有没有办法减小我的SQL查询的大小?

时间:2019-09-12 07:49:47

标签: sql postgresql

我正在尝试通过1个SQL查询从Stores,Offers,Products和Jobs表中获取所需的所有详细信息。

数据库关系:

  • 提供belongs_to商店
  • 提供has_many产品
  • 提供has_many职位

问题是:我想要具有不同职位状态的职位计数。

下的查询确实有用,但我不喜欢它的大小,并且该部分有多个重复项:WHERE "jobs"."offer_id" = "offers"."id"。有什么方法可以使此查询得到优化,甚至可能更小?

        SELECT
        "offers".*,
        "stores"."name" AS store_name,
        (
          SELECT COUNT(*)
          FROM "products"
          WHERE "products"."offer_id" = "offers"."id"
        ) AS products_count,
        ( 
          SELECT COUNT(*)
          FROM "jobs"
          WHERE "jobs"."offer_id" = "offers"."id"
        ) AS jobs_count,
        (
          SELECT COUNT(*)
          FROM "jobs"
          WHERE "jobs"."offer_id" = "offers"."id" AND "jobs"."status" = 1
        ) AS jobs_in_progress_count,
        (
          SELECT COUNT(*)
          FROM "jobs"
          WHERE "jobs"."offer_id" = "offers"."id" AND "jobs"."status" = 2
        ) AS jobs_done_count,
        (
          SELECT COUNT(*)
          FROM "jobs"
          WHERE "jobs"."offer_id" = "offers"."id" AND "jobs"."status" = 3
        ) AS jobs_error_count
        FROM "offers"
        LEFT JOIN "stores" ON "stores"."id" = "offers"."store_id"
        GROUP BY "offers"."id", "stores"."name"
        ORDER BY "offers"."created_at" DESC
        SELECT
          "offers".*,
          "stores"."name" AS store_name,
          products.products_count,
          jobs.jobs_count,
          jobs.jobs_in_progress_count,
          jobs.jobs_done_count,
          jobs.jobs_error_count
        FROM "offers" 
        LEFT JOIN "stores" ON "stores"."id" = "offers"."store_id"
        LEFT JOIN (
          SELECT p.offer_id,
            COUNT(*) AS products_count
          FROM "products" p
          GROUP BY p.offer_id
        ) "products" ON "products"."offer_id" = "offers"."id"
        LEFT JOIN (
          SELECT j.offer_id,
            COUNT(*) AS jobs_count,
            COUNT(*) FILTER (WHERE j.status = 1) AS jobs_in_progress_count,
            COUNT(*) FILTER (WHERE j.status = 2) AS jobs_done_count,
            COUNT(*) FILTER (WHERE j.status = 3) AS jobs_error_count
          FROM "jobs" j
          GROUP BY j.offer_id
        ) "jobs" ON "jobs"."offer_id" = "offers"."id"
        GROUP BY "offers"."id", "stores"."name", "products"."products_count", "jobs"."jobs_count", "jobs"."jobs_in_progress_count", "jobs"."jobs_done_count", "jobs"."jobs_error_count"
        ORDER BY "offers"."created_at" DESC

1 个答案:

答案 0 :(得分:1)

您可以使用条件聚合来计算不同的作业:

SELECT "offers".*,
       "stores"."name" AS store_name,
        (
          SELECT COUNT(*)
          FROM "products"
          WHERE "products"."offer_id" = "offers"."id"
        ) AS products_count,
        count(jobs.offer_id) as jobs_count,
        count(jobs.offer_id) filter (where jobs.status = 1) as jobs_in_progress_count,
        count(jobs.offer_id) filter (where jobs.status = 2) as jobs_done_count,
        count(jobs.offer_id) filter (where jobs.status = 3) as jobs_error_count
FROM "offers"
  LEFT JOIN "stores" ON "stores"."id" = "offers"."store_id"
  LEFT JOIN jobs ON jobs.offer_id = offers.id 
GROUP BY "offers"."id", "stores"."name"
ORDER BY "offers"."created_at" DESC;

先聚合,然后加入聚合结果可能会更快:

SELECT "offers".*,
       "stores"."name" AS store_name,
        (
          SELECT COUNT(*)
          FROM "products"
          WHERE "products"."offer_id" = "offers"."id"
        ) AS products_count,
        jobs.jobs_count,
        jobs.jobs_in_progress_count,
        jobs.jobs_done_count,
        jobs.jobs_error_count
FROM "offers"
  LEFT JOIN "stores" ON "stores"."id" = "offers"."store_id"
  LEFT JOIN (
    SELECT j.offer_id,
          count(*) as jobs_count,
          count(*) filter (where j.status = 1) as jobs_in_progress_count,
          count(*) filter (where j.status = 2) as jobs_done_count,
          count(*) filter (where j.status = 3) as jobs_error_count
    FROM jobs j
    group by j.offer_id
  ) jobs ON jobs.offer_id = offers.id 
ORDER BY "offers"."created_at" DESC;

也可以对产品进行计数:

SELECT "offers".*,
       "stores"."name" AS store_name,
        prod.products_count,
        jobs.jobs_count,
        jobs.jobs_in_progress_count,
        jobs.jobs_done_count,
        jobs.jobs_error_count
FROM "offers"
  LEFT JOIN "stores" ON "stores"."id" = "offers"."store_id"
  LEFT JOIN (
    SELECT j.offer_id,
          count(*) as jobs_count,
          count(*) filter (where j.status = 1) as jobs_in_progress_count,
          count(*) filter (where j.status = 2) as jobs_done_count,
          count(*) filter (where j.status = 3) as jobs_error_count
    FROM jobs j
    group by j.offer_id
  ) jobs ON jobs.offer_id = offers.id 
  LEFT JOIN (
     select p.offer_id, count(*)
     from products p
     group by p.offer_id
  ) prod on prod.offer_id = offers.id
ORDER BY "offers"."created_at" DESC;