不是SQL的专家,但我设法让这个查询工作并提供我需要的结果。诀窍是让它表现出色。这两个表格中都有大约6密耳的记录。它目前运行大约3分钟,这是我需要它的方式。
SELECT p.id,
match.weight
FROM store_promotions p
LEFT JOIN
( SELECT *
FROM
(SELECT id,
(facets.weight::int * 1.5) AS weight
FROM store_promotions promos
JOIN -- this will return all promos, but add weight of 1.5 to the ones that are better matched to customer
(SELECT (jsonb_array_elements(product) ->> 'key') AS barcode,
(jsonb_array_elements(product) ->> 'doc_count') AS weight
FROM customer_transaction_facets
WHERE account_id = '1234567890') facets ON promos.products @> to_jsonb(facets.barcode::text)
UNION SELECT id,
(facets.weight::int * .75) AS weight -- this will return all promos, but add weight of .75 for department matches to the ones that are better matched to customer
FROM store_promotions promos
JOIN
(SELECT (jsonb_array_elements(department) ->> 'key') AS department,
(jsonb_array_elements(department) ->> 'doc_count') AS weight
FROM customer_transaction_facets
WHERE account_id = '1234567890') facets ON promos.departments @> to_jsonb(facets.department::text)) matches) AS MATCH ON p.id = match.id WHERE storeid = '637'
答案 0 :(得分:0)
select id, weight
from
store_promotions promos
left join (
select
(jsonb_array_elements(product) ->> 'key') as key,
(jsonb_array_elements(product) ->> 'doc_count')::int * 1.5 as weight
from customer_transaction_facets
where account_id = '1234567890'
union
select
(jsonb_array_elements(department) ->> 'key') as key,
(jsonb_array_elements(department) ->> 'doc_count')::int * 0.75 as weight
from customer_transaction_facets
where account_id = '1234567890'
) facets on
promos.products @> to_jsonb(facets.key::text)
or
promos.departments @> to_jsonb(facets.key::text)
where storeid = '637'