consultas SQL para unirse a tres mesas

Question

Tengo tres tablas con estructura siguiente información: y

CREATE TABLE customer (
    customer_id mediumint(8) unsigned NOT NULL auto_increment, 
    name varchar(50) NOT NULL, 
    PRIMARY KEY (customer_id) 
); 

INSERT INTO customer VALUES (1, 'Dagmar'); 
INSERT INTO customer VALUES (2, 'Dietmar'); 
INSERT INTO customer VALUES (3, 'Sabine'); 

CREATE TABLE sales_cars (
    sale_id mediumint(8) unsigned NOT NULL auto_increment, 
    customer_id mediumint(8) unsigned NOT NULL, 
    sale_amount decimal(10,2) NOT NULL, 
    PRIMARY KEY (sale_id) 
); 

INSERT INTO sales_cars VALUES (1, 3, 14.40); 
INSERT INTO sales_cars VALUES (2, 1, 28.30); 
INSERT INTO sales_cars VALUES (3, 2, 34.40); 
INSERT INTO sales_cars VALUES (4, 2, 25.60); 

CREATE TABLE sales_parts (
    sale_id mediumint(8) unsigned NOT NULL auto_increment, 
    customer_id mediumint(8) unsigned NOT NULL, 
    sale_amount decimal(10,2) NOT NULL, 
    PRIMARY KEY (sale_id) 
); 

INSERT INTO sales_parts VALUES (1, 2, 68.20); 
INSERT INTO sales_parts VALUES (2, 3, 21.30); 
INSERT INTO sales_parts VALUES (3, 3, 54.40); 
INSERT INTO sales_parts VALUES (4, 1, 35.70); 

sales_car y sales_parts llevan a cabo ventas realizadas por los clientes. La idea es escribir una consulta que sume el "monto de venta" de los automóviles y las piezas para un cliente en particular y agrupa el resultado por identificación.

¿Alguien tiene alguna sugerencia sobre cómo puedo solucionar este problema?

Answer 1

Algo como esto será lo que está después ...

SELECT *, 
     (SELECT SUM(sale_amount) 
      FROM sales_cars 
      WHERE sales_cars.customer_id = customer.customer_id) AS car_sales, 
     (SELECT SUM(sale_amount) 
      FROM sales_parts 
      WHERE sales_parts.customer_id = customer.customer_id) AS part_sales 
    FROM customer; 

Answer 2

Es posible que desee probar algo como lo siguiente:

SELECT c.customer_id, 
     tot_cars.total + tot_parts.total AS total_sales 
FROM customer c 
JOIN (
      SELECT customer_id, SUM(sale_amount) total 
      FROM  sales_cars 
      GROUP BY customer_id 
     ) tot_cars ON (tot_cars.customer_id = c.customer_id) 
JOIN (
      SELECT customer_id, SUM(sale_amount) total 
      FROM  sales_parts 
      GROUP BY customer_id 
     ) tot_parts ON (tot_parts.customer_id = c.customer_id); 

Resultado:

+-------------+-------------+ 
| customer_id | total_sales | 
+-------------+-------------+ 
|   1 |  64.00 | 
|   2 |  128.20 | 
|   3 |  90.10 | 
+-------------+-------------+ 
3 rows in set (0.03 sec) 

---

ACTUALIZACIÓN: relación con sus comentarios a continuación:

Vamos a empezar con el campo sale_date:

CREATE TABLE sales_cars (
    sale_id mediumint(8) unsigned NOT NULL auto_increment, 
    customer_id mediumint(8) unsigned NOT NULL, 
    sale_amount decimal(10,2) NOT NULL, 
    sale_date datetime NOT NULL, 
    PRIMARY KEY (sale_id) 
); 

INSERT INTO sales_cars VALUES (1, 3, 14.40, '2010-07-01 12:00:00'); 
INSERT INTO sales_cars VALUES (2, 1, 28.30, '2010-07-05 12:00:00'); 
INSERT INTO sales_cars VALUES (3, 2, 34.40, '2010-07-10 12:00:00'); 
INSERT INTO sales_cars VALUES (4, 2, 25.60, '2010-07-20 12:00:00'); 

para obtener la fecha de la última venta de cada cliente, puede unirse a la consulta descrita anteriormente con otra tabla derivada, de la siguiente manera :

SELECT c.customer_id, 
     tot_cars.total + tot_parts.total AS total_sales, 
     latest_sales.date AS latest_sale 
FROM customer c 
JOIN (
      SELECT customer_id, SUM(sale_amount) total 
      FROM  sales_cars 
      GROUP BY customer_id 
     ) tot_cars ON (tot_cars.customer_id = c.customer_id) 
JOIN (
      SELECT customer_id, SUM(sale_amount) total 
      FROM  sales_parts 
      GROUP BY customer_id 
     ) tot_parts ON (tot_parts.customer_id = c.customer_id) 
JOIN (
      SELECT customer_id, MAX(sale_date) date 
      FROM  sales_cars 
      GROUP BY customer_id 
     ) latest_sales ON (latest_sales.customer_id = c.customer_id); 

Resultado:

+-------------+-------------+---------------------+ 
| customer_id | total_sales | latest_sale   | 
+-------------+-------------+---------------------+ 
|   1 |  64.00 | 2010-07-05 12:00:00 | 
|   2 |  128.20 | 2010-07-20 12:00:00 | 
|   3 |  90.10 | 2010-07-01 12:00:00 | 
+-------------+-------------+---------------------+ 
3 rows in set (0.07 sec) 

¿Ves el patrón? Existen otros enfoques para abordar el mismo problema, pero unirse a tablas derivadas es una técnica muy fácil y directa.

A continuación, en relación con los cambios en la tabla customer, supongo que te refieres algo como esto:

CREATE TABLE customer (
    customer_id mediumint(8) unsigned NOT NULL auto_increment, 
    first_name varchar(50) NOT NULL, 
    last_name varchar(50) NOT NULL, 
    gender char(1) NOT NULL, 
    PRIMARY KEY (customer_id) 
); 

INSERT INTO customer VALUES (1, 'Joe', 'Doe', 'M'); 
INSERT INTO customer VALUES (2, 'Jane', 'Smith', 'F'); 
INSERT INTO customer VALUES (3, 'Peter', 'Brown', 'M'); 

para concatenar los campos de cadena en MySQL, puede simplemente usar la función CONCAT():

SELECT CONCAT(c.first_name, ' ', c.last_name) as full_name 
FROM customer c; 

Devoluciones:

+-------------+ 
| full_name | 
+-------------+ 
| Jane Smith | 
| Peter Brown | 
| Joe Doe  | 
+-------------+ 
3 rows in set (0.01 sec) 

Para aplicar el El señor "o 'señora' condicionalmente, a continuación, puede utilizar el CASE declaración:

SELECT (CASE c.gender WHEN 'M' THEN 'Mr' WHEN 'F' THEN 'Ms' END) salutaiton, 
     CONCAT(c.first_name, ' ', c.last_name) as full_name 
FROM customer c; 

Devuelve:

+------------+-------------+ 
| salutaiton | full_name | 
+------------+-------------+ 
| Ms   | Jane Smith | 
| Mr   | Peter Brown | 
| Mr   | Joe Doe  | 
+------------+-------------+ 
3 rows in set (0.01 sec) 

También puede concatenar los dos campos juntos:

SELECT CONCAT((CASE c.gender WHEN 'M' THEN 'Mr' WHEN 'F' THEN 'Ms' END), ' ', 
       c.first_name, ' ', c.last_name) as full_name 
FROM customer c; 

Devoluciones :

+----------------+ 
| full_name  | 
+----------------+ 
| Ms Jane Smith | 
| Mr Peter Brown | 
| Mr Joe Doe  | 
+----------------+ 
3 rows in set (0.00 sec) 

Por último, se puede adjuntar esto a nuestra consulta principal, de la siguiente manera:

SELECT c.customer_id, 
     CONCAT((CASE c.gender WHEN 'M' THEN 'Mr' WHEN 'F' THEN 'Ms' END), ' ', 
        c.first_name, ' ', c.last_name) as full_name, 
     tot_cars.total + tot_parts.total AS total_sales, 
     latest_sales.date AS latest_sale 
FROM customer c 
JOIN (
      SELECT customer_id, SUM(sale_amount) total 
      FROM  sales_cars 
      GROUP BY customer_id 
     ) tot_cars ON (tot_cars.customer_id = c.customer_id) 
JOIN (
      SELECT customer_id, SUM(sale_amount) total 
      FROM  sales_parts 
      GROUP BY customer_id 
     ) tot_parts ON (tot_parts.customer_id = c.customer_id) 
JOIN (
      SELECT customer_id, MAX(sale_date) date 
      FROM  sales_cars 
      GROUP BY customer_id 
     ) latest_sales ON (latest_sales.customer_id = c.customer_id); 

Devuelve:

+-------------+----------------+-------------+---------------------+ 
| customer_id | full_name  | total_sales | latest_sale   | 
+-------------+----------------+-------------+---------------------+ 
|   1 | Mr Joe Doe  |  64.00 | 2010-07-05 12:00:00 | 
|   2 | Ms Jane Smith |  128.20 | 2010-07-20 12:00:00 | 
|   3 | Mr Peter Brown |  90.10 | 2010-07-01 12:00:00 | 
+-------------+----------------+-------------+---------------------+ 
3 rows in set (0.02 sec) 

Answer 3

select customer.customer_id,(totalcar + totalparts) as total from customer 
inner join 
(select customer_id ,sum(sale_amount) as totalcar 
     from sales_cars group by customer_id) d 
on customer_id = d.customer_id 
inner join 
(select customer_id , sum(sale_amount) as totalparts 
     from sales_parts group by customer_id) d1 

on customer_id = d1. customer_id 

Answer 4

no pienso subconsultas agregados es buena idea

select customer_id, sum(sale_amount) from (
select customer.customer_id, sale_amount from customer 
join sales_cars on sales_cars.customer_id = customer.customer_id 
union all 
select customer.customer_id, sale_amount from customer 
join sales_parts on sales_parts.customer_id = customer.customer_id 
) totals group by customer_id