I want to increment days when I select a date in SQLite and it points to weekends,
here is the code.
SELECT strftime('%Y-%d-%m','now', '+9 days');
Assuming what you want Saturdays and Sundays to be the following Monday then the following may do what you wish :-
SELECT CASE
WHEN strftime('%w','now') = 0 THEN strftime('%Y-%d-%m','now','+1 days') /* add 1 day when Sunday */
WHEN strftime('%w','now')= 6 THEN strftime('%Y-%d-%m','now','+2 days') /* add 2 days when Saturday */
ELSE strftime('%Y-%d-%m','now')
END AS adjusted_date;
If not then as you can see %w returns the days of the week where 0 is Sunday and 6 is Saturday. and you can use CASE WHEN THEN ELSE clause to conditionally returns values.
Date And Time Functions
Related
I have the following postgresql table;
id | date_slot
------+-------------------------
1 | [2023-02-08,2023-02-15)
2 | [2023-02-20,2023-02-26)
3 | [2023-02-27,2023-03-29)
I want to make a query that return rows contained in these ranges but exclude weekends
for example the query I made return the following but does not exclude the weekends.
SELECT * FROM table where '2023-02-11'::date <# date_slot;
id | date_slot
------+-------------------------
1 | [2023-02-08,2023-02-15)
2023-02-11 is a weekend so it must not return a result. How can I do that?
Selecting workday-only dateranges (without weekends):
You can check what day of the week it is on the first day in the range using extract() and knowing its length from upper()-lower(), determine if it'll cross a weekend: online demo
select *
from test_table
where '2023-02-11'::date <# date_slot
and extract(isodow from lower(date_slot)
+ (not lower_inc(date_slot))::int)
+( (upper(date_slot) - (not upper_inc(date_slot))::int)
-(lower(date_slot) + (not lower_inc(date_slot))::int) ) < 6 ;
Cases where your ranges have varying upper and lower bound inclusivity are handled by lower_inc() and upper_inc() - their boolean result, when cast to int, just adds or subtracts a day to account for whether it's included by the range or not.
The range is on or includes a weekend if it starts on a weekend day or continues for too long from any other day of the week:
4 days, if it starts on a Monday (isodow=1)
3 days, if it starts on a Tuesday (isodow=2)
2 days, if it starts on a Wednesday (isodow=3)
1 days, if it starts on a Thursday (isodow=4)
0 days, if it starts on a Friday (isodow=5)
This means the isodow of range start date and the range length cannot sum up to more than 5 for the range not to overlap a weekend.
You can also enumerate the dates contained by these ranges using generate_series() and see if they include a Saturday (day 6) or a Sunday (0 as dow, 7 as isodow):
select *
from test_table
where '2023-02-11'::date <# date_slot
and not exists (
select true
from generate_series(
lower(date_slot) + (not lower_inc(date_slot))::int,
upper(date_slot) - (not upper_inc(date_slot))::int,
'1 day'::interval) as alias(d)
where extract(isodow from d) in (6,7) );
Selecting records based on workday-only dates:
First comment got it right
select *
from table_with_dateranges dr,
table_with_dates d
where d.date <# dr.date_slot
and extract(isodow from d.date) not in (6,7);
Consider the date "2022-07-02"
For the month July first week only have 3 days in it.
I need to find the number of days in the week for the given date.
In above date the week has 3 days where "2022-07-02" day reside.
Example 2 :
For month June in 2022 first week has 5 days in the week
Therefore if i declare a date as "2022-06-03" it should pass the number of days in the week as 5
I need a query to find the number of days for the specific week.
set datefirst 1 -- assumes Monday is set as start of week
declare #myDate date = getdate();
-- calculate the next last day of week
declare #nextSunday date = dateadd(day, 7 - datepart(weekday, #myDate), #myDate);
select case
-- advancing into the next month implies a partial week
-- datediff(month, #myDate, nextSunday) = 1 would be equivalent
when day(#nextSunday) < day(#myDate) then 7 - day(#nextSunday)
-- else see if still within first week
when day(#nextSunday) < 7 then day(#nextSunday)
else 7 end;
Within a query you might use it this way:
select case
when day(nextSunday) < day(dateColumn) then 7 - day(nextSunday)
when day(nextSunday) < 7 then day(nextSunday)
else 7 end
from
myData cross apply (
values (dateadd(day, 7 - datepart(weekday, dateColumn), dateColumn))
) v(nextSunday);
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=ee5bfb52dabe31dd619cfd136689db59
If you don't want the shorthand form then just replace every instance of nextSunday in the final step with its full expression.
There's nothing in the logic that prevents this from working with another first day of week. I just chose a variable name that helped ellucidate this particular problem.
I have a date in sql which will always fall on a Monday and I'm subtracting 13 weeks to get a weekly report. I am trying to get the same 13 week report but for last year's figures as well.
At the moment, I'm using the following:
calendar_date >= TRUNC(sysdate) - 91
which is working fine.
I need the same for last year.
However, when I split this into calendar weeks, there will also be a partially complete week as it will include 1 or 2 days from the previous week. I need only whole weeks.
e.g. the dates that will be returned for last year will be 14-Feb-2015 to 16-May-2015. I need it to start on the Monday and be 16-Feb-2015. This will change each week as I am only interested in complete weeks...
I would do this:
Get the date by substracting 91 days as you're already doing.
Get the number of the day of the week with TO_CHAR(DATE,'d')
Add the number of days until the next monday to the date.
Something like this:
SELECT TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/YYYY')-91 + MOD(7 - TO_NUMBER(TO_CHAR(TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/RRRR')-91,'d'))+1,7) d
FROM dual
next_day - returns date of first weekday named by char.
with dates as (select to_date('16/05/2015','DD/MM/YYYY') d from dual)
select
trunc(next_day( trunc(d-91) - interval '1' second,'MONDAY'))
from dates;
I want to get next monday from calculated date. In situation when calculated date is monday i have to move back to previous week ( -1 second).
I need to select next friday from current date.
For example:
Today SELECT
Monday 6/24/2013 6/28/2013
Wednesday 6/26/2013 6/28/2013
Friday 6/28/2013 6/28/2013
Saturday 6/29/2013 7/5/2013
I found a few solutions close to what I want but all they do is select fridat of the day's week.
I need to use this date as a Default Value in the table column. When a user inserts a record, this column should automatically be set to the folloing Friday. Say, instead of getdate() in the 'Default Value or Binding', I need to put this statement that selects following Friday.
select DATEADD(day,6-datepart(weekday,the_dt),the_dt)
+ Case when datepart(weekday,the_dt) = 7 then 7 else 0 end
from table
I'm trying to select data grouped by week, which I have working, but I need to be able to specify a different day as the last day of the week. I think something needs to go near INTERVAL (6-weekday('datetime')) but not sure. This kind of SQL is above my pay-grade ($0) :P
SELECT
sum(`value`) AS `sum`,
DATE(adddate(`datetime`, INTERVAL (6-weekday(`datetime`)) DAY)) AS `dt`
FROM `values`
WHERE id = '123' AND DATETIME BETWEEN '2010-04-22' AND '2010-10-22'
GROUP BY `dt`
ORDER BY `datetime`
Thanks!
select
sum(value) as sum,
CASE WHEN (weekday(datetime)<=3) THEN date(datetime + INTERVAL (3-weekday(datetime)) DAY)
ELSE date(datetime + INTERVAL (3+7-weekday(datetime)) DAY)
END as dt
FROM values
WHERE id = '123' and DATETIME between '2010-04-22' AND '2010-10-22'
GROUP BY dt
ORDER BY datetime
This does look pretty evil but, this query will provide you with a sum of value grouped by a week ending on a Thursday (weekday() return of 3).
If you wish to change what day the end of the week is you just need to replace the 3's in the case statement, ie if you wanted Tuesday you would have it say
CASE WHEN (weekday(datetime)<=1) THEN date(datetime + INTERVAL (1-weekday(datetime)) DAY)
ELSE date(datetime + INTERVAL (1+7-weekday(datetime)) DAY)
I hope this helps.
Simple solution that I like. This will return the date for the start of the week assuming the week ends Sunday and starts Monday.
DATE(`datetime`) - INTERVAL WEEKDAY(`datetime`) AS `dt`
This can easily be adjusted to have a week ending on Thursday because Thursday is 3 days earlier than Sunday
DATE(`datetime`) - INTERVAL WEEKDAY(`datetime` + INTERVAL 3 DAY) AS `dt`
this returns for the start of the week that starts on Friday and ends on Thursday.
You can group on this no problem. If you want to use get the end of the week based on the start you do this
DATE(`datetime`) - INTERVAL -6 + WEEKDAY(`datetime` + INTERVAL 3 DAY) AS `dt`
I think you must choose between Sunday and Monday? When you can use DATE_FORMAT for grouping by string format of date, and use %v for grouping by Mondays and %v for grouping by Sundays.
SELECT
sum(`value`) AS `sum`,
DATE_FORMAT(`datetime`,'%v.%m.%Y') AS `dt`
FROM `values`
WHERE id = '123' AND DATETIME BETWEEN '2010-04-22' AND '2010-10-22'
GROUP BY DATE_FORMAT(`datetime`,'%v.%m.%Y')
ORDER BY `datetime`
How to use DATE_FORMAT
I don't remember the exact math, but you can get WEEKDAY to wrap around on different days of the week by adding or subtracting days to its argument. You'll need to tinker with different values of x and y in the expression:
x-weekday(adddate(`datetime`, INTERVAL y DAY))