Weeks between two dates in Postgres - sql

Question

I want to calculate the number of weeks that a customer has been a customer. The goal is to compare this to the distinct number of weeks the customer has made a purchase to determine what % of weeks the customer made a purchase.
With timestamp customer.created_at what is the best way to find the number of weeks a customer has been a customer? Said differently, the difference in weeks between the current week and the week in which the customer signed up.
Calendar weeks starting Monday to be apples-to-apples with distinct calendar weeks a customer has made a purchase.

Answer 1

AFAIK this should work in all cases
-Your calendar week starts on Monday
-Both dates start at least from 05.01.1970 (Monday)
Postgres:
SELECT
(
extract(epoch from date_trunc('week',now()::date))
-
extract(epoch from date_trunc('week','2019-12-20'::date))
)
/604800
(where 604800 is the number of seconds in a week)

Answer 2

SELECT (EXTRACT(days FROM (now() - customer.created_at)) / 7)::int;
First find the difference between current time and customer's created_at time stamp.
Then you tell Postgres to give you the difference in days and then you can divide that by 7 and round it to an integer. Note that if you do not want an integer, you can cast appropriately.

Answer 3

Try this
select (CURRENT_DATE - customer.created_at)/7

Answer 4

This is the code I used. Works.
SELECT
(
(
(
CURRENT_DATE::DATE - (EXTRACT('dow' FROM CURRENT_DATE::DATE) * INTERVAL '1 day')
)::DATE -
(
customers.created_at::DATE - (EXTRACT('dow' FROM customers.created_at::DATE) * INTERVAL '1 day')
)::DATE
) / 7
) + 1

Answer 5

I think you can have that to get the calendar weeks
select to_char(i::date, 'IYYYIW') AS mdate from generate_series('2017-01-01', current_date, '1 week'::interval) i;
and to count
select count(to_char(i::date, 'IYYYIW')) from generate_series('2017-01-01', current_date, '1 week'::interval) i;
and you can replace the date with customer date

Answer 6

#zach's answer is (perhaps) valid, however, it is hard to follow.
I found these two procedures that abstract equivalent logic:
CREATE OR REPLACE FUNCTION first_of_week(date) returns date AS $$
SELECT ($1::date-(extract('dow' FROM $1::date)*interval '1 day'))::date;
$$ LANGUAGE SQL STABLE STRICT;
CREATE OR REPLACE FUNCTION weeks_in_range(date,date) returns int AS $$
SELECT ((first_of_week($2)-first_of_week($1))/7)+1
$$ LANGUAGE SQL STABLE STRICT;
Source:
https://gist.github.com/tkellen/2003309#file-gistfile1-sql-L17-L23
Example:
SELECT weeks_in_range('1900-01-01',CURRENT_DATE);
weeks_in_range
----------------
5854
(1 row)
Note, that the above example assumes Sunday to be the start of the week. Alternatively, if you want Monday as a start of the week, then use date_trunc, e.g.
CREATE OR REPLACE FUNCTION weeks_in_range(date,date) returns int AS $$
SELECT (((EXTRACT(DAY FROM (date_trunc('week', $2::date) - date_trunc('week', $1::date))))/7)+1)::int
$$ LANGUAGE SQL STABLE STRICT;
first_of_week is unnecessary in this case.

Answer 7

select ((in_FinishDate - in_StartDate) / 7 +
case
when ((date_part('dow', in_StartDate) = 6) and (date_part('dow', in_FinishDate) != 6)) or
((date_part('dow', in_FinishDate) = 0) and (date_part('dow', in_StartDate) != 0)) then 1
when (date_part('dow', in_StartDate) = 5) and (date_part('dow', in_FinishDate) in (1, 2, 3, 4)) then 1
when (date_part('dow', in_StartDate) = 4) and (date_part('dow', in_FinishDate) in (1, 2, 3)) then 1
when (date_part('dow', in_StartDate) = 3) and (date_part('dow', in_FinishDate) in (1, 2)) then 1
when (date_part('dow', in_StartDate) = 2) and (date_part('dow', in_FinishDate) in (1)) then 1
else 0
end)::int;
This should do the trick.

Answer 8

Rebuttal to #maleckicoa. The following demonstrates why one cannot user a constant of 52 weeks in a year for difference between dates spanning the Dec-to-Jan time frame.
Instead of hard coding Dec 2019 - Jan 2020 (which works just fine) the following generates the annual dates 29-Dec to 4-Jan for the period 2019 through 2030. In most the maleckicoa's algorithum works, but ...
with recursive annual_dates(dec_dt, jan_dt) as
( select make_date(2019,12,29), make_date(2020,01,04)
union all
select (dec_dt+interval '1 year')::date, (jan_dt+interval '1 year')::date
from annual_dates
where extract(isoyear from dec_dt)::integer < 2030
) --select * from annual_dates;
select dec_dt, jan_dt,
extract( isoyear from jan_dt)*52 + extract( week from jan_dt)
-
extract( isoyear from dec_dt)*52 - extract( week from dec_dt) "Weeks Between"
from annual_dates;
with recursive annual_dates(dec_dt, jan_dt) as
( select make_date(2019,12,29), make_date(2020,01,04)
union all
select (dec_dt+interval '1 year')::date, (jan_dt+interval '1 year')::date
from annual_dates
where extract(isoyear from dec_dt)::integer < 2030
) --select * from annual_dates;
select 'Week for ' || dec_dt || ' is ' || extract( week from dec_dt) || ' And '
'Week for ' || jan_dt || ' is ' || extract( week from jan_dt) || ' Giving Difference of '
|| extract( isoyear from jan_dt)*52 + extract( week from jan_dt)
-
extract( isoyear from dec_dt)*52 - extract( week from dec_dt) || ' Weeks Between'
from annual_dates;

Answer 9

This should work in Postgres for most cases (Week starts with Monday, the year has 52 weeks)
Example for year end 2019:
select
extract( isoyear from '2020-01-04'::date)*52 + extract( week from '2020-01-04'::date)
-
extract( isoyear from '2019-12-29'::date)*52 - extract( week from '2019-12-29'::date)