表结构和样本数据
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, 1, 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, 1, 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 ：
I want to segment records as per customer as following.
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.
客户应该是唯一的。我们编写查询以根据需要获取记录。我的表中没有太多数据，因此我根据数据对条件进行了更改。
我尝试跟随。如果您能通过更好的方法帮助我，我将不胜感激：
查询我试过
select customer_id,rating from (select `customer_id`,5 as rating from orders where `restaurant_id` = 1 and (DATE(`date_created`) between DATE(DATE_SUB(NOW(), INTERVAL 1 DAY)) AND DATE(NOW())) GROUP BY customer_id
UNION ALL
select `customer_id`,4 as rating from orders where `restaurant_id` = 1 and (DATE(`date_created`) BETWEEN DATE(DATE_SUB(NOW(), INTERVAL 4 DAY)) AND DATE(DATE_SUB(NOW(), INTERVAL 2 DAY))) GROUP BY customer_id
UNION ALL
select `customer_id`,3 as rating from orders where `restaurant_id` = 1 and (DATE(`date_created`) BETWEEN DATE(DATE_SUB(NOW(), INTERVAL 8 DAY)) AND DATE(DATE_SUB(NOW(), INTERVAL 5 DAY))) GROUP BY customer_id
UNION ALL
select `customer_id`,2 as rating from orders where `restaurant_id` = 1 and (DATE(`date_created`) BETWEEN DATE(DATE_SUB(NOW(), INTERVAL 20 DAY)) AND DATE(DATE_SUB(NOW(), INTERVAL 9 DAY))) GROUP BY customer_id
UNION ALL
select `customer_id`,1 as rating from orders where `restaurant_id` = 1 and (DATE(`date_created`) BETWEEN DATE(DATE_SUB(NOW(), INTERVAL 40 DAY)) AND DATE(DATE_SUB(NOW(), INTERVAL 21 DAY))) GROUP BY customer_id) as temp group by customer_id order by rating desc;
嗯，我想您只想要最新订单的日期和大案子声明。像这样的东西：
select o.customer_id,
(case when max(date_created) >= date_sub(now(), interval 2 week) then 1
when max(date_created) >= date_sub(now(), interval 4 week) then 2
when max(date_created) >= date_sub(now(), interval 8 week) then 3
. . .
end) as rating
from orders o
group by o.customer_id;
请注意，case评估按顺序评估条件。这就是为什么between没有必要的原因。第一个匹配条件结束的处理case。
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