I am using Oracle SQL Developer and I need to pull counts from a very large database.
I need to count both the number of visits and the number of unique visitors at each of 5 sites (and in total) for each of 10 quarters (and in total) - resulting in 120 counts. Each row in this database represents one visit, and the visitors each have a unique visitor_ID.
Currently, I have one line for each count, but I need to make modifications and I don't want to do it in such an inefficient way this time.
select
sum(case when visit_date between '01-JAN-2019 00:00:00' and '31-MAR-2019 00:00:00' and site=site1 then 1 else 0 end) as 19Q1_visits_site1,
count(distinct case when visit_date between '01-JAN-2019 00:00:00' and '31-MAR-2019 00:00:00' and site=site1 then visitor_id) as 19Q1_unique_site1,
[...]
from visitdata
where [additional qualifiers];
If possible, I would like to create something along the lines of:
19Q1 = visit_date between '01-JAN-2019 00:00:00' and '31-MAR-2019 00:00:00'
19Q2 = visit_date between '01-APR-2019 00:00:00' and '30-JUN-2019 00:00:00'
[...]
allQ = visit_date between '01-JAN-2019 00:00:00' and '30-SEP-2021 00:00:00'
S1 = site in (site1)
[...]
allS = site in (site1, site2, site3, site4, site5)
sites = [S1, S2, S3, S4, S5, allS]
quarters = [19Q1, 19Q2, ..., allQ]
for s in sites:
for q in quarters:
select
sum(case when q and s then 1 else 0 end) as (str(q) + 'visits' + str(s)),
count(distinct case when q and s then visitor_id) as (str(q) + 'unique' + str(s))
from visitdata
where [additional qualifiers];
I know SQL doesn't do for loops. Any advice would be fantastic so I don't have to create another embarrassing script with almost 200 lines. Thanks!
Very briefly, use a built-in or user-defined function that takes a date and returns the quarter to which that date belongs. (SQL Server supports "quarter" as a datepart, but you can probably write this yourself if Oracle doesn't support it. You could also add visit_quarter to your table as a materialized computed column and even index it if you use this a lot.) Then you can write a single grouping query along the lines of
SELECT
site,
quarter(visit_date) as Q,
COUNT(visitor_id) as numvisits,
COUNT(DISTINCT visitor_id) AS numDistinctVisitors
FROM T
WHERE <additional conditions>
GROUP BY site, quarter(visit_date)
ORDER BY site, Q
For finding quarters you can use TO_CHAR() on dates, and for changing the start date and end date of a quarter, you can use ADD_MONTHS() eg
Table
create table randomdates( date_ )
as
select sysdate - dbms_random.value( 1, 600 )
from dual
connect by level <= 50 ;
Query
select date_
, to_char( date_, 'YYYY-Q' ) quarter
from randomdates
;
-- result
DATE_ QUARTER
09/23/2021 2021-3
09/24/2020 2020-3
03/23/2020 2020-1
03/29/2021 2021-1
11/29/2020 2020-4
03/05/2021 2021-1
04/08/2021 2021-2
...
GROUP BY should also be possible - as #Steve Kass suggested.
select to_char( date_, 'YYYY-Q' ), count(*)
from randomdates
group by to_char( date_, 'YYYY-Q' )
order by 1 desc
;
TO_CHAR(DATE_,'YYYY-Q') COUNT(*)
2021-4 6
2021-3 13
2021-2 22
2021-1 13
2020-4 20
2020-3 12
2020-2 10
2020-1 4
...
From your comment:
I'm using a modified fiscal year anyway. Do you know how I can define
quarters when the visit_date is stored as yyyymmdd?
If you need different start/end days for the, ADD_MONTHS() could help eg the modified_quarters in the following query start a month later than the "standard" quarters. Regarding the dates: in Oracle, they contain the century, the year within the century, the month, the day of the month, the hour, the minute and the second (7 bytes). You can just use TO_CHAR() and pick up whichever component (of the date) you need using the "Format Model" eg 'Q' in the example below.
-- Query executed in APEX.
-- Column date_ : no formatting (compare the output to the same query in the dbfiddle).
select date_
, to_char( date_, 'YYYY-Q' ) quarter
, to_char( date_, 'YY' ) || 'Q' || to_char( date_, 'Q' ) quarter_
, to_char( add_months( date_, 1 ), 'YYYY-Q' ) modified_quarter
from randomdates
;
DATE_ QUARTER QUARTER_ MODIFIED_QUARTER
09/23/2021 2021-3 21Q3 2021-4
09/24/2020 2020-3 20Q3 2020-4
03/23/2020 2020-1 20Q1 2020-2
03/29/2021 2021-1 21Q1 2021-2
11/29/2020 2020-4 20Q4 2020-4
03/05/2021 2021-1 21Q1 2021-2
For calculating subtotals and total (counts) per site, you could use GROUP BY ROLLUP() eg
Table & data
-- Caution: dates in this table are not the same as in the randomdates table.
create table sitesanddates( site_, date_ )
as
select
trunc( dbms_random.value( 1, 6 ) )
, sysdate - dbms_random.value( 1, 600 )
from dual
connect by level <= 50 ;
-- group by site and quarter
select site_, to_char( date_, 'YYYY-Q' ), count(*)
from sitesanddates
group by site_, to_char( date_, 'YYYY-Q' )
order by 1, 2
;
SITE_ TO_CHAR(DATE_,'YYYY-Q') COUNT(*)
1 2020-1 1
1 2020-4 3
1 2021-1 1
1 2021-2 1
1 2021-3 2
2 2020-1 1
2 2020-2 1
GROUP BY ROLLUP
select site_, to_char( date_, 'YYYY-Q' ) q_, count(*)
from sitesanddates
group by rollup( site_, to_char( date_, 'YYYY-Q' ) )
order by 1, 2
;
SITE_ Q_ COUNT(*)
1 2020-1 1
1 2020-4 3
1 2021-1 1
1 2021-2 1
1 2021-3 2
1 - 8 -- <- subtotal for site 1
2 2020-1 1
2 2020-2 1
...
5 2020-4 2
5 2021-1 2
5 2021-2 1
5 2021-4 1
5 - 10 -- <- subtotal for site 5
- - 50 -- <- grand total
link to dbfiddle
Related
This question already has answers here:
How to get the previous working day from Oracle?
(4 answers)
Closed 1 year ago.
need help for some oracle stuff ..
I need to get Day-1 from sysdate, holiday and weekend will be excluded .
And for holiday, we need to get the range to get the last workday before holiday.
The start date and end date will coming from my holiday table.
ex :
Holiday Table
HolidayName
Start_date
End_Date
holiday1
5th Aug'21
6th Aug'21
condition :
this query run on 9th Aug 2021
expected result :
4th Aug'21
I've tried some query and function but I just can't get what I need.
Thanks a lot for help!
Here's one way to do it.
select max(d) as last_workday
from (select trunc(sysdate)-level as d from dual connect by level < 30) prior_month
where to_char(d, 'DY') not in ('SAT','SUN')
and not exists (select holidayname from holiday_table
where prior_month.d between start_date and end_date)
;
Without seeing your Holiday table, it's hard to say how many days back you would need to look to find the last workday. If you have a holiday that lasts for more than 30 days, you'll need to change the 30 to a larger number.
You can use a simple case expression to determine what day of the week the start of your holiday is, then subtract a number of days based on that.
WITH
holiday (holidayname, start_date, end_date)
AS
(SELECT 'holiday1', DATE '2021-8-5', DATE '2021-8-6' FROM DUAL
UNION ALL
SELECT 'Christmas', DATE '2021-12-25', DATE '2021-12-26' FROM DUAL
UNION ALL
SELECT 'July 4th', DATE '2021-7-4', DATE '2021-7-5' FROM DUAL)
SELECT holidayname,
start_date,
end_date,
start_date - CASE TO_CHAR (start_date, 'Dy') WHEN 'Mon' THEN 3 WHEN 'Sun' THEN 2 ELSE 1 END AS prior_business_day
FROM holiday;
HOLIDAYNAME START_DATE END_DATE PRIOR_BUSINESS_DAY
______________ _____________ ____________ _____________________
holiday1 05-AUG-21 06-AUG-21 04-AUG-21
Christmas 25-DEC-21 26-DEC-21 24-DEC-21
July 4th 04-JUL-21 05-JUL-21 02-JUL-21
You can use a recursive sub-query factoring clause from this answer:
WITH start_date (dt) AS (
SELECT DATE '2021-05-02' FROM DUAL
),
days ( dt, day, found ) AS (
SELECT dt,
TRUNC(dt) - TRUNC(dt, 'IW'),
0
FROM start_date
UNION ALL
SELECT dt - CASE day WHEN 0 THEN 3 WHEN 6 THEN 2 ELSE 1 END,
CASE WHEN day IN (0, 6, 5) THEN 4 ELSE day - 1 END,
CASE WHEN h.start_date IS NULL THEN 1 ELSE 0 END
FROM days d
LEFT OUTER JOIN holidays h
ON ( dt - CASE day WHEN 0 THEN 3 WHEN 6 THEN 2 ELSE 1 END
BETWEEN h.start_date AND h.end_date )
WHERE found = 0
)
SELECT dt
FROM days
WHERE found = 1;
Which, for the sample data:
CREATE TABLE holidays (HolidayName, Start_date, End_Date) AS
SELECT 'holiday1', DATE '2021-08-05', DATE '2021-08-06' FROM DUAL;
Outputs:
DT
2021-08-04 00:00:00
db<>fiddle here
Don't know if it's very efficient. Did it just for fun
create table holidays (
holiday_name varchar2(100) primary key,
start_date date not null,
end_date date not null
)
/
Table created
insert into holidays (holiday_name, start_date, end_date)
values ('holiday1', date '2021-08-05', date '2021-08-06');
1 row inserted
with days_before(day, wrk_day) as
(select trunc(sysdate - 1) d,
case
when h.holiday_name is not null then 0
when to_char(trunc(sysdate - 1), 'D') in ('6', '7') then 0
else 1
end work_day
from dual
left join holidays h
on trunc(sysdate - 1) between h.start_date and h.end_date
union all
select db.day - 1,
case
when h.holiday_name is not null then 0
when to_char(db.day - 1, 'D') in ('6', '7') then 0
else 1
end work_day
from days_before db
left join holidays h
on db.day - 1 between h.start_date and h.end_date
where db.wrk_day = 0) search depth first by day set order_no
select day from days_before where wrk_day = 1;
DAY
-----------
04.08.2021
Using Oracle 12c, I need to run a script on an existing summary table of projects. The summary table has a project, a start date, and an end date. I need to break this data out into the number of days per month for each project.
An example is Project A has a start date of 2/10/2016 and an end date of 3/10/2016. My ending result for this example should be:
Project A, February, 19
Project A, March, 10
This was an easier one as some dates may span 2 or 3 months. This doesn't seem too difficult but for some reason I'm having trouble wrapping my head around it and overthinking it. Does someone have an quick and easy solution to this? I would like to run this as a SQL statement but a PL/SQL script would also work.
In this solution we don't assume any prior knowledge of the time period covered. Also, this solution does not use joins (which may be important for performance).
with
-- begin test data (this section can be deleted)
inputs ( project, start_date, end_date ) as (
select 'A', date '2014-10-03', date '2014-12-15' from dual union all
select 'B', date '2015-03-01', date '2015-03-31' from dual union all
select 'C', date '2015-11-30', date '2016-03-01' from dual
),
-- end test data; solution begins here (it includes the word "with" from the first line)
prep ( project, end_date, dt ) as (
select project, end_date, start_date from inputs union all
select project, end_date, end_date + 1 from inputs union all
select project, end_date, add_months( trunc(start_date, 'mm'), level )
from inputs
connect by add_months (trunc(start_date, 'mm'), level) <= end_date
and prior project = project
and prior sys_guid() is not null
),
computations ( project, dt, month, day_count ) as (
select project, dt, to_char(dt, 'Mon-yyyy'),
lead(dt) over (partition by project order by dt) - dt
from prep
where dt <= end_date + 1
)
select project, month, day_count
from computations
where day_count > 0
order by project, dt
;
OUTPUT:
PROJECT MONTH DAY_COUNT
------- -------- ---------
A Oct-2014 29
A Nov-2014 30
A Dec-2014 15
B Mar-2015 31
C Nov-2015 1
C Dec-2015 31
C Jan-2016 31
C Feb-2016 29
C Mar-2016 1
9 rows selected
If you can do an assumption on the maximum number of days for a project (1000 in my example), you can use the following:
with yourTable(project, startDate, endDate) as
(
select 'Project a' as project,
date '2016-02-10' as startDate,
date '2016-03-10' as endDate
from dual
UNION ALL
select 'Project XX',
date '2016-01-01',
date '2016-01-10'
from dual
)
select project, to_char(startDate + n, 'MONTH'), count(1)
from yourTable
inner join (
select level n
from dual
connect by level <= 1000
)
on (startDate + n <= endDate)
group by project, to_char(startDate + n, 'MONTH')
The part with the CONNECT BY is used as a date generator, assuming that every project is at maximum 1000 days long; the external query uses the date generator to split the row of a project in many rows, one for each day between start and end date, and then aggregates by month and project to build the output.
A slightly different way, based on months and not days, could be:
select project, to_char(add_months(startDate, n ), 'MONTH'),
case
when trunc(add_months(startDate, n ), 'MONTH') = trunc(add_months(endDate, n ), 'MONTH')
then endDate - startDate +1
when trunc(add_months(startDate, n ), 'MONTH') <= startDate
then last_day(add_months(startDate, n)) - startDate
when last_day(add_months(startDate, n )) >= endDate
then endDate - trunc(add_months(startDate, n ), 'MONTH') +1
else
last_day(add_months(startDate, n )) - trunc(last_day(add_months(startDate, n )), 'MONTH')
end as numOfDays
from yourTable
inner join (
select level -1 n
from dual
connect by level <= 1000
)
on trunc(add_months(startDate, n ), 'MONTH') <= trunc(endDate, 'MONTH')
This is a bit more complicated, to handle the different cases, but more efficient, given that it works at month level, not day level
I think you're after something like:
WITH sample_data AS (SELECT 'A' PROJECT, to_date('10/02/2016', 'dd/mm/yyyy') start_date, to_date('10/03/2016', 'dd/mm/yyyy') end_date FROM dual UNION ALL
SELECT 'B' PROJECT, to_date('10/02/2016', 'dd/mm/yyyy') start_date, to_date('10/06/2016', 'dd/mm/yyyy') end_date FROM dual UNION ALL
SELECT 'C' PROJECT, to_date('10/02/2016', 'dd/mm/yyyy') start_date, to_date('18/02/2016', 'dd/mm/yyyy') end_date FROM dual)
SELECT PROJECT,
to_char(add_months(trunc(start_date, 'mm'), LEVEL -1), 'fmMonth yyyy', 'nls_date_language=english') mnth,
CASE WHEN trunc(end_date, 'mm') = add_months(trunc(start_date, 'mm'), LEVEL -1)
THEN end_date
ELSE add_months(trunc(start_date, 'mm'), LEVEL) -1
END - CASE WHEN trunc(start_date, 'mm') = add_months(trunc(start_date, 'mm'), LEVEL -1)
THEN start_date + 1
ELSE add_months(trunc(start_date, 'mm'), LEVEL -1)
END + 1 num_days
FROM sample_data
CONNECT BY PRIOR PROJECT = PROJECT
AND PRIOR sys_guid() IS NOT NULL
AND add_months(trunc(start_date, 'mm'), LEVEL -1) <= TRUNC(end_date, 'mm');
PROJECT MNTH NUM_DAYS
------- -------------- ----------
A February 2016 19
A March 2016 10
B February 2016 19
B March 2016 31
B April 2016 30
B May 2016 31
B June 2016 10
C February 2016 8
This uses the multi-row connect-by-level technique (the presence of the and prior sys_guid() is not null enables the connect by to loop through each row separately) to loop through each project row in the sample_data table (you presumably have the project information in a table already, so you wouldn't need to have the sample_data subquery at all; you could just reference your table directly in the main SQL).
We then compare the month of the start date with the month of the row being generated by the connect by, and if it's the same month, then we know we need to use the start date, otherwise we use the first of the month of the generated row; we do similarly for the end date.
That way, we can now subtract one from the other and make adjustments to make the calculation correct. You may need to tweak this yourself if you need a start and end date of the same day to count as 1 day, rather than 0 - it'll probably need an extra case statement to take account of when the start and end date are in the same month.
Using this approach won't limit your project length; it could be as long as you liked.
ETA: Looks like Mathguy posted an answer whilst I was typing out my answer, and whilst our basic methods are the same, mine doesn't use an analytic function to determine the difference in the number of days. You may or may not find their answer more performant than mine - you should test both to see which one works best with your data.
I am woking on a query which can give back the count divided by month about the offices that will be closed this summer.
SELECT
qa.tmonth,
COUNT(qa.tmonth) AS qtn
FROM
(
SELECT TO_CHAR(CLOSURE_DATE, 'yyyymm') AS tmonth
FROM Holidays
WHERE CLOSURE_DATE >= TO_DATE('20160501', 'YYYY-MM-DD') AND
CLOSURE_DATE <= TO_DATE('20160901', 'YYYY-MM-DD')
) qa
GROUP BY qa.tmonth;
Since the months: May, June, August and September no office will be closed the output is the following:
TMONTH|QTN
201607|80
But I need a thing like this
TMONTH|QTN
201605|0
201606|0
201607|80
201608|0
201609|0
How could I achieve that?
Thanks to all!
You can try with something like this:
SQL> with holidays(closure_date) as
2 (
3 select date '2016-07-01' from dual union all
4 select date '2016-07-02' from dual union all
5 select date '2016-07-03' from dual union all
6 select date '2016-07-04' from dual union all
7 select date '2016-07-05' from dual
8 )
9 select count(closure_date) as closure_days, to_char(day, 'yyyymm') as month
10 from (
11 select date '2016-05-01' + level -1 as day
12 from dual
13 connect by date '2016-05-01' + level -1 <= date '2016-09-30'
14 ) days
15 left outer join holidays
16 on (day = closure_date)
17 group by to_char(day, 'yyyymm') ;
CLOSURE_DAYS MONTH
------------ ------
0 201608
5 201607
0 201606
0 201605
0 201609
SQL>
This uses a query to build the list of all the days between a starting and an ending date; I used 01/05 and 30/09 and called it days.
Then it queries days with the holidays table in outer join; this way you can count only the days for which there is a corrensponding value in the closure days list, thus counting the closure days for each day, month year; the aggregation for year and month completes the job
A similar approach like above. Tip: You can execute the two sub-queries separately, to analyse the logic.
select to_char (m.month, 'yyyymm') as TMONTH, m.month,
nvl (h.qtn, 0) as QTN
from
(
SELECT add_months(trunc (SYSDATE, 'MONTH'), -(LEVEL-1)) as MONTH
FROM dual
CONNECT BY LEVEL <= 12 -- generate a list of the last 12 month
) m
left join
(
SELECT trunc (closure_date, 'MONTH') as MONTH,
count (*) as QTN
FROM Holidays
group by trunc (closure_date, 'MONTH')
) h
on m.MONTH = h.MONTH
where m.month between DATE '2016-01-01' and sysdate
order by TMONTH desc;
The question I need to answer is this "What is the maximum number of page requests we have ever received in a 60 minute period?"
I have a table that looks similar to this:
date_page_requested date;
page varchar(80);
I'm looking for the MAX count of rows in any 60 minute timeslice.
I thought analytic functions might get me there but so far I'm drawing a blank.
I would love a pointer in the right direction.
You have some options in the answer that will work, here is one that uses Oracle's "Windowing Functions with Logical Offset" feature instead of joins or correlated subqueries.
First the test table:
Wrote file afiedt.buf
1 create table t pctfree 0 nologging as
2 select date '2011-09-15' + level / (24 * 4) as date_page_requested
3 from dual
4* connect by level <= (24 * 4)
SQL> /
Table created.
SQL> insert into t values (to_date('2011-09-15 11:11:11', 'YYYY-MM-DD HH24:Mi:SS'));
1 row created.
SQL> commit;
Commit complete.
T now contains a row every quarter hour for a day with one additional row at 11:11:11 AM. The query preceeds in three steps. Step 1 is to, for every row, get the number of rows that come within the next hour after the time of the row:
1 with x as (select date_page_requested
2 , count(*) over (order by date_page_requested
3 range between current row
4 and interval '1' hour following) as hour_count
5 from t)
Then assign the ordering by hour_count:
6 , y as (select date_page_requested
7 , hour_count
8 , row_number() over (order by hour_count desc, date_page_requested asc) as rn
9 from x)
And finally select the earliest row that has the greatest number of following rows.
10 select to_char(date_page_requested, 'YYYY-MM-DD HH24:Mi:SS')
11 , hour_count
12 from y
13* where rn = 1
If multiple 60 minute windows tie in hour count, the above will only give you the first window.
This should give you what you need, the first row returned should have
the hour with the highest number of pages.
select number_of_pages
,hour_requested
from (select to_char(date_page_requested,'dd/mm/yyyy hh') hour_requested
,count(*) number_of_pages
from pages
group by to_char(date_page_requested,'dd/mm/yyyy hh')) p
order by number_of_pages
How about something like this?
SELECT TOP 1
ranges.date_start,
COUNT(data.page) AS Tally
FROM (SELECT DISTINCT
date_page_requested AS date_start,
DATEADD(HOUR,1,date_page_requested) AS date_end
FROM #Table) ranges
JOIN #Table data
ON data.date_page_requested >= ranges.date_start
AND data.date_page_requested < ranges.date_end
GROUP BY ranges.date_start
ORDER BY Tally DESC
For PostgreSQL, I'd first probably write something like this for a "window" aligned on the minute. You don't need OLAP windowing functions for this.
select w.ts,
date_trunc('minute', w.ts) as hour_start,
date_trunc('minute', w.ts) + interval '1' hour as hour_end,
(select count(*)
from weblog
where ts between date_trunc('minute', w.ts) and
(date_trunc('minute', w.ts) + interval '1' hour) ) as num_pages
from weblog w
group by ts, hour_start, hour_end
order by num_pages desc
Oracle also has a trunc() function, but I'm not sure of the format. I'll either look it up in a minute, or leave to see a friend's burlesque show.
WITH ranges AS
( SELECT
date_page_requested AS StartDate,
date_page_requested + (1/24) AS EndDate,
ROWNUMBER() OVER(ORDER BY date_page_requested) AS RowNo
FROM
#Table
)
SELECT
a.StartDate AS StartDate,
MAX(b.RowNo) - a.RowNo + 1 AS Tally
FROM
ranges a
JOIN
ranges b
ON a.StartDate <= b.StartDate
AND b.StartDate < a.EndDate
GROUP BY a.StartDate
, a.RowNo
ORDER BY Tally DESC
or:
WITH ranges AS
( SELECT
date_page_requested AS StartDate,
date_page_requested + (1/24) AS EndDate,
ROWNUMBER() OVER(ORDER BY date_page_requested) AS RowNo
FROM
#Table
)
SELECT
a.StartDate AS StartDate,
( SELECT MIN(b.RowNo) - a.RowNo
FROM ranges b
WHERE b.StartDate > a.EndDate
) AS Tally
FROM
ranges a
ORDER BY Tally DESC
I have a query in Oracle for a report that looks like this:
SELECT TRUNC (created_dt) created_dt
, COUNT ( * ) AllClaims
, SUM(CASE WHEN filmth_cd in ('T', 'C') THEN 1 ELSE 0 END ) CT
, SUM(CASE WHEN filmth_cd = 'W' THEN 1 ELSE 0 END ) Web
, SUM(CASE WHEN filmth_cd = 'I' THEN 1 ELSE 0 END ) Icon
FROM claims c
WHERE c.clsts_cd NOT IN ('IN', 'WD')
AND TRUNC (created_dt) between
to_date('1/1/2006', 'dd/mm/yyyy') AND
to_date('1/1/2100', 'dd/mm/yyyy')
GROUP BY TRUNC (created_dt)
ORDER BY TRUNC (created_dt) DESC;
It returns data like this:
Create_Dt AllClaims CT Web Icon
1/26/2011 675 356 285 34
1/25/2011 740 322 379 39
...
What I need is a result set that sums all of the daily values into a weekly value. I am pretty new to PL/SQL and not sure where to begin.
Something like
SELECT TRUNC (created_dt, 'IW') created_dt
, COUNT ( * ) AllClaims
, SUM(CASE WHEN filmth_cd in ('T', 'C') THEN 1 ELSE 0 END ) CT
, SUM(CASE WHEN filmth_cd = 'W' THEN 1 ELSE 0 END ) Web
, SUM(CASE WHEN filmth_cd = 'I' THEN 1 ELSE 0 END ) Icon
FROM claims c
WHERE c.clsts_cd NOT IN ('IN', 'WD')
AND TRUNC (created_dt) between
to_date('1/1/2006', 'dd/mm/yyyy') AND
to_date('1/1/2100', 'dd/mm/yyyy')
GROUP BY TRUNC (created_dt, 'IW')
ORDER BY TRUNC (created_dt, 'IW') DESC;
will aggregate the data based on the first day of the ISO week.
Here is a simple example that I quite often use:
SELECT
TO_CHAR(created_dt,'WW'),
max(created_dt),
COUNT(*)
from MY_TABLE
group by
TO_CHAR(created_dt,'WW');
The to_char(created_dt,'WW') create the groups, and the max(created_dt) displays the last day of the week. Use min(created_dt) if you want to display the first day of the week.