Group by Week Number and Week Ending - sql

I am trying to show sql query by grouping by week number and to show week ending of the week rather beginning of week but thus far have been futile in achieving this. How can I do this?
select extract(week from actual_sale_date) as week_number,
to_char(date_trunc('week', actual_sale_date) as date, 'MM/dd/yyyy'), count(*)
from data
where project_id = 'ABC'
and actual_sale_date >= date_trunc('year',current_date)
group by rollup( (actual_sale_date))
Result:
week_number date count
1 01/02/2017 2
1 01/02/2017 1
2 01/09/2017 1
2 01/09/2017 1
2 01/09/2017 1
3 01/16/2017 3
3 01/16/2017 1
10
Requested:
week_number week_ending count
1 01/08/2017 3
2 01/15/2017 3
3 01/22/2017 4
10

You were grouping by the actual_sale_date hence the results for a week weren't getting aggregated by week. To get week ending date, add 6 days to the start of week. Use week_number and week ending date in rollup.
select extract(week from actual_sale_date) as week_number,
to_char(date_trunc('week', actual_sale_date) + interval '6' day,'MM/dd/yyyy'),
count(*)
from data
where project_id = 'ABC'
and actual_sale_date >= date_trunc('year',current_date)
group by rollup((extract(week from actual_sale_date)
,to_char(date_trunc('week', actual_sale_date) + interval '6' day,'MM/dd/yyyy')))

Related

SQL Statement that will pull MINDATE from dates for current month and year

I need a sql statement that will pull the following
the MINIMUM visit date for the current month and year (May 2021)
emp_id visit_date
1 5/3/2021
1 5/4/2021
1 5/10/2021
1 6/5/2021
2 5/2/2021
2 5/6/2021
2 6/7/2021
3 5/9/2021
3 7/8/2021
Result:
1 5/3/2021
2 5/2/2021
3 5/9/2021
SELECT emp_id, visit_date
FROM
THS_ClientVisit
WHERE
THS_ClientVisit.visit_date =
(
SELECT MIN(THS_ClientVisit.visit_date)
FROM dbo.THS_ClientVisit
WHERE MONTH YEAR from current month[i.e. 05-2021] = MONTH YEAR from visit_date {i.e. 05-2021]
);
Thank you for your assistance!
SELECT
emp_id,
MIN(CONVERT(date,rev_timein)) as MinVisitDate
FROM
THS_ClientVisit
WHERE
DATEPART(mm, THS_ClientVisit.rev_timein)= MONTH(CURRENT_TIMESTAMP) AND DATEPART(yy, THS_ClientVisit.rev_timein)=YEAR(CURRENT_TIMESTAMP)
GROUP BY
THS_ClientVisit.emp_id

Split single row into multiple rows based on week of the month in date

I have a table like this
Id Valid_From Valid_To
9744 24/06/2019 07/07/2019
9745 12/08/2019 31/12/9999
I would like to split this table into multiple rows based on the week like this by joining to the date table
Id Valid_from Valid_To Month Week
9744 24/06/2019 07/07/2019 June 4
9744 24/06/2019 07/07/2019 July 1
9744 24/06/2019 07/07/2019 July 2
9745 12/08/2019 31/12/9999 August 2
9745 12/08/2019 31/12/9999 August 3
9745 12/08/2019 31/12/9999 August 4
In this case there will be 3 rows as the valid from and valid two falls between these 3 weeks for ID - 9744
For ID - 9745 the Valid_to date is infinity so we need to just take all the week in the current month from the valid_from date
I then just need to append the output with Month and the Week number
Can someone help me to write a query to have this output?
Thanks
You mention a "date" table. If you have one then you can use a join like this:
select distinct t.id, valid_from, t.valid_to, d.month, d.week
from yourtable t join
date d
on d.date >= t.valid_from and
d.date <= t.valid_to;
If I understand your question right, you need to list all month names and week numbers of these months' existing between valid_from and valid_to dates. I did it by following query:
SELECT
Q.ID,
Q.VALID_FROM,
Q.VALID_TO,
Q.MONTH_NAME,
WEEK_NUMBER
FROM
(
SELECT
CEIL((Q.DATES_BETWEEN_INTERVAL - FIRST_DAY_OF_MONTH + 1) / 7) WEEK_NUMBER,
TO_CHAR(Q.DATES_BETWEEN_INTERVAL, 'MONTH', 'NLS_DATE_LANGUAGE = American') MONTH_NAME,
Q.*
FROM
(
SELECT
LEVEL + S.VALID_FROM DATES_BETWEEN_INTERVAL,
TRUNC(LEVEL + S.VALID_FROM, 'MONTH') FIRST_DAY_OF_MONTH,
S.* FROM
(
SELECT T.*,
(CASE WHEN EXTRACT(YEAR FROM T.VALID_TO) = 9999 THEN LAST_DAY(T.VALID_FROM) ELSE T.VALID_TO END) - T.VALID_FROM DAYS_COUNT
FROM AAA_TABLE T
) S
CONNECT BY LEVEL <= S.DAYS_COUNT
) Q
) Q
GROUP BY
Q.ID,
Q.VALID_FROM,
Q.VALID_TO,
Q.MONTH_NAME,
WEEK_NUMBER
ORDER BY
Q.ID,
Q.VALID_FROM,
Q.VALID_TO,
Q.MONTH_NAME,
WEEK_NUMBER;
But there must be 5th week if the date greater than 28th day of month. Hope this will help you.

day of week function for week day aggregates

I currently have a query that reads from a table and aggregates based on category. It gives me what I need but I"m trying to add another column that looks at all records for that category/employee combo for the days of this past week. SO if the job with this query runs on Wednesday Night, it needs to get a total of all category/employee records for Monday and Tuesday Night as well.
The query:
SELECT employee,
sum(case when category = 'Shoes' and date_of_report >= current_date - 1 days then daily_total else 0 end) as Shoes_DAILY,
sum(case when category = 'Shoes' and date_of_report >= ( current date - ( dayofweek(current date) - 1 ) days ) then sum(daily_total) else 0 end) as dailyTotalWeek
from shoeTotals
where date_of_report >= current_date
group by employee;
So the third column there is what's messing me up saying function use not valid. here's what I want:
The source table has these records for this past week:
employee | daily_total | date_of_report
--------------------------------------------------
123 14 2019-08-26
123 1 2019-08-27
123 56 2019-08-28
123 6 2019-08-29
123 8 2019-08-30 * today
My desired output would get (based on employee and category) the total for today (8) and then the sum of all the employees' records for that category on each preceding weekday. Running on Monday night would only count that days records, friday night would count monday through friday's as shown above.
employee | shoes_daily | dailyTotalWeek
--------------------------------------------------
123 8 85
What am I doing wrong with the dayofweek function?
You cannot nest aggregation functions. I think you simply want:
select employee,
sum(case when category = 'Shoes' and date_of_report >= current_date - 1 days
then daily_total else 0
end) as Shoes_DAILY,
sum(case when category = 'Shoes' and date_of_report >= ( current date - ( dayofweek(current date) - 1 ) days )
then daily_total else 0
end) as dailyTotalWeek
from shoeTotals
where date_of_report >= current date - ( dayofweek(current date) - 1 ) days
group by employee;

How do I compare a current partial month vs a previous partial month with postgres?

I'm building some basic reports and I want to see if I'm on track to surpass last month's metrics without waiting for the month to end. Basically I want to compare June 1 (start of current month) through June 23 (current_date) against May 1 (start of previous month) through May 23 (current_date - 1 month).
My goal is to show a count of distinct users that did event1 and event2.
Here's what I have so far:
CREATE VIEW events AS
(SELECT *
FROM public.event
WHERE TYPE in ('event1',
'event2')
AND created_at > now() - interval '1 months' );
CREATE VIEW MAU AS
(SELECT EXTRACT(DOW
FROM created_at) AS month,
DATE_TRUNC('week', created_at) AS week,
COUNT(*) AS total_engagement,
COUNT(DISTINCT user_id) AS total_users
FROM events
GROUP BY 2,
1
ORDER BY week DESC);
SELECT month,
week,
SUM(total_engagement) OVER (PARTITION BY month
ORDER BY week) AS total_engagment
FROM MAU
ORDER BY 1 DESC,
2
Here's an example of what that returns:
Month Week Unique Engagement
6 2017-05-22 00:00:00 165
6 2017-05-29 00:00:00 355
6 2017-06-05 00:00:00 572
6 2017-06-12 00:00:00 723
5 2017-05-22 00:00:00 757
5 2017-05-29 00:00:00 1549
5 2017-06-05 00:00:00 2394
5 2017-06-12 00:00:00 3261
5 2017-06-19 00:00:00 3592
Expected return
Month Day Total Engagement
6 1 50
6 2 100
6 3 180
5 1 89
5 2 213
5 3 284
5 4 341
Can you point out where I've got this wrong or if there's an easier way to do it?
You are confusing days, weeks and months in your question but from the expected output I assume that you want month number, week number within a month and a count of those pairs.
SELECT
month,
week,
count(*) as total_engagement
FROM (
SELECT
extract(month from created_at) as month,
extract('day' from date_trunc('week', created_at::date) -
date_trunc('week', date_trunc('month', created_at::date))) / 7 + 1 as week
FROM public.event
WHERE type IN ('event1', 'event2')
AND created_at > now() - interval '1 month'
) t
GROUP BY 1,2
The most interesting part could be getting the week number within a month and for that you can check this answer.

SQL count number of users every 7 days

I am new to SQL and I need to find count of users every 7 days. I have a table with users for every single day starting from April 2015 up until now:
...
2015-05-16 00:00
2015-05-16 00:00
2015-05-17 00:00
2015-05-17 00:00
2015-05-17 00:00
2015-05-17 00:00
2015-05-17 00:00
2015-05-18 00:00
2015-05-18 00:00
...
and I need to count the number of users every 7 days (weekly) so I have data weekly.
SELECT COUNT(user_id), Activity_Date FROM TABLE_NAME
I need output like this:
TotalUsers week1 week2 week3 ..........and so on
82 80 14 16
I am using DB Visualizer to query Oracle database.
You should try following,
Select
sum(Week1) + sum(Week2) + sum(Week3) + sum(Week4) + sum(Week5) as Total,
sum(Week1) as Week1,
sum(Week2) as Week2,
sum(Week3) as Week3,
sum(Week4) as Week4,
sum(Week5) as Week5
From (
select
case when week = 1 then 1 else 0 end as Week1,
case when week = 2 then 1 else 0 end as Week2,
case when week = 3 then 1 else 0 end as Week3,
case when week = 4 then 1 else 0 end as Week4,
case when week = 5 then 1 else 0 end as Week5
from
(
Select
CEILING(datepart(dd,visitdate)/7+1) week,
user_id
from visitor
)T
)D
Here is Fiddle
You need to add month & year in the result as well.
SELECT COUNT(user_id), Activity_Date FROM TABLE_NAME WHERE Activity_Date > '2015-06-31';
That would get the amount of users for the last 7 days.
This is my test table:
user_id act_date
1 01/04/2015
2 01/04/2015
3 04/04/2015
4 05/04/2015
..
This is my query:
select week_offset, count(*) nb from (
select trunc((act_date-to_date('01042015','DDMMYYYY'))/7) as week_offset from test_date)
group by week_offset
order by 1
and this is the output:
week_offset nb
0 6
1 3
4 5
5 7
6 3
7 1
18 1
Week offset is the number of the week from 01/04/2015, and we can show the first day of the week.
See here for live testing.
How do you define your weeks? Here's an approach for SQL Server that starts each seven-day block relative to the start of April. The expressions will vary according to your specific needs:
select
dateadd(
dd,
datediff(dd, cast('20150401' as date), Activity_Date) / 7 * 7,
cast('20150401' as date)
) as WeekStart,
count(*)
from T
group by datediff(dd, cast('20150401' as date), Activity_Date) / 7
Oracle:
select
trunc(Activity_date, 'DAY') as WeekStart,
count(*)
from T
group by trunc(Activity_date, 'DAY') /* D and DAY are the same thing */