Is there a way to find the last working day of last month? I know I can get last day of last month with SELECT (date_trunc('month', now())::date - 1), but how do I get the last "weekday"?
I believe this does what you want:
select (date_trunc('month', current_date) + interval '1 month' -
(case extract(dow from date_trunc('month', current_date) + interval '1 month')
when 0 then 2
when 1 then 3
else 1
end) * interval '1 day'
) as last_weekday_in_month
Such requests often suggest the need for a calendar table.
Piggy-backing off of Gordon's answer, I think this is what you want:
SELECT (
date_trunc('month', current_date) - interval '1 day' -
(case extract(dow from date_trunc('month', current_date) - interval '1 day')
when 0 then 2
when 6 then 1
else 0
end) * interval '1 day'
)::date as last_weekday_in_last_month;
Assuming your weekends are 0 (Sunday) and 6 (Saturday). It uses OP's original logic to find the last date of the last month, then Gordon's CASE logic to subtract more days if the last date is 0 or 6.
Related
I currently have 2 queries that I am trying to merge into 1. Basically I want to pull previous days sales if it's Tuesday - Friday and the previous 3 days sales if it's Monday. My queries are below - is there a way to do a conditional select for those date based on the day of the week?
Monday's version
SELECT *
FROM A
WHEN DATE_TRUNC('day', timestamp) IN (CURRENT_DATE - 1, CURRENT_DATE - 2, CURRENT_DATE - 3)
AND DATE_PART(weekday, current_date) = 1
Tuesday - Friday version
SELECT *
FROM A
WHEN DATE_TRUNC('day', timestamp) = CURRENT_DATE - 1
AND DATE_PART(weekday, current_date) = 1
one solution is :
SELECT *
FROM A
WHERE DATE_TRUNC('date', timestamp) <= DATE_TRUNC('date', CURRENT_DATE)
AND DATE_TRUNC('date', timestamp) >= case when DATE_TRUNC('day of week',CURRENT_DATE) = 'Monday' then DATE_TRUNC('date', CURRENT_DATE - 3) else DATE_TRUNC('date',CURRENT_DATE) end
You can use `OR:
SELECT *
FROM A
WHERE (DATE_TRUNC('day', timestamp) IN (CURRENT_DATE - 1, CURRENT_DATE - 2, CURRENT_DATE - 3) AND
DATE_PART(weekday, current_date) = 1
) OR
(DATE_TRUNC('day', timestamp) = CURRENT_DATE - 1 AND
DATE_PART(weekday, current_date) <> 1
)
This can be slightly simplified to:
WHERE timestamp > CURRENT_DATE -
(CASE WHEN DATE_PART(weekday, current_date) = 1 INTERVAL '3 DAY'
ELSE INTERVAL '1 DAY'
END)
Note: This assumes that there are no future timestamps.
I have below table
ABC Date
200 2019-02-22
-200 2019-02-23
1200 2019-02-24
-500 2019-02-25
'
'
'
'
-889 2015-01-11
I need to get values for from ABC for every day of last week of every month
select ABC
from table 1
where date between '2019-03-26' and '2019-03-30'
this is for month of march 2019. How do i create a loop such that it displays value for everyday of last week of every month for 3 years
You can use date arithmetic to get the last week of each month. In Terdata, I think this is one solution:
select abc
from table1
where (date >= (current_date - extract(day from date) * interval '1 day') - interval '6 day' and
date <= current_date - extract(day from date) * interval '1 day'
) or
(date >= (current_date - extract(day from date) * interval '1 day') - interval '1 month' - interval '6 day' and
date <= current_date - extract(day from date) * interval '1 day' - interval '1 month'
) or
(date >= (current_date - extract(day from date) * interval '1 day') - interval '2 month' - interval '6 day' and
date <= current_date - extract(day from date) * interval '1 day' - interval '12month'
);
SELECT ABC, DATE FROM table_1 WHERE DATEPART(wk, DATE) =
DATEPART(wk, EOMONTH(DATE)) AND DATE <= DATEADD(year,3,GETDATE())
DATEPART(wk, DATE) gives me the week number of that date, DATEPART(wk,EOMONTH(DATE)) gives me the week number of (the last day of the corresponding date's month). So, when I check this, I will only select dates belonging to the last week of every month. The next filter is to select only those dates which are lesser than 3 years from now (GETDATE()).
I'm trying to grab all the appointments that have been booked for this current week I initially tried this:
select *
from appointments
where staff_id = 'id'
and (to_timestamp(appointments.start_time)) > (current_date - interval '1 week')
and (to_timestamp(appointments.start_time)) < (current_date + interval '1 week')
This just selects all the appointments that equal +/- 7 days, so not the current week. This was my next try. I know its broken but I think it gets the point across.
select *
from appointments
where staff_id = 'id'
and (to_timestamp(appointments.start_time)) > (current_date - interval concat(extract(dow from current_date), ' days'))
and (to_timestamp(appointments.start_time)) < (current_date + interval concat(7 - extract(dow from current_date), ' days'))
How would I go about building this query?
If you agree with Postgres's definition of a week, then you can do:
select a.*
from appointments a
where staff_id = 'id' and
to_timestamp(appointments.start_time) >= date_trunc('week', current_date) and
to_timestamp(appointments.start_time) < date_trunc('week', current_date) + interval '1 week';
I have a recursive query that provides the number of orders placed week over week (week_no, week_start, and week_end). I'd like to create a similar breakdown for a month over month analysis.
WITH recursive weeks (week_start, week_end, time_end, weekno) AS (
VALUES ('2015-12-27'::date, '2016-01-02'::date, '2016-04-02'::date, 1)
UNION ALL
SELECT (week_end + interval '1 day')::date,
(CASE
WHEN (week_end + interval '7 days')::date > time_end THEN time_end
ELSE (week_end + interval '7 days')::date
END)::date,
time_end,
weekno+1
FROM weeks
WHERE time_end > week_end)
Any help would be greatly appreciated.
Why would you use a recursive query for this? Use generate_series():
select g.week_start, g.week_start + interval '6 day' as week_end,
row_number() over (order by g.week_start) as weeknum
from generate_series('2015-12-27'::timestamp,
'2016-01-02'::timestamp,
interval '1 week'
) g(week_start);
The equivalent for months would be like:
select g.month_start, g.month_start + interval '1 month' - interval '1 day' as month_end,
row_number() over (order by g.month_start) as monthnum
from generate_series('2015-12-01'::timestamp,
'2016-01-01'::timestamp,
interval '1 month'
) g(month_start);
i do have a query which works fine but I was just wondering if there are other ways or alternate method to bettter this.
I have a table where i am fetching those records exceeding or do not fall between 1 year time interval however there is only the year and ISO week number column in the table (integer values).
basically the logic is to check ISO WEEK - YEAR falls between 'current_date - interval '1 year' AND current_date.
My query is as below :
select * from raj_weekly_records where
(date_dimension_week > extract(week from current_date) and date_dimension_year = extract(year from current_date) )
or (date_dimension_week < extract(week from current_date) and (extract(year from current_date)-date_dimension_year=1) )
or(extract(year from current_date)-date_dimension_year>1);
Here date_dimension_week and date_dimension_year are the only integer parameters by which I need to check is there any other alternate or better way?.This code is working fine no issues here.
Here is an idea. Convert the year/week to a numeric format: YYYYWW. That is, the year times 100 plus the week number. Then you can do the logic with a single comparison:
select *
from raj_weekly_records
where date_dimension_year * 100 + date_dimension_week
not between (extract(year from current_date) - 1) * 100 + extract(week from current_date) and
extract(year from current_date) * 100 + extract(week from current_date)
(There might be an off-by one error, depending on whether the weeks at the ends are included or excluded.)
select *
from raj_weekly_records
where
date_trunc('week',
'0001-01-01 BC'::date + date_dimension_year * interval '1 year'
)
+ (date_dimension_week + 1) * interval '1 week'
- interval '1 day'
not between
current_date - interval '1 year' and current_date