表格结构和示例数据
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');
要求：
我想按照以下客户细分记录。
根据客户上次购买获得评分
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.
根据客户订单数量获得评分
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.
根据客户的总交易量获得评分
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.
客户应该是独一无二的。我们根据需要编写三种不同的查询来获取记录。
我试过以下。有没有办法在单个查询中获得结果。如果你能帮助我更好地做同样的事情，我将不胜感激：
1.）查询上次购买
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;
输出
customer_id rating
1 5
2 5
5 5
2.。查询订单数量
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
输出
customer_id rating
1 5
2 1
5 1
3.）按客户查询总交易
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
输出
customer_id rating
1 5
2 2
5 2
预期输出
customer_id R1 R2 R3
1 5 5 5
2 5 1 2
3 5 1 2
答案 0 :(得分：2)
您可以对查询的这些不同结果集使用联接。 http://sqlfiddle.com/#!9/192b0/3
SELECT * FROM (
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 R1
from orders o where o.restaurant_id = 1
group by o.customer_id) AS lastPurchase
LEFT JOIN
(
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 R2
from orders o where o.restaurant_id = 1
group by o.customer_id
) AS orderQuery USING(customer_id)
LEFT JOIN
(
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 R3
from orders o where o.restaurant_id = 1
group by o.customer_id
) AS totalTransactions USING(customer_id)
答案 1 :(得分：2)
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
试试这个。它比上面的答案更快。无需使用连接。请检查此http://sqlfiddle.com/#!9/192b0/5