Find From/To Dates across multiple rows - SQL Postgres - sql

Question

I want to be able to "book" within range of dates, but you can't book across gaps of days. So booking across multiple rates is fine as long as they are contiguous.
I am happy to change data structure/index, if there are better ways of storing start/end ranges.
So far I have a "rates" table which contains Start/End Periods of time with a daily rate.
e.g. Rates Table.
ID Price From To
1 75.00 2015-04-12 2016-04-15
2 100.00 2016-04-16 2016-04-17
3 50.00 2016-04-18 2016-04-30
For the above data I would want to return:
From To
2015-04-12 2016-4-30
For simplicity sake it is safe to assume that dates are safely consecutive. For contiguous dates To is always 1 day before from.
For the case there is only 1 row, I would want it to return the From/To of that single row.
Also to clarify if I had the following data:
ID Price From To
1 75.00 2015-04-12 2016-04-15
2 100.00 2016-04-17 2016-04-18
3 50.00 2016-04-19 2016-04-30
4 50.00 2016-05-01 2016-05-21
Meaning where there is a gap >= 1 day it would count as a separate range.
In which case I would expect the following:
From To
2015-04-12 2016-04-15
2015-04-17 2016-05-21
Edit 1
After playing around I have come up with the following SQL which seems to work. Although I'm not sure if there are better ways/issues with it?
WITH grouped_rates AS
(SELECT
from_date,
to_date,
SUM(grp_start) OVER (ORDER BY from_date, to_date) group
FROM (SELECT
gite_id,
from_date,
to_date,
CASE WHEN (from_date - INTERVAL '1 DAY') = lag(to_date)
OVER (ORDER BY from_date, to_date)
THEN 0
ELSE 1
END grp_start
FROM rates
GROUP BY from_date, to_date) AS start_groups)
SELECT
min(from_date) from_date,
max(to_date) to_date
FROM grouped_rates
GROUP BY grp;

Answer 1

This is identifying contiguous overlapping groups in the data. One approach is to find where each group begins and then do a cumulative sum. The following query adds a flag indicating if a row starts a group:
select r.*,
(case when not exists (select 1
from rates r2
where r2.from < r.from and r2.to >= r.to or
(r2.from = r.from and r2.id < r.id)
)
then 1 else 0 end) as StartFlag
from rate r;
The or in the correlation condition is to handle the situation where intervals that define a group overlap on the start date for the interval.
You can then do a cumulative sum on this flag and aggregate by that sum:
with r as (
select r.*,
(case when not exists (select 1
from rates r2
where (r2.from < r.from and r2.to >= r.to) or
(r2.from = r.from and r2.id < r.id)
)
then 1 else 0 end) as StartFlag
from rate r
)
select min(from), max(to)
from (select r.*,
sum(r.StartFlag) over (order by r.from) as grp
from r
) r
group by grp;

Answer 2

CREATE TABLE prices( id INTEGER NOT NULL PRIMARY KEY
, price MONEY
, date_from DATE NOT NULL
, date_upto DATE NOT NULL
);
-- some data (upper limit is EXCLUSIVE)
INSERT INTO prices(id, price, date_from, date_upto) VALUES
( 1, 75.00, '2015-04-12', '2016-04-16' )
,( 2, 100.00, '2016-04-17', '2016-04-19' )
,( 3, 50.00, '2016-04-19', '2016-05-01' )
,( 4, 50.00, '2016-05-01', '2016-05-22' )
;
-- SELECT * FROM prices;
-- Recursive query to "connect the dots"
WITH RECURSIVE rrr AS (
SELECT date_from, date_upto
, 1 AS nperiod
FROM prices p0
WHERE NOT EXISTS (SELECT * FROM prices nx WHERE nx.date_upto = p0.date_from) -- no preceding segment
UNION ALL
SELECT r.date_from, p1.date_upto
, 1+r.nperiod AS nperiod
FROM prices p1
JOIN rrr r ON p1.date_from = r.date_upto
)
SELECT * FROM rrr r
WHERE NOT EXISTS (SELECT * FROM prices nx WHERE nx.date_from = r.date_upto) -- no following segment
;
Result:
date_from | date_upto | nperiod
------------+------------+---------
2015-04-12 | 2016-04-16 | 1
2016-04-17 | 2016-05-22 | 3
(2 rows)