I am trying to write a query which gives the number of days in each month between two specified dates.
Example:
date 1: 2018-01-01
date 2: 2018-05-23
Expected Output:
month days
2018-01-01, 31
2018-02-01, 28
2018-03-01, 31
2018-04-01, 30
2018-05-01, 23
Use generate_series and group by date_trunc
SELECT date_trunc('month',dt) AS month,
COUNT(*) as days
FROM generate_series( DATE '2018-01-01',DATE '2018-05-23',interval '1 DAY' )
as dt group by date_trunc('month',dt)
order by month;
Demo
Related
I'd like to have a range of day 20th - 25th in each month in BigQuery but i dont know what syntax should i use. For ex:
Jan 20 - 25
Feb 20 - 25
and so on
I only can think of creating a CTE for every month then union all those.
Consider below query.
SELECT DATE_ADD(month, INTERVAL day - 1 DAY) date_range,
FROM UNNEST(GENERATE_DATE_ARRAY('2022-01-01', '2022-03-01', INTERVAL 1 MONTH)) month,
UNNEST(GENERATE_ARRAY(20, 25)) day;
Query results
Below seem to be more simple than my original answer and you could adjust date range by specifying condition on WHERE clause.
SELECT *
FROM UNNEST(GENERATE_DATE_ARRAY('2022-01-01', '2022-12-31', INTERVAL 1 DAY)) date_range
WHERE EXTRACT(DAY FROM date_range) BETWEEN 21 AND 25
For the usecase that you commented,
WHERE EXTRACT(DAY FROM date_range) >= 21 OR EXTRACT(DAY FROM date_range) = 1
db fiddle
run select *, return_date - pickup_date as total from order_history order by id; return the following result:
id pickup_date return_date date_ranges total
1 2020-03-01 2020-03-12 [2020-03-01,2020-04-01) 11
2 2020-03-01 2020-03-22 [2020-03-01,2020-04-01) 21
3 2020-03-11 2020-03-22 [2020-03-01,2020-04-01) 11
4 2020-02-11 2020-03-22 [2020-02-01,2020-03-01) 40
5 2020-01-01 2020-01-22 [2020-01-01,2020-02-01) 21
6 2020-01-01 2020-04-22 [2020-01-01,2020-02-01) 112
for example:
--id=6. total = 112. 112 = 22+ 31 + 29 + 30
--therefore toal should split: jan2020: 30, feb2020:29, march2020: 31, 2020apr:22.
first split then aggregate. aggregate based over range min(pickup_date), max(return_date) then tochar cast to 'YYYY-MM'; In this case the aggregate should group by 2020-01, 2020-02, 2020-03,2020-04.
but if pickup_date in the same month with return_date then compuate return_date - pickup_date then aggregate/sum the result, group by to_char(pickup_date,'YYYY-MM')
step-by-step demo: db<>fiddle
Not quite perfect, but a sketch:
SELECT
id,
ARRAY_AGG( -- 4
LEAST(return_date, gs + interval '1 month - 1 day') -- 2
- GREATEST(pickup_date, gs) -- 3
+ interval '1 day'
)
FROM order_history,
generate_series( -- 1
date_trunc('month', pickup_date),
date_trunc('month', return_date),
interval '1 month'
) gs
GROUP BY id
Generate a set of months that are included in the given date range
a) Calculate the last day of the month (first of a month + 1 month is first of the next month; minus 1 day is last of the current month). This is the max day for returning in this month. b) if it happened earlier, then take the earler day (LEAST())
Same for pickup day. Afterwards calculate the difference of the days kept in one month.
Aggregate the values for one month.
Open questions / Potential enhancements:
You said:
jan2020: 30, feb2020:29, march2020: 31, 2020apr:22.
Why is JAN given with 30 days? On the other hand you count APR 22 days (1st - 22nd). Following the logic, JAN should be 31, shouldn't it?
If you don't want to count the very first day, then you can change (3.) to
GREATEST(pickup_date + interval '1 day', gs)
There's a problem with day saving time in March (30 days, 23 hours instead of 31 days). This can be faced by some rounding, for example.
There is one table:
ID DATE
1 2017-09-16 20:12:48
2 2017-09-16 20:38:54
3 2017-09-16 23:58:01
4 2017-09-17 00:24:48
5 2017-09-17 00:26:42
..
The result I need is the last 7-days of data with hourly aggregated count of rows:
COUNT DATE
2 2017-09-16 21:00:00
0 2017-09-16 22:00:00
0 2017-09-16 23:00:00
1 2017-09-17 00:00:00
2 2017-09-17 01:00:00
..
I tried different stuff with EXTRACT, DISTINCT and also used the generate_series function (most stuff from similar stackoverflow questions)
This try was the best one currently:
SELECT
date_trunc('hour', demotime) as date,
COUNT(demotime) as count
FROM demo
GROUP BY date
How to generate hourly series for 7 days and fill-in the count of rows?
SQL DEMO
SELECT dd, count("demotime")
FROM generate_series
( current_date - interval '7 days'
, current_date
, '1 hour'::interval) dd
LEFT JOIN Table1
ON dd = date_trunc('hour', demotime)
GROUP BY dd;
To work from now and now - 7 days:
SELECT dd, count("demotime")
FROM generate_series
( date_trunc('hour', NOW()) - interval '7 days'
, date_trunc('hour', NOW())
, '1 hour'::interval) dd
LEFT JOIN Table1
ON dd = date_trunc('hour', demotime)
GROUP BY dd;
How to generate 52 weeks from current date using postgresql
for example: from current_date(i.e todays date) to 52 weeks .
You can use generate_series which allows you to define range (start, stop) as well as the step interval like: generate_series(startDate, endDate, stepBy) so depending on output format you are after you could do something like:
SELECT generate_series(
current_date,
current_date + interval '52 weeks',
interval '1 week'
) weeks;
which would generate something like this:
weeks
2017-05-24 00:00:00
2017-05-31 00:00:00
2017-06-07 00:00:00
...
I have a column as date. In that column I have a value as '2016-05-06' I want a result in such manner that it will add the complete one month into this column. But it should return a one day before result.
So when i execute the query like:
select date,(date + interval '1 month') as new_column
from batchproduct_info;
it give me the result as:
date new_column
2016-05-06 2016-06-06 00:00:00
2016-05-07 2016-06-07 00:00:00
But I want result in this format:
date new_column
2016-05-06 2016-06-05 00:00:00
2016-05-07 2016-06-06 00:00:00
i.e it should subtract the one day from one month.
This is a solution to your problem:
select date, (date + '1 month'::interval - '1 day'::interval) as new_column
from batchproduct_info;