Mysql calculating percentage of repeat rows - mysql

Question

So I am trying to calculate the amount of repeat orders in my system per restaurant. This is defined as the number of users (based on their email address, eo_email) that have ordered more than once from that restaurant. Examples under the schema
Here is the table that represents my restaurants
CREATE TABLE IF NOT EXISTS `lf_restaurants` (
`r_id` int(8) NOT NULL AUTO_INCREMENT,
`r_name` varchar(128) DEFAULT NOT NULL,
PRIMARY KEY (`r_id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 ;
INSERT INTO `lf_restaurants` (`eo_id`, `eo_ref_id`) VALUES
('1', 'Restaurant X'),
('2', 'Cafe Y');
And this is my orders table
CREATE TABLE IF NOT EXISTS `ecom_orders` (
`eo_id` mediumint(9) NOT NULL AUTO_INCREMENT,
`eo_ref_id` varchar(12) DEFAULT NOT NULL,
`eo_email` varchar(255) DEFAULT NOT NULL,
`eo_order_parent` int(11) NOT NULL,
PRIMARY KEY (`eo_id`),
) ENGINE=InnoDB DEFAULT CHARSET=latin1 ;
INSERT INTO `ecom_orders` (`eo_id`, `eo_ref_id`, `eo_email`, `eo_order_parent`) VALUES
('', '', 'a#a.com', '1'),
('', '', 'a#a.com', '1'),
('', '', 'a#a.com', '1'),
('', '', 'a#a.com', '1'),
('', '', 'a#a.com', '1'),
('', '', 'b#b.com', '1'),
('', '', 'b#b.com', '1'),
('', '', 'c#c.com', '1'),
('', '', 'd#d.com', '1'),
('', '', 'e#e.com', '1'),
('', '', 'a#a.com', '2'),
('', '', 'c#c.com', '2'),
('', '', 'c#c.com', '2'),
('', '', 'e#e.com', '2');
So Restaurant X (r_id 1) has 10 orders. Users a#a.com and b#b.com have ordered from that restaurant multiple times, and c#c.com, d#d.com, and e#e.com have only ordered once, so it would need to return 40%
Cafe Y (r_id 2) has 4 orders. User c#c.com has ordered twice, users a#a.com and e#e.com have only ordered once, so it would need to return 33%
I am not sure posting what I have got already will be much good, as I keep running into 'Subquery has more than 1 result' or if I wrap that subquery in its own dummy query with a count, it wont let me use fields I need from the main query such as r_id. But here goes:
SELECT r_name,
(SELECT COUNT(*) AS cnt_users
FROM (
SELECT *
FROM ecom_orders
WHERE eo_order_parent = r_id
GROUP BY eo_email
) AS cnt_dummy
) AS num_orders,
(SELECT COUNT(*) AS cnt
FROM ecom_orders
WHERE eo_order_parent = r_id
GROUP BY eo_order_parent, eo_email
) AS num_rep_orders
FROM lf_restaurants
ORDER BY num_orders DESC
The num_orders subquery is saying it doesnt recognise r_id, as I am guessing this is due to the order in which things are executed
The num_rep_orders subquery is coming back as multiple rows, but really i want that to come back with just a single value, which I could do if I made it like the num_orders subquery but then would run into the r_id doesnt exist problem.
So my question is: How do I get these values that I need without running into subquery has more than 1 row, and r_id does not exist?
Then from those 2 values I can work out the percentage and all should be gravy :) Any help much appreciated!

Answer 1

So Restaurant X (r_id 1) has 10 orders. Users a#a.com and b#b.com have
ordered from that restaurant multiple times, and c#c.com, d#d.com, and
e#e.com have only ordered once, so it would need to return 40%
Cafe Y (r_id 2) has 4 orders. User c#c.com has ordered twice, users
a#a.com and e#e.com have only ordered once, so it would need to return
33%
Okay. So let's start with getting the number of repeating customers.
SELECT eo_order_parent, eo_email, COUNT(eo_email) AS orders FROM ecom_orders
GROUP BY eo_order_parent, eo_email
HAVING orders > 1;
And the total number of different customers
SELECT eo_order_parent, COUNT(eo_email) FROM ecom_orders
GROUP BY eo_order_parent;
But we can do this in one go:
SELECT eo_order_parent,
SUM(CASE WHEN orders > 1 THEN 1 ELSE 0 END) AS repeats,
SUM(1) AS total FROM
(
SELECT eo_order_parent, eo_email, COUNT(*) AS orders FROM ecom_orders
GROUP BY eo_order_parent, eo_email
) AS eo_group_1
GROUP BY eo_order_parent;
This gives:
+-----------------+---------+-------+
| eo_order_parent | repeats | total |
+-----------------+---------+-------+
| 1 | 2 | 5 |
| 2 | 1 | 3 |
+-----------------+---------+-------+
2 rows in set (0.00 sec)
Then 2/5 is your 40%, and 1/3 is 33%.

Answer 2

The following query computes the number of repeat customers and the total number of customers per restaurant
SELECT
u.r_id,
u.r_name,
SUM(u.no_orders > 1) AS repeats,
SUM(u.no_orders) AS orders,
COUNT(u.eo_email) AS customers
FROM (
SELECT
r.*,
o.eo_email,
COUNT(o.eo_id) AS no_orders
FROM lf_restaurants r
LEFT JOIN ecom_orders o ON o.eo_order_parent = r.r_id
GROUP BY o.eo_email
) u
GROUP BY
r.r_id;
The subquery first computes the number of orders per customer/restaurant pair. The outer query computes from this the number of customers, the number of repeating customers and the total number of customers per restaurant. You can also compute the percentage (but this does not have to be done in the query).