Navjot Singh Navjot Singh - 5 months ago 14
MySQL Question

Get result in single query rather then three different query

Table structure and sample data

CREATE TABLE IF NOT EXISTS `orders` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`customer_id` int(11) NOT NULL,
`restaurant_id` int(11) NOT NULL,
`bill_id` int(11) NOT NULL,
`source_id` int(1) NOT NULL,
`order_medium_id` int(11) NOT NULL,
`purchase_method` varchar(255) NOT NULL,
`totalamount` int(11) NOT NULL,
`delivery_charg` int(11) NOT NULL,
`discount` int(11) NOT NULL,
`vat` int(11) NOT NULL,
`total_price` int(11) NOT NULL DEFAULT '0',
`date_created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `customer_id` (`customer_id`),
KEY `source_id` (`source_id`),
KEY `restaurant_id` (`restaurant_id`),
KEY `bill_id` (`bill_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=22 ;

--
-- Dumping data for table `orders`
--

INSERT INTO `orders` (`id`, `customer_id`, `restaurant_id`, `bill_id`, `source_id`, `order_medium_id`, `purchase_method`, `totalamount`, `delivery_charg`, `discount`, `vat`, `total_price`, `date_created`) VALUES
(1, 1, 1, 1, 1, 0, 'cash', 1600, 0, 0, 0, 1600, '2016-05-29 13:05:40'),
(2, 2, 1, 2, 2, 1, 'cash', 1820, 0, 0, 0, 1820, '2016-06-27 07:21:25'),
(4, 1, 1, 3, 3, 0, 'cash', 1770, 0, 0, 0, 1770, '2016-05-31 13:05:56'),
(5, 3, 1, 4, 2, 1, 'cash', 1300, 0, 0, 0, 1300, '2016-06-27 07:21:31'),
(6, 1, 1, 5, 1, 0, 'cash', 950, 0, 0, 0, 950, '2016-06-02 13:06:15'),
(7, 1, 1, 6, 1, 0, 'cash', 1640, 0, 0, 0, 1640, '2016-06-03 13:06:24'),
(8, 1, 1, 7, 2, 2, 'cash', 1600, 0, 0, 0, 1600, '2016-06-27 07:21:36'),
(9, 1, 1, 8, 2, 2, 'cash', 1575, 0, 0, 0, 1575, '2016-06-27 07:21:40'),
(10, 1, 1, 9, 3, 0, 'cash', 1125, 0, 0, 0, 1125, '2016-06-06 13:06:48'),
(11, 1, 1, 10, 2, 3, 'cash', 1920, 0, 0, 0, 1920, '2016-06-27 07:21:51');


Requirement :

I want to segment records as per customer as following.

Get Rating on the basis of last purchase by customer

1. customers who ordered in last 2 week then give ratingflag 5
2. customers who ordered between 2 weeks to 4 week then give ratingflag 3
3. customers who ordered between 4 weeks to 8 week then give ratingflag 2
and so on.


Get Rating on the basis of number of order by customer

1. Customer who ordered more then 5 in a month then give rating 5
2. Customer who ordered less then 5 and more then in a month then 4 give rating 4
and so on.


Get Rating on the basis of total transaction by customer

1. Customer who ordered more then 5000 rs in a month then give rating 5
2. Customer who ordered less then 5000 rs and more then in a month then 4000 give rating 4
and so on.


Customer should be unique. We write three different query for getting records according to requirement.

I tried following . Is there any way to get result in single query. I would appreciate if you could help me with better approach of doing the same :

1.) Query for last purchase

select o.customer_id,
(case when max(date_created) >= date_sub(now(), interval 2 week) then 5
when max(date_created) >= date_sub(now(), interval 4 week) then 4
when max(date_created) >= date_sub(now(), interval 8 week) then 3
when max(date_created) >= date_sub(now(), interval 10 week) then 2
when max(date_created) >= date_sub(now(), interval 12 week) then 1
end) as rating
from orders o where o.restaurant_id = 1
group by o.customer_id;


Output

customer_id rating
1 5
2 5
5 5


2.) Query for number of order

select o.customer_id,
(case when count(bill_id) >= 6 then 5
when count(bill_id) >= 4 and count(bill_id) < 6 then 4
when count(bill_id) >= 3 and count(bill_id) < 4 then 3
when count(bill_id) >= 2 and count(bill_id) < 3 then 2
when count(bill_id) >= 1 then 1
end) as rating
from orders o where o.restaurant_id = 1
group by o.customer_id


Output

customer_id rating
1 5
2 1
5 1


3.) Query for total transaction by customer

select o.customer_id,
(case when sum(total_price) >= 5000 then 5
when sum(total_price) >= 3000 and sum(total_price) < 5000 then 4
when sum(total_price) >= 2000 and sum(total_price) < 3000 then 3
when sum(total_price) >= 1000 and sum(total_price) < 2000 then 2
when sum(total_price) < 1000 then 1
end) as rating
from orders o where o.restaurant_id = 1
group by o.customer_id


Output

customer_id rating
1 5
2 2
5 2


Expected Output

customer_id R1 R2 R3
1 5 5 5
2 5 1 2
3 5 1 2

Answer
select o.customer_id,
       (case when max(date_created) >= date_sub(now(), interval 2 week) then 5
             when max(date_created) >= date_sub(now(), interval 4 week) then 4
             when max(date_created) >= date_sub(now(), interval 8 week) then 3
             when max(date_created) >= date_sub(now(), interval 10 week) then 2
             when max(date_created) >=  date_sub(now(), interval 12 week) then 1
        end) as rating1, 
        (case when count(bill_id) >= 6 then 5
             when count(bill_id) >= 4  and count(bill_id) < 6 then 4
             when count(bill_id) >= 3  and count(bill_id) < 4 then 3
             when count(bill_id) >= 2  and count(bill_id) < 3 then 2
             when count(bill_id) >= 1 then 1
        end) as rating2,
        (case when sum(total_price) >= 5000 then 5
             when sum(total_price) >= 3000  and sum(total_price) < 5000 then 4
             when sum(total_price) >= 2000  and sum(total_price) < 3000 then 3
             when sum(total_price) >= 1000  and sum(total_price) < 2000 then 2
             when sum(total_price) < 1000  then 1
        end) as rating3
from orders o where o.restaurant_id = 1
group by o.customer_id

Try this. It is faster than above answer. No need to use joins. Check this http://sqlfiddle.com/#!9/192b0/5