I want to find all of the employees whose birthday is upcoming in the next 14 days.
I've tried using WHERE DATEDIFF(day, birthday_dt, CURRENT_DATE) <=14
SELECT
full_name,
birthday_dt,
FROM
Employees
WHERE
DATEDIFF(day, birthday_dt, CURRENT_DATE) <=14;
I expect the results to show as:
Walter White | 08-29-1957
Mickey Mouse | 09-01-1999
But I received empty results because no employee is 14 days old.
SELECT
full_name,
birthday_dt,
FROM
Employees
WHERE
365 - DATEDIFF(day, birthday_dt, current_date)%365 <= 14;
Related
I want to count %days when a user was active. A query like this
select
a.id,
a.created_at,
CURRENT_DATE - a.created_at::date as days_since_registration,
NOW() as current_d
from public.accounts a where a.id = 3257
returns
id created_at days_since_registration current_d tot_active
3257 2022-04-01 22:59:00.000 1 2022-04-02 12:00:0.000 +0400 2
The person registered less than 24 hours ago (less than a day ago), but there are two distinct dates between the registration and now. Hence, if a user was active one hour before midnight and one hour after midnight, he is two days active in less than a day (active 200% of days)
What is the right way to count distinct dates and get 2 for a user, who registered at 23:00:00 two hours ago?
WITH cte as (
SELECT 42 as userID,'2022-04-01 23:00:00' as d
union
SELECT 42,'2022-04-02 01:00:00' as d
)
SELECT
userID,
count(d),
max(d)::date-min(d)::date+1 as NrOfDays,
count(d)/(max(d)::date-min(d)::date+1) *100 as PercentageOnline
FROM cte
GROUP BY userID;
output:
userid
count
nrofdays
percentageonline
42
2
2
100
I have a pickupDate and returnDate in my OrderHistory table. I want to extract the sum of rental days of all OrderHistory entries, grouped/ordered by month. A cte seems to be the solution but I don´t get how to implement it in my query since the cte´s i saw were refering to themselves where it says "FROM cte".
I tried something like this:
SELECT
SUM((EXTRACT (DAY FROM("OrderHistory"."returnDate")-("OrderHistory"."pickupDate")))) as traveltime
, to_char("OrderHistory"."pickupDate"::date, 'YYYY-MM') as M
FROM
"OrderHistory"
GROUP BY
M
ORDER BY
M
But the outcome doesn´t split bookings btw two months (e.g. pickupDate=27th march 2022 and returnDate=03rd of april 2022) but will assign the whole 7 days to the month of march, since the returndate is in it. It should show 4 days in march and 3 in april.
Sorry for the probably very stupid question but I am a beginner. (my code is written in postgresql btw)
PostgreSQL naming conventions
Are PostgreSQL column names case-sensitive?
use legal, lower-case names exclusively so double-quoting is not
needed.
Final result in db fiddle
Add daterange column.
alter table order_history add column date_ranges daterange;
update order_history
with a(m_begin, m_end, pickup_date) as
(select date_trunc('month', pickup_date)::date,
(date_trunc('month', pickup_date) + interval '1 month - 1 day')::date,
pickup_date from order_history)
update order_history set date_ranges =
daterange(a.m_begin, a.m_end,'[]') from a
where a.pickup_date = order_history.pickup_date;
then final query:
WITH A AS(
select
pickup_date,
return_date,
return_date - pickup_date as total,
case when return_date <# date_ranges then (return_date - pickup_date)
else ( date_trunc('month', pickup_date) + interval '1 month - 1 day')::date - pickup_date
end partial_mth
from order_history),
b as (SELECT *, a.total - partial_mth parital_not_mth FROM a)
select *,
case when to_char(pickup_date,'YYYY-MM') = to_char(return_date,'YYYY-MM')
then
sum(partial_mth) over(partition by to_char(pickup_date,'YYYY-MM')) +
sum(parital_not_mth) over (partition by to_char(return_date,'YYYY-MM'))
else sum(partial_mth) over(partition by to_char(pickup_date,'YYYY-MM'))
end
from b;
After trying different things I think I found the best answer to my question, that I want to share with the community:
WITH hier as (
SELECT
"OrderHistory"."pickupDate" as start_date
, "OrderHistory"."returnDate" as end_date
, to_char("OrderHistory"."pickupDate"::date, 'YYYY-MM') as M
FROM
"OrderHistory"
GROUP BY
1, 2, 3
ORDER BY
3
), calendar as (
select date '2022-01-01' + (n || ' days')::interval calendar_date
from generate_series(0, 365) n
)
select
to_char(calendar_date::date, 'YYYY-MM')
, count(*) as tage_gebucht
from calendar
inner join hier on calendar.calendar_date between start_date and end_date
where calendar_date between '2022-01-01' and '2022-12-31'
group by 1
order by 1;
I think this is the simplest solution I came up with.
I users table and a jobs. User has many jobs and jobs have a start_date and end_date:
Column | Type | Modifiers
----------------+-----------------------------+---------------------------------------------------
id | integer | not null default nextval('jobs_id_seq'::regclass)
title | character varying |
employer | character varying |
start_date | date |
end_date | date |
user_id | integer |
I need to calculate the total number of months that a person has spent working within the past X years.
I've looked at OVERLAPS and played with intervals a bit but I can't quite figure out what I need. I want to make sure that even it the start_date is outside the X years range that I still count the months that are inside the range.
Here is what I have so far:
select sum(EXTRACT(YEAR FROM months) * 12 + EXTRACT(MONTH FROM months))
as working_months
from (
select CASE current
WHEN true THEN
age(current_date, start_date)
ELSE age(end_date, start_date)
END as months
from jobs inner join users on jobs.user_id = users.id
where users.id = 4
) as employment_time;
with jobs (start_date, end_date, user_id) as ( values
('2000-01-01'::date, '2005-12-31'::date, 1),
('2007-10-01', '2008-09-30', 1),
('2010-09-01', '2014-10-20', 1)
)
select
user_id,
extract(year from work_time) * 12 + extract(month from work_time) as months
from (
select
user_id,
sum(age(upper(period), lower(period))) as work_time
from (
select
user_id,
daterange(start_date, end_date, '[]') *
daterange((current_date - interval '10 years')::date, current_date)
as period
from jobs
) s
group by user_id
) s
;
user_id | months
---------+--------
1 | 70
Range type -
Range functions
The basic query would be this:
SELECT sum(extract(year from months) * 12 + extract(month from months)) AS working_months
FROM (
SELECT
age(CASE (start_date, start_date) OVERLAPS (current_date, interval '-5 years')
WHEN true THEN start_date
ELSE current_date - interval '5 years'
END AS strt::timestamp,
CASE current
WHEN true THEN current_date
ELSE end_date
END AS fin::timestamp) AS months
FROM jobs
WHERE user_id = 4) AS employment_time;
You may also put this in a SQL function with parameters for the number of years and user_id. Note that you throw away partial months from individual jobs. You can add extract(day from months) / 30 to the top SELECT to harvest those partial months into full months.
This assumes that jobs cannot overlap. If they do, then the query becomes much more complex.
I got the information I needed from my last post about Postgres: Defining the longest streak (in days) per developer.
However now I want know the longest streak per developer regardless of Saturdays or Sundays. For instance, Bob worked from Thursday 18, Friday 19, Monday 22 and Tuesday 23, hence Bob streak is 4 days.
I understand I can use the DOW window function, which gives me 0 as Sunday , 1 Monday and so on. But
I don’t see how I can apply DOW function in the last solution proposed by Gordon Linoff.
Can some of you help me in this matter? Cheers,
WITH
working_limits AS (
SELECT
MIN(mr_date) AS start_date,
MAX(mr_date) AS end_date
FROM
xxx
),
working_days AS (
SELECT
ROW_NUMBER() OVER () AS day_number,
s.d::date AS date
FROM
GENERATE_SERIES((SELECT start_date FROM working_limits),
(SELECT end_date FROM working_limits),
'1 day') AS s(d)
WHERE
EXTRACT(dow FROM s.d) BETWEEN 1 AND 5),
worked_days AS (
SELECT
ROW_NUMBER() OVER () AS day_number,
developer,
mr_date AS date
FROM
xxx
ORDER BY
developer,
mr_date
)
SELECT
y.developer,
MAX(y.days)
FROM (
SELECT
x.developer,
COUNT(*) AS days
FROM (
SELECT
wngd.date,
wd.developer,
wngd.day_number - wd.day_number AS delta
FROM
working_days wngd INNER JOIN worked_days wd
ON
wngd.date = wd.date) AS x
GROUP BY
x.developer,
x.delta) AS y
GROUP BY
y.developer;
I have a table with a column REGDATE, a registration date (YYYY-MM-DD HH:MM:SS). I would like to show an histogram (ExtJS) in order to understand in which period of the years users are signing up. I would like to do this for the past twelve months with respect to the current date and to group dates by week.
Any hints?
FWIW in PostgreSQL, Karaszi has an answer that works, but there is a faster query:
SELECT date_trunc('week', REGDATE) AS "Week" , count(*) AS "No. of users"
FROM <<TABLE>>
WHERE REGDATE > now() - interval '12 months'
GROUP BY 1
ORDER BY 1;
I based this off the work of Ben Goodacre
in MySQL:
SELECT COUNT(*), DATE_FORMAT(regdate, "%X%V") AS regweek FROM table GROUP BY regweek;
or
SELECT COUNT(*), YEARWEEK(NOW(), 2) as regweek FROM table GROUP BY regweek;
in PostgreSQL:
SELECT COUNT(*), EXTRACT(YEAR FROM regdate)::text || EXTRACT(WEEK FROM regdate)::text AS regweek FROM table GROUP BY regweek;
Maybe this?
select to_char(REGDATE,'WW') "Week number",
count(*) "number of signups",
from YOUR_TABLE
where REGDATE > current_date-365
group by to_char(REGDATE,'WW')
order by to_char(REGDATE,'WW')
Hint: (SQL)
SELECT CONVERT (VARCHAR(7), REGDATE, 120) AS [RegistrationMonth]
FROM ...
GROUP BY CONVERT (VARCHAR(7), REGDATE, 120)
ORDER BY CONVERT (VARCHAR(7), REGDATE, 120)