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SQL Queries to join three tables

Tags:

sql

I have three tables with following structure and info:

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 and sales_parts hold sales made by customers. The idea is to write a query that sums the "sale_amount" of both cars and parts for a particular customer and groups the result by id.

Does someone hast a suggestion how I can go about this problem?

like image 372
user224645 Avatar asked Jul 20 '10 12:07

user224645


3 Answers

You may want to try something like the following:

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);

Result:

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

UPDATE: Further to comments below:

Let's start with the sale_date field:

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');

To get the date of the latest sale of each customer, you can join the query described previously with another derived table, as follows:

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);

Result:

+-------------+-------------+---------------------+
| 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)

Do you see the pattern? There are other approaches to tackle the same problem, but joining with derived tables is a very easy and straightforward technique.

Then regarding the changes in the customer table, I assume you mean something like this:

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');

To concatenate string fields in MySQL, you can simply use the CONCAT() function:

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

Returns:

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

To apply the 'Mr' or 'Ms' conditionally, you can then use the CASE statement:

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;

Returns:

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

You can also concatenate the two fields together:

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;

Returns:

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

Finally, we can attach this to our main query, as follows:

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);

Returns:

+-------------+----------------+-------------+---------------------+
| 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)
like image 74
Daniel Vassallo Avatar answered Oct 06 '22 08:10

Daniel Vassallo


Something like this will be what you are after...

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;
like image 38
Brian Hooper Avatar answered Oct 06 '22 09:10

Brian Hooper


  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
like image 30
Pranay Rana Avatar answered Oct 06 '22 08:10

Pranay Rana