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Calculating a date in Postgres by adding months?
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Closed 7 years ago.
How to implement add_months in PostgreSQL ??
Like oracle ADD_MONTHS returns the date.
Ex. ADD_MONTHS(hire_date,1)
use
hire_date + interval '1 month'
this will return exactly one month added to hire_date.
For More References on date and time functions in postgre Date time function
CREATE FUNCTION add_months(start DATE, months INT) RETURNS DATE AS
$$
SELECT (start + (months || ' months')::INTERVAL)::DATE
$$
LANGUAGE sql IMMUTABLE
In PostgreSQL you can create a function to do the job
create or replace function ADD_MONTHS(var_dte date,cnt int) returns setof date as
$$
declare
qry text;
begin
qry = format( 'select (''%s''::date + interval ''%s'')::date',var_dte,cnt||' month') ;
RETURN QUERY
EXECUTE qry;
end
$$
language plpgsql
and call this function
select ADD_MONTHS('2015-11-27',1)
Result:
add_months
date
----------
2015-12-27
in your case
select hire_date
,ADD_MONTHS(hire_date,1)
from table_name
Related
I'm trying to figure out how to use the Postgresql EXTRACT function to convert a given date_variable into its equivalent day of the week. I understand that it will convert the date_variable into a numbering from 0 - 6 (0 is Sunday, 6 is Saturday etc)
I've created a simple table to test my queries. Here I will attempt to convert the start_date into its DOW equivalent.
DROP TABLE IF EXISTS test;
CREATE TABLE test(
start_date date PRIMARY KEY,
end_date date
);
INSERT INTO test (start_date, end_date) VALUES ('2021-03-31', '2021-03-31'); -- Today (wed), hence 3
INSERT INTO test (start_date, end_date) VALUES ('2021-03-30', '2021-03-30'); -- Yesterday (tues), hence 2
INSERT INTO test (start_date, end_date) VALUES ('2021-03-29', '2021-03-29'); -- Day before (mon), hence 1
If I were to run the query below
SELECT (EXTRACT(DOW FROM t.start_date)) AS day FROM test t;
It works fine, and returns the result as intended (returns a single column table with values (3, 2, 1) respectively.)
However, when I attempt to write a function to return the exact same query
CREATE OR REPLACE FUNCTION get_day()
RETURNS TABLE (day integer) AS $$
BEGIN
RETURN QUERY
SELECT (EXTRACT(DOW FROM t.start_date)) as day
FROM test t;
END;
$$ LANGUAGE plpgsql;
SELECT * FROM get_day(); -- throws error "structure of query does not match function result type"
I get an error instead. I cant seem to find the issue and don't know what is causing it.
extract() returns a double precision value, but your function is declared to return an integer. You need to cast the value:
CREATE OR REPLACE FUNCTION get_day()
RETURNS TABLE (day integer) AS $$
BEGIN
RETURN QUERY
SELECT EXTRACT(DOW FROM t.start_date)::int as day
FROM test t;
END;
$$ LANGUAGE plpgsql;
But you don't really need PL/pgSQL for this, a language SQL function would also work.
CREATE OR REPLACE FUNCTION get_day()
RETURNS TABLE (day integer) AS $$
SELECT EXTRACT(DOW FROM t.start_date)::int as day
FROM test t;
$$
LANGUAGE sql
stable;
As you are not passing any parameters, I would actually use a view for this:
create or replace view day_view
AS
SELECT EXTRACT(DOW FROM t.start_date)::int as day
FROM test t;
The extract function returns values of type double precision.
You declare result to be integer.
You should cast the result of EXTRACT to integer:
CREATE OR REPLACE FUNCTION get_day()
RETURNS TABLE (day integer) AS $$
BEGIN
RETURN QUERY
SELECT EXTRACT(DOW FROM t.start_date)::integer as day
FROM test t;
END;
$$ LANGUAGE plpgsql;
I've got a plpgsql function. I need to take the date 5 days from today, and then divide month into "fives" to takte the start of "last five". The problem is thay make_date does not exist in the posgres version that is used on the server....
create or replace function getFirstDayOfFive()
returns timestamp with time zone as $$
declare
firstDay timestamp;
startOp timestamp;
begin
startOp = now() - interval '5 day';
SELECT
make_date(
date_part('year', startOp)::int,
date_part('month', startOp)::int,
greatest(
floor(date_part('day', startOp) / 5) * 5,
1
)::int
)
INTO firstDay;
RETURN firstDay;
end;
$$
language plpgsql;
It worked fine last week, but now I got an error when I call it
ERROR: BŁĄD: function make_date(integer, integer, integer) does not exist
LINE 2: make_date(
^
HINT: There is no function matching provided name and arguments. Maybe you should cast data.
QUERY: SELECT
make_date(
date_part('year', startOp)::int,
date_part('month', startOp)::int,
greatest(
floor(date_part('day', startOp) / 5) * 5,
1
)::int
)
CONTEXT: PL/pgSQL function "getfirstdayoffive" line 7 at wyrażenie SQL
SQL state: 42883
What happened that earlier it worked and now it gives error?
[Edit]
I found out that make_date is available from postgresQL 9.4, but on the server there is posthresQL 9.1 is there any way to do the same in this old version od DB? I'm trying to replace the make_date with something like
create or replace function getFirstDayOfFive()
returns timestamp with time zone as $$
declare
firstDay timestamp;
startOp timestamp;
begin
startOp = now() - interval '5 day';
SELECT
date to_char(startOp, 'YYYY-MM-')||to_char(greatest(
floor(date_part('day', startOp) / 5) * 5,
1
)::int)
INTO firstDay;
RETURN firstDay;
end;
$$
language plpgsql;
I think you can simplify this by simply adding the desired number of days to the start of the month. Apparently you only want a date so I would also recommend to change the return type to date
create or replace function getfirstdayoffive()
returns date
as
$$
select date_trunc('month', current_date - 5)::date
+ (greatest(floor(extract(day from current_date - 5) / 5) * 5, 1))::int - 1;
$$
language sql
stable;
I have a database function as below:
drop function test(month_interval text)
create or replace function test (month_interval text) returns date as
$$
select ('2020-07-01'::date - interval month_interval)::date;
$$ language sql;
select * from test('2 months')
I have a scenario where I want to dynamically compute month intervals and want to have one database query that can be used by passing month intervals as function parameters. However when i do this it gives me the following error :
ERROR: syntax error at or near "month_interval"
You could cast the text to an interval:
create or replace function test (month_interval text) returns date as
$$
select ('2020-07-01'::date - month_interval::interval)::date;
$$ language sql;
select test('2 months');
But why not pass an interval directly?
create or replace function test (month_interval interval) returns date as
$$
select ('2020-07-01'::date - month_interval)::date;
$$ language sql;
select test(interval '2 months');
Alternatively you can pass the number of months, then use make_interval:
create or replace function test (num_months int) returns date as
$$
select ('2020-07-01'::date - make_interval(months => num_months))::date;
$$ language sql;
select test(2);
I want to create a function to get the right week number of year.
I already posted here to find a 'native' solution, but apparently there is not.
I tryed to create funcrtion based on this mysql example
Here is the code translated to postgresql:
CREATE OR REPLACE FUNCTION week_num_year(_date date)
RETURNS integer AS
$BODY$declare
_year integer;
begin
select date_part('year',_date) into _year;
return ceil((to_char(_date,'DDD')::integer+(to_char(('01-01-'||_year)::date,'D')::integer%7-7))/7);
end;$BODY$
LANGUAGE plpgsql VOLATILE
COST 100;
But it gives wrong result, can someone help me ?
My config: PostgreSQL 9.2
If you want proper week numbers use:
select extract(week from '2012-01-01'::date);
This will produce the result 52, which is correct if you look on a calendar.
Now, if you actually want to define week numbers as "Every 7 days starting with the first day of the year" that's fine, though it doesn't match the week numbers anyone else uses and has some odd quirks:
select floor((extract(doy from '2011-01-01'::date)-1)/7)+1;
By the way, parsing date strings and hacking them up with string functions is almost always a really bad idea.
create or replace function week_num_year(_date date)
returns integer as
$body$
declare
_year date;
_week_number integer;
begin
select date_trunc('year', _date)::date into _year
;
with first_friday as (
select extract(doy from a::date) ff
from generate_series(_year, _year + 6, '1 day') s(a)
where extract(dow from a) = 5
)
select floor(
(extract(doy from _date) - (select ff from first_friday) - 1) / 7
) + 2 into _week_number
;
return _week_number
;
end;
$body$
language plpgsql immutable
You can retrieve the day of the week and also the week of the year by running:
select id,extract(DOW from test_date),extract(week from test_date), testdate,name from yourtable
What about the inbuild extract function?
SELECT extract (week from current_timestamp) FROM A_TABLE_FROM_YOUR_DB;
Is there any function(s) in postgres equivalent to Oracle function LAST_DAY().
I need to get last day in postgres (including month and year)
Well, In postgres, it seems there's no such function equivalent to LAST_DAY() available in oracle.
If you need to, you can have your own in the following ways as a
Select Query
SELECT (date_trunc('MONTH', now()) + INTERVAL '1 MONTH - 1 day')::date;
plsql Function
CREATE OR REPLACE FUNCTION last_day(date)
RETURNS date AS
$$
SELECT (date_trunc('MONTH', $1) + INTERVAL '1 MONTH - 1 day')::date;
$$ LANGUAGE 'sql'
IMMUTABLE STRICT;
Hope this helps.
create or replace funCtion last_day(fromdt anyelement)
returns date as
$BODY$
SELECT (date_trunc('MONTH', cast(fromdt as date)) + INTERVAL '1 MONTH - 1 day')::date;
$BODY$
LANGUAGE sql VOLATILE
COST 100;
ALTER FUNCTION last_day(anyelement)
OWNER TO postgres;