Question

我正在使用AWS RDS上托管的带有PostGIS 2.3的Posrgres 9.6。我正在尝试优化一些地理半径查询，以查询来自不同表的数据。

我正在考虑两种方法：具有多个联接的单个查询或两个单独但更简单的查询。

从总体上讲，为了简化结构，我的架构是：

CREATE EXTENSION "uuid-ossp";
CREATE EXTENSION IF NOT EXISTS postgis;


CREATE TABLE addresses (
    id bigint NOT NULL,
    latitude double precision,
    longitude double precision,
    line1 character varying NOT NULL,
    "position" geography(Point,4326),
    CONSTRAINT enforce_srid CHECK ((st_srid("position") = 4326))
);

CREATE INDEX index_addresses_on_position ON addresses USING gist ("position");

CREATE TABLE locations (
    id bigint NOT NULL,
    uuid uuid DEFAULT uuid_generate_v4() NOT NULL,
    address_id bigint NOT NULL
);

CREATE TABLE shops (
    id bigint NOT NULL,
    name character varying NOT NULL,
    location_id bigint NOT NULL
);

CREATE TABLE inventories (
    id bigint NOT NULL,
    shop_id bigint NOT NULL,
    status character varying NOT NULL
);

addresses表保存地理数据。 position列是在插入或更新行时从lat-lng列计算得出的。

每个address与一个location相关联。

每个address可能有多个shops，每个shop将有一个inventory。

为简洁起见，我省略了它们，但是所有表在参考列上都有适当的外键约束和btree索引。

表中有数十万行。

有了这个查询，我的主要用例就可以通过单个查询来满足，该查询在距中心地理位置（addresses）1000米以内的10.0, 10.0中进行搜索，并返回所有表中的数据：

SELECT
    s.id AS shop_id,
    s.name AS shop_name,
    i.status AS inventory_status,
    l.uuid AS location_uuid,
    a.line1 AS addr_line,
    a.latitude AS lat,
    a.longitude AS lng
FROM addresses a
JOIN locations l ON l.address_id = a.id
JOIN shops s ON s.location_id = l.id
JOIN inventories i ON i.shop_id = s.id
WHERE ST_DWithin(
    a.position,                             -- the position of each address
    ST_SetSRID(ST_Point(10.0, 10.0), 4326), -- the center of the circle
    1000,                                   -- radius distance in meters
    true
);

此查询有效，并且EXPLAIN ANALYZE表明它确实使用了GIST索引。

但是，我也可以将此查询分为两部分，并在应用程序层中管理中间结果。例如，这也可以工作：

--- only search for the addresses
SELECT
    a.id as addr_id,
    a.line1 AS addr_line,
    a.latitude AS lat,
    a.longitude AS lng
FROM addresses a
WHERE ST_DWithin(
    a.position,                             -- the position of each address
    ST_SetSRID(ST_Point(10.0, 10.0), 4326), -- the center of the circle
    1000,                                   -- radius distance in meters
    true
);

--- get the rest of the data
SELECT
    s.id AS shop_id,
    s.name AS shop_name,
    i.status AS inventory_status,
    l.id AS location_id,
    l.uuid AS location_uuid
FROM locations l
JOIN shops s ON s.location_id = l.id
JOIN inventories i ON i.shop_id = s.id
WHERE
    l.address_id IN (1, 2, 3, 4, 5)  -- potentially thousands of values
;

l.address_id IN (1, 2, 3, 4, 5)中的值来自第一个查询。

两个拆分查询的查询计划看起来比第一个查询查询的查询简单，但我想知道这本身是否意味着第二个解决方案更好。

我知道内部联接的优化非常好，最好是单次往返数据库。

那内存使用情况如何？还是表上的资源争用？（例如锁）

Answer 1

我使用IN(...)（重新）将第二个代码合并为一个查询：

--- get the rest of the data
SELECT
    s.id AS shop_id,
    s.name AS shop_name,
    i.status AS inventory_status,
    l.id AS location_id,
    l.uuid AS location_uuid
FROM locations l
JOIN shops s ON s.location_id = l.id
JOIN inventories i ON i.shop_id = s.id
WHERE l.address_id IN ( --- only search for the addresses
        SELECT a.id
        FROM addresses a
        WHERE ST_DWithin(a.position, ST_SetSRID(ST_Point(10.0, 10.0), 4326), 1000 true)
        );

或者类似地，使用EXISTS(...)：

--- get the rest of the data
SELECT
    s.id AS shop_id,
    s.name AS shop_name,
    i.status AS inventory_status,
    l.id AS location_id,
    l.uuid AS location_uuid
FROM locations l
JOIN shops s ON s.location_id = l.id
JOIN inventories i ON i.shop_id = s.id
WHERE EXISTS ( SELECT * --- only search for the addresses
        FROM addresses a
        WHERE a.id = l.address_id 
        AND ST_DWithin( a.position, ST_SetSRID(ST_Point(10.0, 10.0), 4326), 1000, true)
        );

Postgres：一个查询具有多个JOIN与多个查询

1 个答案: