from locations_view;
location | Atlanta | London | New York | Sydney | Tokyo
----------+---------+--------+----------+--------+-------
Atlanta | 0 | 1 | 0 | 0 | 0
London | 0 | 0 | 0 | 0 | 0
New York | 0 | 0 | 0 | 0 | 1
Sydney | 0 | 1 | 0 | 0 | 0
Tokyo | 0 | 0 | 0 | 2 | 0
(5 rows)
这种方法我用过几次，但我没有处理过大数据的经验，所以我不能保证它是有效的。