I have a sales table and want to get the day of week for the orders and the number of orders placed on each day of the week. Everything seems to work but no matter what I try the 'week' is non standard. I've tried SET DATEFIRST 1; but still do not get the results I want.
SELECT DATENAME(Weekday,orderdate) AS Weekday, COUNT(orderid) AS NumOrders
FROM Sales.Orders
GROUP BY DATENAME(Weekday,orderdate);
The results:
Weekday | NumOrders
1. Wednesday 25
2. Saturday 33
3. Monday 100
4. Sunday 115
5. Thursday 87
6. Tuesday 42
Could the no orders for Friday be causing the output not to order by start of week? Thanks.
I think you just need an ORDER BY:
SELECT DATENAME(Weekday, orderdate) AS Weekday, COUNT(orderid) AS NumOrders
FROM Sales.Orders
GROUP BY DATENAME(Weekday, orderdate), DATEPART(Weekday, orderdate)
ORDER BY DATEPART(Weekday, orderdate);
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
In PostgreSQL (I'm on version 9.6.6), what's the simplest way to get the week number, starting on Sunday?
DATE_PART('week',x) returns:
The number of the ISO 8601 week-numbering week of the year. By definition, ISO weeks start on Mondays and the first week of a year contains January 4 of that year. In other words, the first Thursday of a year is in week 1 of that year. (doc)
Say my query is like:
WITH dates as (SELECT generate_series(timestamp '2014-01-01',
timestamp '2014-01-31',
interval '1 day'
)::date AS date
)
SELECT
date,
TO_CHAR(date,'Day') AS dayname,
DATE_PART('week',date) AS weekofyear
FROM dates
Returns:
date dayname weekofyear
--------------------------------
2014-01-01 Wednesday 1
2014-01-02 Thursday 1
2014-01-03 Friday 1
2014-01-04 Saturday 1
2014-01-05 Sunday 1 <- I want this to be 2
2014-01-06 Monday 2
2014-01-07 Tuesday 2
2014-01-08 Wednesday 2
So far I have tried:
SELECT
date,
TO_CHAR(date,'Day') AS dayname,
DATE_PART('week',date) AS week_iso,
DATE_PART('week',date + interval '1 day') AS week_alt
FROM dates
which won't quite work if the year begins on a Sunday.
Also, I want week 1 to contain January 1 of that year. So if January 1 is a Saturday, I want week 1 to be one day long (instead of being week 53 in the ISO style). This behavior is consistent with the Excel WEEKNUM function.
To get the week number of the year, with weeks starting on Sunday, we need to know how many Sundays between the first day of the year and the target date.
I adapted the solution here by #Erwin Brandstetter. This solution counts Sundays inclusive of the first day of the year and exclusive of the target date.
Then, because I want the first (partial) week to be week one (not zero), I need to add 1 unless the first day of the year is a Sunday (in which case it's already week one).
WITH dates as (SELECT generate_series(timestamp '2014-01-01',
timestamp '2014-01-31',
interval '1 day'
)::date AS date
)
SELECT
date,
TO_CHAR(date,'Day') AS dayname,
DATE_PART('week',date) AS week_iso,
((date - DATE_TRUNC('year',date)::date) + DATE_PART('isodow', DATE_TRUNC('year',date)) )::int / 7
+ CASE WHEN DATE_PART('isodow', DATE_TRUNC('year',date)) = 7 THEN 0 ELSE 1 END
AS week_sundays
FROM dates
Returns
date dayname weekofyear week_sundays
--------------------------------
2014-01-01 Wednesday 1 1
2014-01-02 Thursday 1 1
2014-01-03 Friday 1 1
2014-01-04 Saturday 1 1
2014-01-05 Sunday 1 2
2014-01-06 Monday 2 2
2014-01-07 Tuesday 2 2
To show how this works for years starting on Sunday:
2017-01-01 Sunday 52 1
2017-01-02 Monday 1 1
2017-01-03 Tuesday 1 1
2017-01-04 Wednesday 1 1
2017-01-05 Thursday 1 1
2017-01-06 Friday 1 1
2017-01-07 Saturday 1 1
2017-01-08 Sunday 1 2
The task is not as daunting as it first appears. It mainly requires finding the first Sun on or after the 1-Jan. That date becomes the last day of the first week. From there calculation of subsequent weeks is merely. a matter of addition. The other significant point is with week definition there will always be 53 week per year and the last day of the last week is 31-Dec. The following generates an annual calendar for this week definition.
create or replace function non_standard_cal(year_in integer)
returns table (week_number integer, first_day_of_week date, last_day_of_week date)
language sql immutable leakproof strict rows 53
as $$
with recursive cal as
(select 1 wk, d1 start_of_week, ds end_of_week, de stop_date
from (select d1+substring( '0654321'
, extract(dow from d1)::integer+1
, 1)::integer ds
, d1, de
from ( select make_date (year_in, 1,1) d1
, make_date (year_in+1, 1,1) -1 de
) a
) b
union all
select wk+1, end_of_week+1, case when end_of_week+7 > stop_date
then stop_date
else end_of_week+7
end
, stop_date
from cal
where wk < 53
)
select wk, start_of_week, end_of_week from cal;
$$ ;
As a general rule I avoid magic numbers, but sometimes they're useful; as in this case. In magic number (actually a string) '0654321' each digit represents the number of days needed to reach the first Mon on or after 1-Jan when indexed by the standard day numbering system (0-6 as Sun-Sat). The result is the Mon being the last day of the first week. That generatess the 1st row of the recursive CTE. The remaining rows just add the appropriate number days for each week until the 53 weeks have been generated. The following shows the years needed to ensure each day of week gets it's turn to 1-Jan (yea some days duplicate). Run individual years to validate its calendar.
do $$
declare
cal record;
yr_cal cursor (yr integer) for
select * from non_standard_cal(2000+yr) limit 1;
begin
for yr in 18 .. 26
loop
open yr_cal(yr);
fetch yr_cal into cal;
raise notice 'For Year: %, week: %, first_day: %, Last_day: %, First day is: %'
, 2000+yr
,cal.week_number
,cal.first_day_of_week
,cal.last_day_of_week
,to_char(cal.first_day_of_week, 'Day');
close yr_cal;
end loop;
end; $$;
Following may work - tested with two cases in mind:
WITH dates as (SELECT generate_series(timestamp '2014-01-01',
timestamp '2014-01-10',
interval '1 day'
)::date AS date
union
SELECT generate_series(timestamp '2017-01-01',
timestamp '2017-01-10',
interval '1 day'
)::date AS date
)
, alt as (
SELECT
date,
TO_CHAR(date,'Day') AS dayname,
DATE_PART('week',date) AS week_iso,
DATE_PART('week',date + interval '1 day') AS week_alt
FROM dates
)
select date, dayname,
week_iso, week_alt, case when week_alt <> week_iso
then week_alt
else week_iso end as expected_week
from alt
order by date
Output:
date dayname week_iso week_alt expected_week
2014-01-01 Wednesday 1 1 1
2014-01-02 Thursday 1 1 1
2014-01-03 Friday 1 1 1
2014-01-04 Saturday 1 1 1
2014-01-05 Sunday 1 2 2
2014-01-06 Monday 2 2 2
2014-01-07 Tuesday 2 2 2
....
2017-01-01 Sunday 52 1 1
2017-01-02 Monday 1 1 1
2017-01-03 Tuesday 1 1 1
2017-01-04 Wednesday 1 1 1
2017-01-05 Thursday 1 1 1
2017-01-06 Friday 1 1 1
2017-01-07 Saturday 1 1 1
2017-01-08 Sunday 1 2 2
This query works perfectly replacing monday with sunday as the start of the week.
QUERY
SELECT CASE WHEN EXTRACT(day from '2014-01-05'::date)=4 AND
EXTRACT(month from '2014-01-05'::date)=1 THEN date_part('week',
'2014-01-05'::date) ELSE date_part('week', '2014-01-05'::date + 1)
END;
OUTPUT
date_part
-----------
2
(1 row)
I am trying to write a query which gives the number of days in each month between two specified dates.
Example:
date 1: 2018-01-01
date 2: 2018-05-23
Expected Output:
month days
2018-01-01, 31
2018-02-01, 28
2018-03-01, 31
2018-04-01, 30
2018-05-01, 23
Use generate_series and group by date_trunc
SELECT date_trunc('month',dt) AS month,
COUNT(*) as days
FROM generate_series( DATE '2018-01-01',DATE '2018-05-23',interval '1 DAY' )
as dt group by date_trunc('month',dt)
order by month;
Demo
I would like to insert subquery a date based on it day. Plus, each date can only be used four times. Once it reached fourth times, the fifth value will use another date of same day. In other word, use date of Monday of next week. Example, Monday with 6 JUNE 2016 to Monday with 13 JUNE 2016 (you may check the calendar).
I have a query of getting a list of date based on presentationdatestart and presentationdateend from presentation table:
select a.presentationid,
a.presentationday,
to_char (a.presentationdatestart + delta, 'DD-MM-YYYY', 'NLS_CALENDAR=GREGORIAN') list_date
from presentation a,
(select level - 1 as delta
from dual
connect by level - 1 <= (select max (presentationdateend - presentationdatestart)
from presentation))
where a.presentationdatestart + delta <= a.presentationdateend
and a.presentationday = to_char(a.presentationdatestart + delta, 'fmDay')
order by a.presentationdatestart + delta,
a.presentationid; --IMPORTANT!!!--
For example,
presentationday presentationdatestart presentationdateend
Monday 01-05-2016 04-06-2016
Tuesday 01-05-2016 04-06-2016
Wednesday 01-05-2016 04-06-2016
Thursday 01-05-2016 04-06-2016
The query result will list all possible dates between 01-05-2016 until 04-06-2016:
Monday 02-05-2016
Tuesday 03-05-2016
Wednesday 04-05-2016
Thursday 05-05-2016
....
Monday 30-05-2016
Tuesday 31-05-2016
Wednesday 01-06-2016
Thursday 02-06-2016 (20 rows)
This is my INSERT query :
insert into CSP600_SCHEDULE (studentID,
studentName,
projectTitle,
supervisorID,
supervisorName,
examinerID,
examinerName,
exavailableID,
availableday,
availablestart,
availableend,
availabledate)
select '2013816591',
'mong',
'abc',
'1004',
'Sue',
'1002',
'hazlifah',
2,
'Monday', //BASED ON THIS DAY
'12:00:00',
'2:00:00',
to_char (a.presentationdatestart + delta, 'DD-MM-YYYY', 'NLS_CALENDAR=GREGORIAN') list_date //FOR AVAILABLEDATE
from presentation a,
(select level - 1 as delta
from dual
connect by level - 1 <= (select max (presentationdateend - presentationdatestart)
from presentation))
where a.presentationdatestart + delta <= a.presentationdateend
and a.presentationday = to_char(a.presentationdatestart + delta, 'fmDay')
order by a.presentationdatestart + delta,
a.presentationid;
This query successfully added 20 rows because all possible dates were 20 rows. I would like modify the query to be able to insert based on availableDay and each date can only be used four times for each different studentID.
Possible outcome in CSP600_SCHEDULE (I am removing unrelated columns to ease readability):
StudentID StudentName availableDay availableDate
2013 abc Monday 01-05-2016
2014 def Monday 01-05-2016
2015 ghi Monday 01-05-2016
2016 klm Monday 01-05-2016
2010 nop Tuesday 02-05-2016
2017 qrs Tuesday 02-05-2016
2018 tuv Tuesday 02-05-2016
2019 wxy Tuesday 02-05-2016
.....
2039 rrr Monday 09-05-2016
.....
You may check the calendar :)
I think what you're asking for is to list your students and then batch them up in groups of 4 - each batch is then allocated to a date. Is that right?
In which case something like this should work (I'm using a list of tables as the student names just so I don't need to insert any data into a custom table) :
WITH students AS
(SELECT table_name
FROM all_tables
WHERE rownum < 100
)
SELECT
table_name
,SYSDATE + (CEIL(rownum/4) -1)
FROM
students
;
I hope that helps you
...okay, following your comments, I think this might be a better solution :
WITH students AS
(SELECT table_name student_name
FROM all_tables
WHERE rownum < 100
)
, dates AS
(SELECT TRUNC(sysdate) appointment_date from dual UNION
SELECT TRUNC(sysdate+2) from dual UNION
SELECT TRUNC(sysdate+4) from dual UNION
SELECT TRUNC(sysdate+6) from dual UNION
SELECT TRUNC(sysdate+8) from dual UNION
SELECT TRUNC(sysdate+10) from dual UNION
SELECT TRUNC(sysdate+12) from dual UNION
SELECT TRUNC(sysdate+14) from dual
)
SELECT
s.student_name
,d.appointment_date
FROM
--get a list of students each with a sequential row number, ordered by student name
(SELECT
student_name
,ROW_NUMBER() OVER (ORDER BY student_name) rn
FROM students
) s
--get a list of available dates with a sequential row number, ordered by date
,(SELECT
appointment_date
,ROW_NUMBER() OVER (ORDER BY appointment_date) rn
FROM dates
) d
WHERE 1=1
--allocate the first four students to date rownumber1, next four students to date rownumber 2...
AND CEIL(s.rn/4) = d.rn
;
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;