时间戳

首页 > 解决方案 > 在 PostgreSQL 中聚合和连接两个表

postgresql - 在 PostgreSQL 中聚合和连接两个表

问题描述

我正在尝试使用来自两个不同表的聚合来生成聚合输出表。我不清楚如何加入这两个结果。这两个表，一个列出每个商店中的所有产品，另一个列出每个产品的价格变化如下所示。

| product_id | daily_price | date       |
|------------|-------------|------------|
| 1          | 1.25$       | 01-01-2000 |
| 1          | ...         | ...        |
| 1          | 1$          | 31-12-2000 |
| 2          | 4.5$        | 01-01-2000 |
| 2          | ...         | ...        |
| 2          | 4.25$       | 31-12-2000 |

| store_id | product_id |
|----------|------------|
| 1        | 1          |
| 1        | 2          |
| 2        | 1          |
| 2        | 3          |
| 3        | 2          |

第一个聚合获得所有产品的平均每日价格（它会有所不同）。

SELECT product_id, ROUND((AVG(price)),2) as average_price FROM product_dailyprices
    GROUP BY product_id;

| product_id | average_price |
|------------|---------------|
| 1          | 50            |
| 2          | 100           |
| 3          | 250           |

第二个查询让我知道每个商店中可用的不同产品的数量

SELECT store, COUNT(product_id) as product_count FROM products
    GROUP BY store;

| store_id | product_count |
|----------|---------------|
| 1        | 200           |
| 2        | 250           |
| 3        | 225           |

我对如何执行查询以产生以下内容有点迷茫：

| store_id | product_count | average_price_at_store |
|----------|---------------|------------------------|
| 1        | 34            | 6.51$                  |
| 2        | 45            | 3.23$                  |
| 3        | 36            | 5.37$                  |

谢谢您的帮助！

标签： postgresqljoinselectaggregate

解决方案

由于您没有为表提供 SQL，因此让我们使用以下裸骨结构：

CREATE TABLE products
(
    id   SERIAL NOT NULL,
    name text   NOT NULL,

    CONSTRAINT products_pk PRIMARY KEY (id)
);

CREATE TABLE stores
(
    id   SERIAL NOT NULL,
    name text   NOT NULL,

    CONSTRAINT stores_pk PRIMARY KEY (id)
);

CREATE TABLE daily_prices
(
    product_id  INTEGER          NOT NULL,
    daily_price DOUBLE PRECISION NOT NULL,
    date        timestamptz,

    CONSTRAINT daily_prices_product FOREIGN KEY (product_id) REFERENCES products (id)
);

CREATE TABLE locations
(
    store_id   INTEGER NOT NULL,
    product_id INTEGER NOT NULL,

    CONSTRAINT products_product_fk FOREIGN KEY (product_id) REFERENCES products (id),
    CONSTRAINT products_store_fk FOREIGN KEY (store_id) REFERENCES stores (id)
);

让我们输入一些样本数据来帮助验证查询是否有效：

INSERT INTO products(name)
VALUES ('product 1');

INSERT INTO products(name)
VALUES ('product 2');


INSERT INTO products(name)
VALUES ('product 3');

INSERT INTO stores(name)
VALUES ('store 1');

INSERT INTO stores(name)
VALUES ('store 2');

insert into locations (store_id, product_id)
values (1, 1),
       (1, 2),
       (2, 2),
       (2, 3);



INSERT INTO daily_prices(product_id, daily_price, date)
VALUES (1, 2.0, '01-01-2020');

INSERT INTO daily_prices(product_id, daily_price, date)
VALUES (1, 4.0, '02-01-2020');

INSERT INTO daily_prices(product_id, daily_price, date)
VALUES (2, 3.0, '01-01-2020');

INSERT INTO daily_prices(product_id, daily_price, date)
VALUES (2, 5.0, '02-01-2020');

INSERT INTO daily_prices(product_id, daily_price, date)
VALUES (3, 10.0, '01-01-2020');

INSERT INTO daily_prices(product_id, daily_price, date)
VALUES (3, 20.0, '02-01-2020');

然后生成所需表的查询如下所示：

select l.store_id                   as store_id,
       count(distinct l.product_id) as number_of_products,
       avg(dp.daily_price)          as average_price
from locations l
         join daily_prices dp on dp.product_id = l.product_id
group by l.store_id;

我们可以手动验证它是否计算出预期的结果：

+--------+------------------+-------------+
|store_id|number_of_products|average_price|
+--------+------------------+-------------+
|1       |2                 |3.5          |
|2       |2                 |9.5          |
+--------+------------------+-------------+

推荐阅读