Hotel Room Rates for different seasons - mysql

Question

I have a database (MySQL) with a table containing date ranges (as startdate and enddate) and a rate field. The date range implies different seasons (low, high etc.). The scenario is such that a person checks in the hotel and his duration of stay is in two seasons. A sample data is like below:
SeasonName SartDate EndDate Rate
Low 01-01-2007 30-04-2007 100.00
High 01-05-2007 31-08-2007 150.00
Peak 01-09-2007 31-12-2007 200.00
The client's Check In Date is 29-04-2007 and Check Out Date is 03-05-2007. I need to calculate the exact number of nights for each season and also calculate the total amount.
The IDE is VB6. Any help will be extremely appreciated.
Thanks
Tom
Thanks for the response. I need the SQL to extract the information. As for the date validity, lets assume the rate applies till midnight (00:00). Hope i have clarified.
Tom

Answer 1

Having worked in a hotel and written the reservation system, hourly time is irrelevant as far
as billing goes. Everything is always charged by night. (Unless you plan to run a place that charges by the hour! ;-)) Check-in and check-out are operational considerations.
Do not use stored procuedures if you actually want to write a real reservation system.
It defeats the purpose of having a database.
Also, writing out dates like this is 2007-04-29 is really great way because not every one is from the same place and this is an international standard. Also notice, if you were to turn this into a string it will still be sorted correctly!
You need make a calandar table as MySQL does not have in built in functions to do it.
This procedure will build up dates for you.
drop table if exists calendar;
create table calendar
(
date_ date primary key
);
drop procedure fill_calendar;
delimiter $$
create procedure fill_calendar(start_date date, end_date date)
begin
declare date_ date;
set date_=start_date;
while date_ < end_date do
insert into calendar values(date_);
set date_ = adddate(date_, interval 1 day);
end while;
end $$
delimiter ;
call fill_calendar('2007-1-1', '2007-12-31');
from: http://www.ehow.com/how_7571744_mysql-calendar-tutorial.html
drop table if exists rates;
create table rates
(
season varchar(100) primary key,
start_date date references calendar(date_),
end_date date references calendar(date_),
rate float
);
insert into rates values ('Low', '2007-01-01', '2007-04-30', 100.00);
insert into rates values ('High', '2007-05-01', '2007-08-31', 150.00);
insert into rates values ('Peak', '2007-09-01', '2007-12-21', 200.00);
select * from rates;
season start_date end_date rate
Low 2007-01-01 2007-04-30 100
High 2007-05-01 2007-08-31 150
Peak 2007-09-01 2007-12-21 200
I'm going to ignore the dates you have given in your question and the assume the client is not travelling backwards in time.
select
date_, rate
from calendar
join rates
on date_ >= start_date and date_ <= end_date
where date_ between '2007-04-29' and '2007-5-01'
;
date_ rate
2007-04-29 100
2007-04-30 100
2007-05-01 150
select
sum(rate)
from calendar
join rates
on date_ >= start_date and date_ <= end_date
where date_ between '2007-04-29' and '2007-5-01'
sum(rate)
350
And, as you can see the sql is quite concise and readable without resorting to functions or procedures. This will be able to scale properly and handle more complex questions. Also, it enables referential checking to be used since the data is table based.

Answer 2

I used the DATEDIFF function to know the number of days between 2 dates. But since this function returns the number of days excluding the last day (e.g. DATEDIFF(d, '2007-04-29', '2007-04-30') return 1 instead of 2), I used a rate table like this:
SeasonName StartDate EndDate Rate
Low 01-01-2007 01-05-2007 100.00
High 01-05-2007 01-09-2007 150.00
Peak 01-09-2007 01-01-2008 200.00
This is the query I used. The inner of the two SELECTs calculates an effective end date for the customer's stay according to the season (either the end date of the season or it's date prior to it's check out date).
I used 02-05-2007 as the customer's stay end date instead of 03-05-2007 since usually a customer does does not pay for the day he checks out.
SELECT SeasonName, DATEDIFF(d, '2007-04-29', EffectiveEndDate) as NumberOfDays, Rate
FROM (
SELECT SeasonName,
CASE
WHEN EndDate < '2007-05-02' THEN EndDate
ELSE '2007-05-02'
END AS EffectiveEndDate,
Rate
FROM HotelRate
WHERE (StartDate <= '2007-04-29' and EndDate > '2007-04-29')
or (StartDate <= '2007-05-02' and EndDate > '2007-05-02')
) as SubSelect
This gives me this result:
SeasonName NumberOfDays Rate
Low 2 100.00
High 3 150.00
I hope it helps :)

Answer 3

Disclaimer: This is not the most efficient one, but it's the more clear, and if you have the price list cached in an array, the performance hit will be meaningless.
Dim numberOfDays As Integer, i As Integer
Dim CheckInDate As Date, CheckOutDate As Date, CurrDate As Date
Dim TotalRate As Currency
TotalRate = 0
CheckInDate = DateSerial(2007, 4, 29)
CheckOutDate = DateSerial(2007, 5, 3)
''// -1 asumming the last day is checkout day
numberOfDays = DateDiff("d", CheckInDate, CheckOutDate) - 1
For i = 0 To numberOfDays
CurrDate = DateAdd("d", i, CheckInDate)
TotalRate = TotalRate + CalculateRateForDay(CurrDate)
Next

Answer 4

Use an auxiliary calendar table:
SELECT S1.client_ID, H1.Rate, C1.dt
FROM HotelRates AS H1
INNER JOIN Calendar AS C1
ON C1.dt BETWEEN H1.SartDate AND H1.EndDate
INNER JOIN Stays AS S1
ON C1.dt >= check_in_date AND
C1.dt < check_out_date;