找不到正确的 SQL 函数（SUM/COUNT - LEFT JOIN）

Question

我对 SQL 函数的问题是，我经常混合使用 SUM / COUNT，甚至可以在其中放置 WHERE 子句和 GROUP BY 的简短版本，SQL 是我的缺陷之一。

好的，现在这是我当前的问题，我需要创建一个 SQL 函数（在我的 PHP 模型中）来获取我“今天”的订单数量及其总成本，但我需要按国家/地区分隔结果（对不起我的英语），例如我想要的结果应该是这样的（假设我今天有 20 个订单）：
美国：15 个订单（共 288 美元），加拿大：5 个订单（共 94 美元）

我尝试了很多东西，唯一没有给我错误的两个是这两个（但与我需要的结果相去甚远）：

SELECT sum(o.amount) AS totalmoney, p.name, p.id, p.name_clean, p.iso_code 
FROM orders AS o 
LEFT JOIN customers AS c ON o.id_customer = c.id 
LEFT JOIN countries AS p ON c.country = p.id    
WHERE DATE(o.date_create) = CURDATE()

SELECT sum(o.amount) AS totalmoney, o.date_create, p.name, COUNT(p.id) AS totalpays,
p.name_clean, p.iso_code, c.country 
FROM orders AS o 
LEFT JOIN customers AS c ON o.id_customer = c.id 
LEFT JOIN countries AS p ON c.country = p.id 
GROUP BY o.date_create

我的三个表是（表名：列有用）：

国家：id - 名称

客户：id - country_id

订单：id - id_customer - 金额 ($) - date_create

你能帮帮我吗？

感谢您的宝贵时间。

------------- 使用我使用的答案进行编辑 ---------------

感谢所有参与并帮助我更好地理解 SQL 的人。

SELECT sum(orders.amount) total, countries.name
FROM orders 
INNER JOIN customers ON orders.id_customer = customers.id
INNER JOIN countries ON countries.id = customers.country_id
WHERE orders.date_create = ?
GROUP BY customers.id

这正是我需要的。

Answer 1

您想在某个日期按国家/地区分组，因此您的分组依据是国家，日期是哪里：

SELECT sum(orders.amount) total, countries.name
FROM orders 
INNER JOIN customers ON orders.id_customer = customers.id
INNER JOIN countries ON countries.id = customers.country_id
WHERE orders.date_create = ?
GROUP BY customers.id

这里适合使用内部连接，因为连接应该只为每个订单的客户和国家/地区生成一行。

Answer 2

尝试一下： -- PostgreSQL

drop table orders;
drop table countries;
drop table customers;

CREATE TABLE countries
(
    id   integer not null,
    name varchar(200)
);

insert into countries(id, name) values (1, 'MEXICO');
insert into countries(id, name) values (2, 'US');
insert into countries(id, name) values (3, 'CANADA');

create table customers
(
    id   integer not null,
    name varchar(200),
    country integer not null
);

insert into customers(id, name, country) values (1, 'Huey',1);
insert into customers(id, name, country) values (2, 'Dewey',2);
insert into customers(id, name, country) values (3, 'Louie',3);

create table orders
(
    id          integer not null,
    id_customer integer not null,
    amount      double precision,
    date_create date
);

insert into orders values (1, 1,  500.0, '20200103');
insert into orders values (2, 1,  1000, '20200103');
insert into orders values (3, 2, 500, '20200103');
insert into orders values (4, 3, 500, '20200103');
insert into orders values (5, 1,  500, '20200103');

SELECT COUNT(*) AS num_orders,  sum(o.amount) AS totalmoney, p.name
FROM orders AS o
         INNER JOIN customers AS c ON o.id_customer = c.id
         INNER JOIN countries AS p ON c.country = p.id
WHERE DATE(o.date_create) = '20200103'
 group by   p.name
order by num_orders desc;

Answer 3

您可以使用over 和sum 和count

select sum(amount) over(partition by contry_id) as totalamount,
       count(contry_id) over(partition by contry_id) as totalcount,
       name, ord.id 
from orders ord left join customers cus on ord.customer_id = cus.id
                left join countries con on cus.contry_id = con.id
where date_create = [date]  --using string as a pseudo date when testing, so change is needed

这是一个带有一些伪数据的db<>fiddle。

这还将按创建日期获取与 OP 匹配的数据组，需要更多 imo。

select distinct sum(amount) over(partition by contry_id,date_create) as totalamount,
       count(contry_id) over(partition by contry_id,date_create) as totalcount,
       name, date_create
from orders ord left join customers cus on ord.customer_id = cus.id
                left join countries con on cus.contry_id = con.id

具有窗口函数的优点：over 是您可以使用具有不同“条件”的不同聚合函数，并且不需要 group by 每个非聚合列，这通常会导致很多子查询。

Answer 4

这应该做你想做的事：

SELECT 
    c.name,
    sq.totalmoney
FROM
    countries as c
JOIN
    (SELECT 
        SUM(o.amount) AS totalmoney,
        c.id
    FROM
        (SELECT 
            *
        FROM 
            orders 
        WHERE 
            DATE(date_create) = CURDATE()
        ) AS o 
    JOIN 
        customers AS c 
    ON 
        o.id_customer = c.id
    GROUP BY c.id) AS sq
on c.id = sq.id;