Question

我目前陷入任务7 http://sqlzoo.net/wiki/Guest_House_Assessment_Medium任务包括额外内容。计算预订总额5128，包括额外费用。

这就是我得到的

select sum(b.nights*r.amount+e.amount)
from booking b inner join rate r 
ON room_type_requested=r.room_type
and
b.occupants=r.occupancy
inner join extra e on b.booking_id=e.booking_id
where b.booking_id=5128

回答我得到额外的是451.80，这显然是错误的。如果我只计算这些夜晚的房价金额

select sum(b.nights*r.amount)
from booking b inner join rate r 
ON room_type_requested=r.room_type
and
b.occupants=r.occupancy

where b.booking_id=5128

我得到216.00这对于3晚没有额外费用的预订似乎是正确的。但是，在任务下建议的正确答案是118.56，附加内容。怎么会这样？

有人可以向我解释为什么它是118.56以及我做错了什么。谢谢。

Answer 1

你应该单独计算房间和额外账单的账单，然后将它们加在一起。

select bill_1 + bill_2
from
(select booking_id, sum(amount*nights) as bill_1 from 
booking b JOIN rate r ON b.room_type_requested = r.room_type and b.occupants = r.occupancy
group by booking_id
having booking_id = 5128) t1

join

(select booking_id, sum(amount) as bill_2 from extra
group by booking_id
having booking_id = 5128) t2

on t2.booking_id = t1.booking_id;

Answer 2

我刚刚经历过这些教程，这是我在SQLZOO上发现的一些不正确的答案之一，因此我在SO上找到了这篇文章，因为我试图弄清楚我在做什么错。

我认为值得强调的是，对于像这样的实例，如果您不同意网站的答案，您可以尝试解决问题，您可以输入更简单的查询来获取基础数据。在这种情况下，它确实表明网站的答案是错误的。

问：多少晚上的费率是多少？

SELECT amount, nights
FROM booking JOIN rate ON
    (room_type=room_type_requested AND occupancy=occupants)
WHERE booking_id=5128 

Result = 3 nights at 72.00 = 216.00

问：有什么额外的东西？

SELECT * FROM extra WHERE booking_id=5128

Result is two items (breakfast and phone calls), 18.00 + 1.80 = 19.80

所以根据数据的正确答案是216.00 + 19.80 = 235.80。

Answer 3

使用逻辑，SQL应为：

select sum(r2.amount * b1.nights)+ sum(e3.amount) as total 
from booking b1
inner join rate  r2
on r2.room_type = b1.room_type_requested
and r2.occupancy = b1.occupants
left join (select booking_id, sum(amount) as amount 
           from extra
           group by booking_id) e3
on e3.booking_id = b1.booking_id
where b1.booking_id = 5128

返回235.80

Answer 4

输出将为235.80，但网站列出了错误的输出118.56。

SELECT sum(b.nights * r.amount) +
  (SELECT sum(e.amount)
   FROM extra AS e
   WHERE e.booking_id = b.booking_id)
FROM booking AS b
INNER JOIN rate AS r ON r.occupancy = b.occupants
AND r.room_type = b.room_type_requested
WHERE b.booking_id = 5128
GROUP BY b.booking_id;

sql计算总晚数*费率+额外费用？ sqlzoo

4 个答案: