Discard existing dates that are included in the result, SQL Server - sql

Question

In my database I have a Reservation table and it has three columns Initial Day, Last Day and the House Id.
I want to count the total days and omit those who are repeated, for example:
+-------------+------------+------------+
| | Results | |
+-------------+------------+------------+
| House Id | InitialDay | LastDay |
+-------------+------------+------------+
| 1 | 2017-09-18 | 2017-09-20 |
| 1 | 2017-09-18 | 2017-09-22 |
| 19 | 2017-09-18 | 2017-09-22 |
| 20 | 2017-09-18 | 2017-09-22 |
+-------------+------------+------------+
If you noticed the House Id with the number 1 has two rows, and each row has dates but the first row is in the interval of dates of the second row. In total the number of days should be 5 because the first shouldn't be counted as those days already exist in the second.
The reason why this is happening is that each house has two rooms, and different persons can stay in that house on the same dates.
My question is: how can I omit those cases, and only count the real days the house was occupied?

Answer 1

In your are using SQL Server 2012 or higher you can use LAG() to get the previous final date and adjust the initial date:
with ReservationAdjusted as (
select *,
lag(LastDay) over(partition by HouseID order by InitialDay, LastDay) as PreviousLast
from Reservation
)
select HouseId,
sum(case when PreviousLast>LastDay then 0 -- fully contained in the previous reservation
when PreviousLast>=InitialDay then datediff(day,PreviousLast,LastDay) -- overlap
else datediff(day,InitialDay,LastDay)+1 -- no overlap
end) as Days
from ReservationAdjusted
group by HouseId
The cases are:
The reservation is fully included in the previous reservation: we only need to compare end dates because the previous row is obtained ordering by InitialDay, LastDay, so the previous start date is always minor or equal than the current start date.
The current reservation overlaps with the previous: in this case we adjust the start and don't add 1 (the initial day is already counted), this case include when the previous end is equal to the current start (is a one day overlap).
There is no overlap: we just calculate the difference and add 1 to count also the initial day.
Note that we don't need extra condition for the reservation of a HouseID because by default the LAG() function returns NULL when there isn't a previous row, and comparisons with null always are false.
Sample input and output:
| HouseId | InitialDay | LastDay |
|---------|------------|------------|
| 1 | 2017-09-18 | 2017-09-20 |
| 1 | 2017-09-18 | 2017-09-22 |
| 1 | 2017-09-21 | 2017-09-22 |
| 19 | 2017-09-18 | 2017-09-27 |
| 19 | 2017-09-24 | 2017-09-26 |
| 19 | 2017-09-29 | 2017-09-30 |
| 20 | 2017-09-19 | 2017-09-22 |
| 20 | 2017-09-22 | 2017-09-26 |
| 20 | 2017-09-24 | 2017-09-27 |
| HouseId | Days |
|---------|------|
| 1 | 5 |
| 19 | 12 |
| 20 | 9 |

Answer 2

select house_id,min(initialDay),max(LastDay)
group by houseId
If I understood correctly!
Try out and let me know how it works out for you.
Ted.

Answer 3

While thinking through your question I came across the wonder that is the idea of a Calendar table. You'd use this code to create one, with whatever range of dates your want for your calendar. Code is from http://blog.jontav.com/post/9380766884/calendar-tables-are-incredibly-useful-in-sql
declare #start_dt as date = '1/1/2010';
declare #end_dt as date = '1/1/2020';
declare #dates as table (
date_id date primary key,
date_year smallint,
date_month tinyint,
date_day tinyint,
weekday_id tinyint,
weekday_nm varchar(10),
month_nm varchar(10),
day_of_year smallint,
quarter_id tinyint,
first_day_of_month date,
last_day_of_month date,
start_dts datetime,
end_dts datetime
)
while #start_dt < #end_dt
begin
insert into #dates(
date_id, date_year, date_month, date_day,
weekday_id, weekday_nm, month_nm, day_of_year, quarter_id,
first_day_of_month, last_day_of_month,
start_dts, end_dts
)
values(
#start_dt, year(#start_dt), month(#start_dt), day(#start_dt),
datepart(weekday, #start_dt), datename(weekday, #start_dt), datename(month, #start_dt), datepart(dayofyear, #start_dt), datepart(quarter, #start_dt),
dateadd(day,-(day(#start_dt)-1),#start_dt), dateadd(day,-(day(dateadd(month,1,#start_dt))),dateadd(month,1,#start_dt)),
cast(#start_dt as datetime), dateadd(second,-1,cast(dateadd(day, 1, #start_dt) as datetime))
)
set #start_dt = dateadd(day, 1, #start_dt)
end
select *
into Calendar
from #dates
Once you have a calendar table your query is as simple as:
select distinct t.House_id, c.date_id
from Reservation as r
inner join Calendar as c
on
c.date_id >= r.InitialDay
and c.date_id <= r.LastDay
Which gives you a row for each unique day each room was occupied. If you need a sum of how many days each room was occupied it becomes:
select a.House_id, count(a.House_id) as Days_occupied
from
(select distinct t.House_id, c.date_id
from so_test as t
inner join Calendar as c
on
c.date_id >= t.InitialDay
and c.date_id <= t.LastDay) as a
group by a.House_id

Answer 4

Create a table of all the possible dates and then join it to the Reservations table so that you have a list of all days between InitialDay and LastDay. Like this:
DECLARE #i date
DECLARE #last date
CREATE TABLE #temp (Date date)
SELECT #i = MIN(Date) FROM Reservations
SELECT #last = MAX(Date) FROM Reservations
WHILE #i <= #last
BEGIN
INSERT INTO #temp VALUES(#i)
SET #i = DATEADD(day, 1, #i)
END
SELECT HouseID, COUNT(*) FROM
(
SELECT DISTINCT HouseID, Date FROM Reservation
LEFT JOIN #temp
ON Reservation.InitialDay <= #temp.Date
AND Reservation.LastDay >= #temp.Date
) AS a
GROUP BY HouseID
DROP TABLE #temp