我有一个管理狗玩具库的数据库。
dog_toy_history
表记录了狗检出的玩具
玩具库。当玩具归还时,狗只能有一个新玩具,
一只狗一次只能有一个玩具。
CREATE TABLE `dog_toy_history` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`dog_id` int(10) unsigned NOT NULL,
`toy_id` varchar(8) NOT NULL,
`created_date` datetime NOT NULL,
PRIMARY KEY (`id`)
)
INSERT INTO `dog_toy_history`
(`id`,`dog_id`,`toy_id`,`created_date`)
VALUES
(1,1,'a','2013-07-05 00:00:00'),
(2,2,'b','2013-07-15 00:00:00'),
(3,3,'c','2013-07-20 00:00:00'),
(4,1,'d','2013-07-25 00:00:00'),
(5,2,'a','2013-08-05 00:00:00'),
(6,1,'b','2013-08-10 00:00:00'),
(7,2,'d','2013-08-15 00:00:00'),
(8,1,'a','2013-08-20 00:00:00');
鉴于dog_id和日期范围,我希望能够确定玩具 狗在这个范围内,以及他们有玩具的日期。
set @dog_id = 1;
set @start_date = '2013-07-05';
set @end_date = '2013-07-25';
<run special query> ... desired result:
toy_id, start_date, end_date
a 2013-07-05 2013-07-25
d 2013-07-25 2013-07-25
set @dog_id = 1;
set @start_date = '2013-07-05';
set @end_date = '2013-08-19';
<run special query> ... desired result:
toy_id, start_date, end_date
a 2013-07-05 2013-07-25
d 2013-07-25 2013-08-10
b 2013-08-10 2013-08-19
set @dog_id = 2;
set @start_date = '2013-07-01';
set @end_date = '2013-09-01';
<run special query> ... desired result:
toy_id, start_date, end_date
b 2013-07-15 2013-08-05
a 2013-08-05 2013-08-15
d 2013-08-15 2013-09-01
答案 0 :(得分:0)
这一定对你有用。我使用您提供的数据在我的本地mysql数据库中进行了测试,并且运行良好。
select dth.toy_id, dth.created_date as start_date,
ifnull((select min(dth2.created_date) from dog_toy_history dth2 where dth2.created_date > dth.created_date and dth2.dog_id='2'),'2013-09-01') as end_date
from dog_toy_history dth
where (dth.created_date between '2013-07-01' and '2013-09-01') and dog_id='2' ORDER by start_date;
我将为那些不熟悉mysql的人提供一些解释:
内部选择会带来下一个玩具的 created_date 字段(如果存在)。这将代表当前玩具的 end_date 。
ifnull(exp1,exp2)
指定下一个玩具不存在时的默认值。在这种情况下, end_date 将是输入的结束日期。