How to find first free time in reservations table in PostgreSql - sql

Question

Reservation table contains reservations start dates, start hours and durations.
Start hour is by half hour increments in working hours 8:00 .. 18:00 in work days.
Duration is also by half hour increments in day.
CREATE TABLE reservation (
startdate date not null, -- start date
starthour numeric(4,1) not null , -- start hour 8 8.5 9 9.5 .. 16.5 17 17.5
duration Numeric(3,1) not null, -- duration by hours 0.5 1 1.5 .. 9 9.5 10
primary key (startdate, starthour)
);
table structure can changed if required.
How to find first free half hour in table which is not reserved ?
E.q if table contains
startdate starthour duration
14 9 1 -- ends at 9:59
14 10 1.5 -- ends at 11:29, e.q there is 30 minute gap before next
14 12 2
14 16 2.5
result should be:
starthour duration
11.5 0.5
Probably PostgreSql 9.2 window function should used to find
first row whose starthour is greater than previous row starthour + duration
How to write select statement which returns this information ?

Answer 1

Postgres 9.2 has range type and I would recommend to use them.
create table reservation (reservation tsrange);
insert into reservation values
('[2012-11-14 09:00:00,2012-11-14 10:00:00)'),
('[2012-11-14 10:00:00,2012-11-14 11:30:00)'),
('[2012-11-14 12:00:00,2012-11-14 14:00:00)'),
('[2012-11-14 16:00:00,2012-11-14 18:30:00)');
ALTER TABLE reservation ADD EXCLUDE USING gist (reservation WITH &&);
"EXCLUDE USING gist" creates index which disallows to inset overlapping entries. You can use the following query to find gaps (variant of vyegorov's query):
with gaps as (
select
upper(reservation) as start,
lead(lower(reservation),1,upper(reservation)) over (ORDER BY reservation) - upper(reservation) as gap
from (
select *
from reservation
union all values
('[2012-11-14 00:00:00, 2012-11-14 08:00:00)'::tsrange),
('[2012-11-14 18:00:00, 2012-11-15 00:00:00)'::tsrange)
) as x
)
select * from gaps where gap > '0'::interval;
'union all values' masks out non working times hence you can make reservation between 8am and 18pm only.
Here is the result:
start | gap
---------------------+----------
2012-11-14 08:00:00 | 01:00:00
2012-11-14 11:30:00 | 00:30:00
2012-11-14 14:00:00 | 02:00:00
Documentation links:
- http://www.postgresql.org/docs/9.2/static/rangetypes.html "Range Types"
- https://wiki.postgresql.org/images/7/73/Range-types-pgopen-2012.pdf

Answer 2

Maybe not the best query, but it does what you want:
WITH
times AS (
SELECT startdate sdate,
startdate + (floor(starthour)||'h '||
((starthour-floor(starthour))*60)||'min')::interval shour,
startdate + (floor(starthour)||'h '||
((starthour-floor(starthour))*60)||'min')::interval
+ (floor(duration)||'h '||
((duration-floor(duration))*60)||'min')::interval ehour
FROM reservation),
gaps AS (
SELECT sdate,shour,ehour,lead(shour,1,ehour)
OVER (PARTITION BY sdate ORDER BY shour) - ehour as gap
FROM times)
SELECT * FROM gaps WHERE gap > '0'::interval;
Some notes:
It will be better not to separate time and data of the event. If you have to, then use standard types;
If it is not possible to go with standard types, create function to convert numeric hours into the time format.