5

我想提取具有空 bookingId 的记录并获得最大未预订天数（从第一个空闲日开始）。预期的结果应该是：

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, 2013-08-24, 7 days free

最好的事情是，如果我还可以查询空闲时间段：例如查询 1,2,3,4,5,6,7..14.. 空闲天数。这是我的源数据的一个例子：

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一个好主意，我可以尝试实现。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;
4

3 回答 3

4

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

在这里看到它。

如果您需要搜索所有空闲时段，例如 14 天，您可以添加 HAVING：

group by id, endDate
having count(*) - (endDate is null) >= 14
order by id, startDate
于 2013-02-06T17:39:05.243 回答
0

玩过这个。我可能遗漏了一些明显的东西，但我看不出用一个语句就能做到这一点的简单方法。

但是我想出了这种讨厌的方法。

SELECT z.baseid, z.bookingdate, 
CASE 
    WHEN j.id IS NOT NULL THEN '11+ days free'
    WHEN i.id IS NOT NULL THEN '10 days free'
    WHEN h.id IS NOT NULL THEN '9 days free'
    WHEN g.id IS NOT NULL THEN '8 days free'
    WHEN f.id IS NOT NULL THEN '7 days free'
    WHEN e.id IS NOT NULL THEN '6 days free'
    WHEN d.id IS NOT NULL THEN '5 days free'
    WHEN c.id IS NOT NULL THEN '4 days free'
    WHEN b.id IS NOT NULL THEN '3 days free'
    WHEN a.id IS NOT NULL THEN '2 days free'
    ELSE '1 day free'
END AS DaysFree
FROM pricesbookings z
INNER JOIN pricesbookings y
ON z.baseid = y.baseid AND z.bookingid = 0 AND y.bookingid != 0 AND DATE_ADD(y.bookingdate, INTERVAL 1 DAY) = z.bookingdate
LEFT JOIN pricesbookings a ON z.baseid = a.baseid AND z.bookingid = 0 AND a.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 1 DAY) = a.bookingdate
LEFT OUTER JOIN pricesbookings b ON a.baseid = b.baseid AND b.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 2 DAY) = b.bookingdate
LEFT OUTER JOIN pricesbookings c ON b.baseid = c.baseid AND c.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 3 DAY) = c.bookingdate
LEFT OUTER JOIN pricesbookings d ON c.baseid = d.baseid AND d.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 4 DAY) = d.bookingdate
LEFT OUTER JOIN pricesbookings e ON d.baseid = e.baseid AND e.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 5 DAY) = e.bookingdate
LEFT OUTER JOIN pricesbookings f ON e.baseid = f.baseid AND f.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 6 DAY) = f.bookingdate
LEFT OUTER JOIN pricesbookings g ON f.baseid = g.baseid AND g.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 7 DAY) = g.bookingdate
LEFT OUTER JOIN pricesbookings h ON g.baseid = h.baseid AND h.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 8 DAY) = h.bookingdate
LEFT OUTER JOIN pricesbookings i ON h.baseid = i.baseid AND i.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 9 DAY) = i.bookingdate
LEFT OUTER JOIN pricesbookings j ON i.baseid = j.baseid AND j.bookingid = 0 AND DATE_ADD(z.bookingdate, INTERVAL 10 DAY) = j.bookingdate
ORDER BY z.baseid, z.bookingdate

这最多只能计算 11 天或更多天（如果需要，可以轻松扩展，但确实需要提前知道最大数量），而且可能效率极低。

基本上，表别名 z 是第一天，它与表别名 y 连接以检查前一天是否已预订。然后 LEFT JOIN 针对加载更多的表副本，每个副本都在日期中添加了额外的一天。然后使用 CASE 语句来检查找到的最大的一个，从而为您提供免费天数。

有效，但您的数据库可能不喜欢它！

于 2013-02-06T17:32:28.497 回答
0

请试试这个...

select
  concat_ws(',',(concat("ID=",id)),
  min(startDate),
 (concat((count(*) - (endDate is null))," Days Free"))) as result
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, startDateselect
  concat_ws(',',(concat("ID=",id)),
  min(startDate),
 (concat((count(*) - (endDate is null))," Days Free"))) as result
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
于 2013-02-07T04:43:11.063 回答