postgresql max(count(*)) - php - php

Question

I have a problem in postgresql.
I have one cohorte (gathering of people) and i would like counting the persons in this cohorte.
Begin date : "2014-09-01", End date : "2014-11-30".
I have 5 persons between 09/01 and 09/22
I have 5 persons between 09/20 and 09/25
I have 5 persons between 09/26 and 10/05
I have 5 persons between 10/01 ans 11/30
I want to have the max of accommodation for each month between the begin date and the end date in SQL (or PHP). Expected max person count:
September(09) => 10
October(10) => 10
November(11) => 5

Answer 1

Find the maximum of simultaneously present persons on a single day for every month in a given period.
I suggest generate_series() to produce the series of days in your period. Then aggregate twice:
First to get a count for each day. A single day can be dealt with plain BETWEEN. Your ranges are obviously meant to be with include borders.
Second to get the maximum per month.
SELECT date_trunc('month', day)::date AS month, max(ct) AS max_ct
FROM (
SELECT g.day, count(*) AS ct
FROM cohorte
,generate_series('2014-09-01'::date -- first of Sept.
,'2014-11-30'::date -- last of Nov.
,'1 day'::interval) g(day)
WHERE g.day BETWEEN t_begin AND t_end
GROUP BY 1
) sub
GROUP BY 1
ORDER BY 1;
Returns:
month | max_ct
-----------+--------
2014-09-01 | 10
2014-10-01 | 10
2014-11-01 | 5
Use to_char() to prettify the month output.
SQL Fiddle .. is down ATM. Here is my test case (that you should have provided):
CREATE TEMP TABLE cohorte (
cohorte_id serial PRIMARY KEY
,person_id int NOT NULL
,t_begin date NOT NULL -- inclusive
,t_end date NOT NULL -- inclusive
);
INSERT INTO cohorte(person_id, t_begin, t_end)
SELECT g, '2014-09-01'::date, '2014-09-22'::date
FROM generate_series (1,5) g
UNION ALL
SELECT g+5, '2014-09-20', '2014-09-25'
FROM generate_series (1,5) g
UNION ALL
SELECT g+10, '2014-09-26', '2014-10-05'
FROM generate_series (1,5) g
UNION ALL
SELECT g+15, '2014-10-01', '2014-11-30'
FROM generate_series (1,5) g;
For more complex checks I'd suggest the OVERLAPS operator:
Find overlapping date ranges in PostgreSQL
For more complex scenarios I'd also consider range types:
Preventing adjacent/overlapping entries with EXCLUDE in PostgreSQL

Answer 2

can't you use window function?
I'd try something like this (I've not tested this code, just exposed my thoughts)
SELECT max(count) FROM (
SELECT count(*) OVER (PARTITION BY ???) as count
FROM contract
WHERE daterange(dateStart, dateEnd, '[]') && daterange('2014-09-01', '2014-10-01', '[)')
) as max
Here, my problem remains that I can't find a way to partition for each day of the interval. Maybe this is a wrong approach, but I would be interested by a solution based on windows.
edit: with this request, you have the max of simultaneous present, but over all the time, not only a given month
with presence as (
SELECT id, generate_series(begin_date, end_date, '1 day'::interval) AS date
FROM test
),
presents as (
SELECT count(*) OVER (PARTITION BY date) AS count
FROM presence
)
SELECT max(count) from presents;
Here we come, I think
Imagine your person table has 3 columns :
id
entrance_date
leaving_date
the request would look like
WITH presents as (
SELECT id,
daterange(entrance_date, leaving_date, '[]') * daterange('2014-09-01', '2014-11-30', '[]') as range
FROM person
WHERE daterange(entrance_date, leaving_date, '[]') && daterange('2014-09-01', '2014-11-30', '[]')
),
present_per_day as (
SELECT id,
generate_series(lower(range), upper(range), '1 day'::interval) AS date
FROM presents
),
count_per_day as (
SELECT count(*) OVER (PARTITION BY date) AS count,
date
FROM present_per_day
),
SELECT max(count) OVER (PARTITION BY date_part('year', date), date_part('month', date)) as max,
date_part('year', date),
date_part('month', date)
FROM count_per_day;
(I have to leave, I hope I'll have time to test it later)
In fact, #erwin solution is much much more easy and efficient than this one.