我想提取具有空的bookingId的记录并获得最多未预订的天数(从第一个免费日).预期结果应该是:
id = 1,2013-08-03,7 days free
id = 1,2013-08-24,7 days free
id = 2,2013-08-07,10 days free
id = 2,7 days free
最好的事情是,如果我还可以查询免费时间段:例如查询1,2,3,4,5,6,7..14 ..免费日.这是我的源数据的一个示例:
id bookingDate bookingId
--------------------------------
1 2013-08-03 0
1 2013-08-04 0
1 2013-08-05 0
1 2013-08-06 0
1 2013-08-07 0
1 2013-08-08 0
1 2013-08-09 0
1 2013-08-10 112
1 2013-08-11 112
1 2013-08-12 112
1 2013-08-13 112
1 2013-08-14 112
1 2013-08-15 112
1 2013-08-16 112
1 2013-08-17 112
1 2013-08-18 112
1 2013-08-19 112
1 2013-08-20 112
1 2013-08-21 112
1 2013-08-22 112
1 2013-08-23 112
1 2013-08-24 0
1 2013-08-25 0
1 2013-08-26 0
1 2013-08-27 0
1 2013-08-28 0
1 2013-08-29 0
1 2013-08-30 0
1 2013-08-31 0
2 2013-08-03 78
2 2013-08-04 78
2 2013-08-05 78
2 2013-08-06 78
2 2013-08-07 0
2 2013-08-08 0
2 2013-08-09 0
2 2013-08-10 0
2 2013-08-11 0
2 2013-08-12 0
2 2013-08-13 0
2 2013-08-14 0
2 2013-08-15 0
2 2013-08-16 0
2 2013-08-17 39
2 2013-08-18 39
2 2013-08-19 39
2 2013-08-20 39
2 2013-08-21 39
2 2013-08-22 39
2 2013-08-23 39
2 2013-08-24 0
2 2013-08-25 0
2 2013-08-26 0
2 2013-08-27 0
2 2013-08-28 0
2 2013-08-29 0
2 2013-08-30 0
2 2013-08-31 0
如果有人对更好的数据结构有个好主意,我可以尝试实现.该数据库仍在建设中:-)
编辑:
CREATE TABLE IF NOT EXISTS `pricesBookings` (
`id` int(11) NOT NULL AUTO_INCREMENT,`baseId` int(11) NOT NULL,`bookingDate` date NOT NULL,`bookingId` int(11) NOT NULL,`price` decimal(10,2) NOT NULL,PRIMARY KEY (`id`),UNIQUE KEY `baseId` (`baseId`,`bookingDate`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
最佳答案
这应该给出正确的结果:
select
id,min(startDate) as startFreeDate,count(*) - (endDate is null) numFreeDays
from (
select
pb1.id,pb1.bookingDate startDate,min(pb2.bookingDate) endDate
from
pricesBookings pb1 left join pricesBookings pb2
on pb1.id=pb2.id
and pb2.price>0
and pb2.bookingDate>pb1.bookingDate
where
pb1.price=0
group by
pb1.id,pb1.bookingDate
) s
group by id,endDate
order by id,startDate
看它here.
如果您需要搜索所有免费插槽,例如14天,您可以添加HAVING:
group by id,endDate
having count(*) - (endDate is null) >= 14
order by id,startDate