Oracle sql sort week days by current day - sql

I am trying to sort the days based on the order: Saturday, Sunday, Monday, Tuesday, Wednesday, Thursday, Friday.
I am trying using case:
select day,
CASE day
WHEN 1 THEN 1
WHEN 2 THEN 2
WHEN 3 THEN 3
WHEN 4 THEN 4
WHEN 5 THEN 5
WHEN 6 THEN 6
WHEN 7 THEN 7
else 0
END as day_nr
from week where day in (1,2,3,4,5,6,7)
order by day_nr asc
This is ok when I select all the days of the week. But if I want only for the day 1,5,6 the ordering is not correct. Gets the first day -Monday. How to proceed?

If you're trying to sort a set of dates by day of the week, with Saturday being the first, then consider ordering by a modified date:
create table t1(my_date date);
insert into t1
select trunc(sysdate)+rownum
from dual
connect by level <= 20
select
my_date,
to_char(my_date,'Day'),
to_char(my_date,'D')
from
t1
order by
to_char(my_date + 1,'D');
http://sqlfiddle.com/#!4/5940b/3
The downside is that it's not very intuitive, so add a code comment if you use this method.
Edit: Where you have a list of numbers, order by a case statement with either a list conversion:
case day
when 1 then 3
when 2 then 4
when 3 then 5
when 4 then 6
when 5 then 7
when 6 then 1 -- saturday
when 7 then 2
end
... or the more compact, but not as intuitive:
case
when day <= 5 then day + 2
else day - 5
end
order by case

In Oracle day 1 is Sunday by default.
SELECT * FROM
(
SELECT trunc(sysdate) + LEVEL-1 my_dt
, TO_CHAR(trunc(sysdate) + LEVEL-1, 'DY') Wk_Day
, TO_CHAR(trunc(sysdate) + LEVEL-2, 'D' ) Day#
FROM dual
CONNECT BY LEVEL <= 10
)
WHERE Day# IN (1,5,6)
ORDER BY my_dt, Day#
/
MY_DT WK_DAY DAY#
------------------------
5/10/2013 FRI 5
5/11/2013 SAT 6
5/13/2013 MON 1
5/17/2013 FRI 5
5/18/2013 SAT 6

Related

Oracle - Get all days from month

How create select which return all days from month where month = PARAMETER_MONTH eg. 5 and year = extract(year from sysdate).
1
2
3
..
30
31
"Row generator" is the search keyword. For example:
SQL> with temp (col) as
2 (select to_date(&par_month, 'mm') from dual)
3 select to_number(to_char(col + level - 1, 'dd')) day
4 from temp
5 connect by level <= last_day(col) - col + 1
6 order by day;
Enter value for par_month: 5
DAY
----------
1
2
3
4
5
6
<snip>
29
30
31
31 rows selected.
SQL>
TO_DATE function will convert entered value into the 1st day of that month in current year, so you don't have to worry about "and year = extract(year from sysdate)" you mentioned in question.

Calculate manual week number of year in Oracle SQL

i have a date column and along that i need to calculate another column in oracle for week number of they year, the weeks should be from sunday to saturday, starting first day of the year.
for example for current year
Week 1 : 1 Jan 2020 (Wednesday) - 4 Jan 2020(Saturday)
Week 2 : 5 Jan 2020 (Sunday) - 11 Jan 2020(Saturday)
. . . . .
Week 5 : 26 Jan 2020 (Sunday) - 1 Feb 2020 (Saturday)
and so on...
You need to write your own logic using a hierarchy query.
Something like the following:
SQL> SELECT WEEKNUMBER,
2 WEEK_START,
3 CASE WHEN WEEKNUMBER = 1 THEN FIRST_WEEKEND ELSE WEEK_START + 6 END AS WEEK_END
4 FROM
5 (SELECT
6 CASE
7 WHEN LEVEL = 1 THEN FIRST_DAY
8 ELSE FIRST_WEEKEND + ( LEVEL - 2 ) * 7 + 1
9 END AS WEEK_START,
10 FIRST_WEEKEND,
11 LEVEL AS WEEKNUMBER
12 FROM
13 ( SELECT
14 TRUNC(SYSDATE, 'YEAR') FIRST_DAY,
15 NEXT_DAY(TRUNC(SYSDATE, 'YEAR'), 'SATURDAY') FIRST_WEEKEND
16 FROM DUAL )
17 CONNECT BY
18 CASE WHEN LEVEL = 1 THEN FIRST_DAY
19 ELSE FIRST_WEEKEND + ( LEVEL - 2 ) * 7 + 1
20 END < ADD_MONTHS(TRUNC(SYSDATE, 'YEAR'), 12));
WEEKNUMBER WEEK_STAR WEEK_END
---------- --------- ---------
1 01-JAN-20 04-JAN-20
2 05-JAN-20 11-JAN-20
3 12-JAN-20 18-JAN-20
4 19-JAN-20 25-JAN-20
5 26-JAN-20 01-FEB-20
6 02-FEB-20 08-FEB-20
7 09-FEB-20 15-FEB-20
8 16-FEB-20 22-FEB-20
9 23-FEB-20 29-FEB-20
10 01-MAR-20 07-MAR-20
11 08-MAR-20 14-MAR-20
.......
.......
53 27-DEC-20 02-JAN-21
Cheers!!
One other option would be
with t as
(
select trunc(sysdate,'yyyy')+level-1 as day, to_char(trunc(sysdate,'yyyy')+level-1,'DY','NLS_DATE_LANGUAGE=AMERICAN') as weekday, level - 1 as lvl
from dual
connect by level <= 366
), t1 as
(
select case
when lvl = 0 then
day
else
case
when weekday = 'SUN' then
day
end
end as day1,
lvl
from t
), t2 as
(
select case
when weekday = 'SAT' then
day
end as day2,
lvl
from t
)
select concat( 'Week ', wk1) as week, day1, day2
from
(
select row_number() over (order by lvl) wk1, day1
from t1 where day1 is not null ) t1
left join
(
select row_number() over (order by lvl) wk2, day2
from t2 where day2 is not null ) t2
on wk2 = wk1
by using select .. from dual connect by level syntax and case..when expression to scan all the current year.
Demo

Getting a list of the next n days of the week?

I have two columns of weekdays and the number of days between these weekdays like this:
Monday | Saturday | 13
Tuesday | Friday | 54
Friday | Wednesday | 10
etc. I need to use SQL to make another column that contains a string with all the 13 or 54 weekdays between the two weekdays. So something like "Monday, Tuesday, Wednesday, ... Monday, Tuesday ... Friday, Saturday"
How can this be accomplished?
Thanks.
Below is for BigQuery Standard SQL
#standardSQL
SELECT start, duration,
REGEXP_EXTRACT(
REPEAT('Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,',CAST(duration/7+1 AS INT64)),
CONCAT(r'', start, r'(?:(?:,\w+){', CAST(duration - 1 AS STRING), '})')
) days_list
FROM `project.dataset.table`
you can test / play with it using dummy data from your question as below
#standardSQL
WITH `project.dataset.table` AS (
SELECT 'Monday' start, 'Saturday' finish, 13 duration UNION ALL
SELECT 'Tuesday', 'Friday', 54 UNION ALL
SELECT 'Friday', 'Wednesday', 10
)
SELECT start, duration,
REGEXP_EXTRACT(
REPEAT('Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,', CAST(duration / 7 + 1 AS INT64) ),
CONCAT(r'', start, r'(?:(?:,\w+){', CAST(duration - 1 AS STRING), '})')
) days_list
FROM `project.dataset.table`
with result as
Row start duration days_list
1 Monday 13 Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday
2 Tuesday 54 Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday
3 Friday 10 Friday,Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday

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 */

Counting the number of days excluding sunday between two dates

I am trying to calculate number of days betwen two dates excluding sundays. This is my query,
SELECT F_PLANHM_END_DT
- F_PLANHM_ST_DT
- 2
* (TO_CHAR (F_PLANHM_END_DT, 'WW') - TO_CHAR (F_PLANHM_ST_DT, 'WW'))
FROM VW_S_CURV_PROC
WHERE HEAD_MARK = 'IGG-BLH-BM 221';
SELECT COUNT (*)
FROM (SELECT SYSDATE + l trans_date
FROM ( SELECT LEVEL - 1 l
FROM VW_S_CURV_PROC
CONNECT BY LEVEL <= ( (SYSDATE + 7) - SYSDATE)))
WHERE TO_CHAR (trans_date, 'dy') NOT IN ('sun');
I am retrieving date from a view called VW_S_CURV_PROC with start date : F_PLANHM_ST_DT and end date F_PLANHM_END_DT. Somehow i cant make this to work. Please help me...
You could use the ROW GENERATOR technique to first generate the dates for a given range, and then exclude the SUNDAYs.
For example, this query will give me the total count of days between 1st Jan 2014 and 31st Dec 2014, excluding the Sundays -
SQL> WITH DATA AS
2 (SELECT to_date('01/01/2014', 'DD/MM/YYYY') date1,
3 to_date('31/12/2014', 'DD/MM/YYYY') date2
4 FROM dual
5 )
6 SELECT SUM(holiday) holiday_count
7 FROM
8 (SELECT
9 CASE
10 WHEN TO_CHAR(date1+LEVEL-1, 'DY','NLS_DATE_LANGUAGE=AMERICAN') <> 'SUN'
11 THEN 1
12 ELSE 0
13 END holiday
14 FROM data
15 CONNECT BY LEVEL <= date2-date1+1
16 )
17 /
HOLIDAY_COUNT
-------------
313
SQL>