寻找最近点 - 优化

Question

下面的SQL查询几乎是项目中最常用的部分。它的工作原理完全符合我的要求，但是它的成本(cost=11835.77..11835.82 rows=21 width=137)太高并且消耗服务器资源。

SELECT
  "companies"."id",
  "companies"."name",
  MIN(
    ST_Distance(
      addresses.location,
      ST_SetSRID(ST_Point(28.9856799, 41.0842721), 4326)
    )
  ) as distance
from
  "companies"
  left join "branches" on "companies"."id" = "branches"."company_id"
  and "branches"."active" = true
  inner join "addresses" on "branches"."id" = "addresses"."addressable_id"
  and "addresses"."addressable_type" = 'App\Domains\Company\Models\Branch'
where
  "available" = true
group by
  "companies"."id"
order by
  "distance" asc
limit
  21 offset 0;

如果我必须简要解释一下；每个公司都有很多分支机构。我将分支的位置保留在addresses表中。我的目标是通过分页列出距离发送点最近的公司。

以下查询的成本(cost=0.57..23.12 rows=21 width=137)非常低，但是拥有多个分支机构的公司会出现重复。但该公司必须列为单一公司。

select
  "companies"."id",
  "companies"."name"
from
  "companies"
  left join "branches" on "branches"."company_id" = "companies"."id"
  left join "addresses" on "addresses"."addressable_id" = "branches"."id"
where
  "addresses"."addressable_type" = 'App\Domains\Company\Models\Branch'
  and "branches"."active" = true
  and "available" = true
order by
  "addresses"."location" <-> ST_SetSRID(ST_Point(28.9856799, 41.0842721), 4326)
limit
  21 offset 0

我使用 PostgreSQL 13 和 PostGIS 作为数据库。

您可以在此处找到示例数据。

我想确切地看到结果；

 id |      name       
----+-----------------
  1 | Apple
 13 | Volvo
  9 | Burger King
 18 | Sunexpress
 11 | Togg
 19 | MC Donalds
 14 | THY
 16 | Lufthansa
  6 | Migros
  5 | Carrefour
  4 | Starbucks
  3 | Apartment
 10 | Tesla
  2 | Coffee
 17 | Pegasus
 22 | LG
 15 | British Airways
 12 | Volkswagen
 21 | Samsung
 20 | KFC
  7 | Google

Answer 1

不必计算所有记录到给定几何图形的距离来找到最短距离，只需使用Jim Jones<->所说的距离运算符即可。

所以改变这一行

MIN(
    ST_Distance(
      addresses.location,
      ST_SetSRID(ST_Point(28.9856799, 41.0842721), 4326)
    )
  ) as distance

到

MIN(ST_SetSRID(ST_Point(28.9856799, 41.0842721), 4326) <-> addresses.location) AS distance

这在选择列表中。

SELECT
    "companies"."id",
    "companies"."name",
    MIN(ST_SetSRID (ST_Point (28.9856799, 41.0842721), 4326) <-> addresses.location) AS distance
FROM
    "companies"
    LEFT JOIN "branches" ON "companies"."id" = "branches"."company_id"
        AND "branches"."active" = TRUE
    INNER JOIN "addresses" ON "branches"."id" = "addresses"."addressable_id"
        AND "addresses"."addressable_type" = 'App\Domains\Company\Models\Branch'
GROUP BY
    "companies"."id"
ORDER BY
    "distance" ASC
LIMIT 21 OFFSET 0;

这个查询成本是(cost=183.36..183.42 rows=21 width=30)