'Compare date filed with month and year in Postgres
I have a date field in one of my tables and the column name is from_dt
. Now I have to compare a month and year combination against this from_dt
field and check whether the month has already passed. The current database function uses separate conditions for the month and the year, but this is wrong as it will compare month and year separately. The current code is like this
SELECT bill_rate, currency FROM table_name WHERE
emp_id = employee_id_param
AND EXTRACT(MONTH FROM from_dt) <= month_param
AND EXTRACT(YEAR FROM from_dt) <= year_param
Now the fromt_dt
field has value 2021-10-11
. If I give month_param
as 01
and year_param
as 2022
, this condition will not work as the month 10 is greater than 1, which I have given. Basically, I need to check whether 01-2022
(Jan 2022) is greater than r equal to 2021-10-01
(October 1st, 2021). It would be very much helpful if someone can shed some light here.
Solution 1:[1]
If you just want to check whether one date is >=
then another:
# select '2022-01-01'::date >= '2021-10-11'::date;
?column?
----------
t
If you want to restrict to year/month then:
select date_trunc('month','2022-01-01'::date) >= date_trunc('month', '2021-10-11'::date);
?column?
----------
t
Where the date_trunc
components are:
select date_trunc('month','2022-01-01'::date) ;
date_trunc
------------------------
2022-01-01 00:00:00-08
select date_trunc('month','2021-10-11'::date) ;
date_trunc
------------------------
2021-10-01 00:00:00-07
See Postgres date_trunc for more information.
Solution 2:[2]
Assuming the given year_param
and month_param
are integers you can use the make_date
function to create the first of the year_month and date_trunc
to get the first on the month from the table. Just compare those values. (See date functions) So:
select bill_rate, currency
from table_name
where emp_id = employee_id_param
and date_trunc('month',from_dt) =
make_date( year_param, month_param, 01);
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Adrian Klaver |
Solution 2 | Belayer |