Hello I want to return in my PostgreSQL the difference between two dates:
START: 2016-06-01 00:00:00
END: 2016-06-06 08:35:33
Expected return value: 128:35:33, formatted like format [h]:mm:ss;# in Excel. Hours must be added up if there is more than 24 hours of difference.
Here's my SQL:
SELECT EXTRACT(EPOCH FROM dt_termino::timestamp - dt_inicio::timestamp)/3600 FROM crm.task_interacao WHERE id_task_tarefa = 1
UPDATE!!!
hello now i'm facing another problema I have a table like this:
my table in database like this
start;end
2013-06-01 09:29:33;2016-06-07 14:08:19
2016-06-07 14:22:09;2016-06-07 14:22:43
2016-06-07 14:22:51; null
i need to sum values ....i'm trying as you said (1st awnser).. I cant use function because i'm using inside a php code
SELECT SUM(COALESCE(end::timestamp, now()::timestamp) - start::timestamp) FROM crm.task_interacao WHERE id_task_tarefa = 1
but is returning
1102 days 26:07:54.864879
why 26 hours??? I was supose be te at maximum 24...
no problem now to return (Days HH:MM:SS) and not miliseconds
You can simply subtract timestamps to get interval:
select '2016-06-06 08:35:33'::timestamp- '2016-06-01 00:00:00' result
result
-----------------
5 days 08:35:33
(1 row)
There is no standard function to convert the result to the format you need but you can write one:
create or replace function interval_without_days(interval)
returns interval language sql as $$
select $1- date_part('day', $1)* '1d'::interval+ date_part('day', $1)* '24h'::interval;
$$;
select interval_without_days('2016-06-06 08:35:33'::timestamp- '2016-06-01 00:00:00');
interval_without_days
-----------------------
128:35:33
(1 row)
Question #2. Use the functions date_trunc(text, interval) and justify_hours(interval):
select date_trunc('sec', justify_hours('1102 days 26:07:54.864879'));
date_trunc
--------------------
1103 days 02:07:54
(1 row)
Related
Lets say we have the dates
'2017-01-01'
and
'2017-01-15'
and I would like to get a series of exactly N timestamps in between these dates, in this case 7 dates:
SELECT * FROM
generate_series_n(
'2017-01-01'::timestamp,
'2017-01-04'::timestamp,
7
)
Which I would like to return something like this:
2017-01-01-00:00:00
2017-01-01-12:00:00
2017-01-02-00:00:00
2017-01-02-12:00:00
2017-01-03-00:00:00
2017-01-03-12:00:00
2017-01-04-00:00:00
How can I do this in postgres?
Possibly this can be useful, using the generate series, and doing the math in the select
select '2022-01-01'::date + generate_series *('2022-05-31'::date - '2022-01-01'::date)/15
FROM generate_series(1, 15)
;
output
?column?
------------
2022-01-11
2022-01-21
2022-01-31
2022-02-10
2022-02-20
2022-03-02
2022-03-12
2022-03-22
2022-04-01
2022-04-11
2022-04-21
2022-05-01
2022-05-11
2022-05-21
2022-05-31
(15 rows)
WITH seconds AS
(
SELECT EXTRACT(epoch FROM('2017-01-04'::timestamp - '2017-01-01'::timestamp))::integer AS sec
),
step_seconds AS
(
SELECT sec / 7 AS step FROM seconds
)
SELECT generate_series('2017-01-01'::timestamp, '2017-01-04'::timestamp, (step || 'S')::interval)
FROM step_seconds
Conversion to function is easy, let me know if have trouble with it.
One problem with this solution is that extract epoch always assumes 30-days months. If this is problem for your use case (long intervals), you can tweak the logic for getting seconds from interval.
You can divide the difference between the end and the start value by the number of values you want:
SELECT *
FROM generate_series('2017-01-01'::timestamp,
'2017-01-04'::timestamp,
('2017-01-04'::timestamp - '2017-01-01'::timestamp) / 7)
This could be wrapped into a function if you want to avoid repeating the start and end value.
I am looking for a function in PostgreSQL which help me to generate recurring date after every 90 days from created date
for example: here is a demo table of mine.
id date name
1 "2020-09-08" "abc"
2 "2020-09-08" "xyz"
3 "2020-09-08" "def"
I need furure date like 2020-12-08, 2021-03-08, 2021-06-08, and so on
First it's important to note that, if you happen to have a date represented as text, then you can convert it to a date via:
SELECT TO_DATE('2017-01-03','YYYY-MM-DD');
So, if you happen to have a text as an input, then you will need to convert it to date. Next, you need to know that if you have a date, you can add days to it, like
SELECT CURRENT_DATE + INTERVAL '90 day';
Now, you need to understand that you can use dynamic variables, like:
select now() + interval '1 day' * 180;
Finally, you will need a temporary table to generate several values described as above. Read more here: How to return temp table result in postgresql function
Summary:
create a function
that generates a temporary table
where you insert as many records as you like
having the date shifted
and converting text to date if needed
You can create a function that returns a SETOF dates/timestamps. The below function takes 3 parameters: a timestamp, an interval, the num_of_periods desired. It returns num_of_periods + 1 timestamps, as it returns the original timestamp and the num_of_periods each the specified interval apart.
create or replace
function generate_periodic_time_intervals
( start_date timestamp
, period_length interval
, num_of_periods integer
, out gen_timestamp timestamp
)
returns setof timestamp
language sql
immutable strict
as $$
select (start_date + n * period_length)::timestamp
from generate_series(0,num_of_periods) gs(n)
$$;
For your particular case to timestamp/date as necessary. The same function would work for your case with the interval specified as '3 months' or of '90 days'. Just a note the interval specified can be any valid INTERVAL data type. See here. It also demonstrates the difference between 3 months and 90 days.
I tried to ruh this query in postgres :
Select to_char((select add_months (to_date ('10/10/2019', 'dd/mm/yyyy'), '11/11/2019') ) , 'dd/mm/yyyy') as temp_date
I got an error :
Function add_months (date, unknown) does not exist
Hint: no function matches the given name and argument types. You might need to add explicit type casts.
Please help
As documented in the manual there is no add_months function in Postgres
But you can simply add an interval:
select to_date('10/10/2019', 'dd/mm/yyyy') + interval '10 months'
If you need to format that date value to something:
select to_char(to_date('10/10/2019', 'dd/mm/yyyy') + interval '10 months', 'yyyy-mm-dd')
No one, even running on Oracle, has run the original query- at least not successfully. It appears that query is expecting to add two months together (in this case Oct and Nov). That is not what the function does. It adds an integer number of months to the specified date and returns the resulting date. As indicated in Postgres just adding the desired interval. However, if you have many occurrences ( like converting) of this the following implements a Postgres version.
create or replace function add_months(
date_in date
, n_months_in integer)
returns date
language sql immutable strict
as
$$
-- given a date and an integer for number of months return the calendar date for the specified number of months away.
select (date_in + n_months_in * interval '1 month')::date
$$ ;
-- test
-- +/- 6 months from today.
select current_date "today"
, add_months(current_date,6) "6 months from now"
, add_months(current_date,-6) "6 months ago"
;
I am working recently with postgres and I have to make several calculations. However I have not been able to imitate the HOUR () function of Excel, I read the official information but it did not help me much.
The function receives a decimal and obtains the hour, minutes and seconds of the same, example the decimal 0,99988426 returns 11:59:50. Try doing this in postgres (i use PostgreSQL 10.4) with the to_timestamp function: select to_char (to_timestamp (0.99988426), 'HH24: MI: SS'); this return 19:00:00. Surely I am omitting something, some idea of how to solve this?
24:00:00 or 86400 seconds = 1
Half day(12:00 noon) or 43200 seconds = 43200/86400 = 0.5
11:59:50 or 86390 seconds = 86390/86400 = 0.99988426
So to convert your decimal value to time, all you have to do is multiply it with 86400 which will give you seconds and convert it to your format in following ways:
SELECT TO_CHAR((0.99988426 * 86400) * '1 second'::interval, 'HH24:MI:SS');
SELECT (0.99988426 * 86400) * interval '1 sec';
There are two major differences to handle:
Excel does not consider the time zone. The serial date 0 starts at 0h00, but Postgres uses the time zone so it becomes 19h. You would need to use UTC in Postgres result to have the same as in Excel.
select to_char (to_timestamp (0), 'HH24: MI: SS'),to_char (to_timestamp (0) AT TIME ZONE 'UTC', 'HH24: MI: SS');
to_char | to_char
------------+------------
19: 00: 00 | 00: 00: 00
Excel considers that 1 is one day, while Postgres considers 1 as 1 second. To get the same behavior, multiply your number by the 86400, i.e. the number of seconds in a day
select to_char (to_timestamp (0.99988426*86400) AT TIME ZONE 'UTC', 'HH24: MI: SS');
to_char
------------
23: 59: 50
(1 row)
I would like to take a string in the format 'YYYYQQ' and parse it into a date. Specifically I would like to parse it into the first date of the quarter. For example I would like to parse '2016Q2' into '2016-04-01'.
Per the Postgres documentation, "Q (quarter) is ignored by to_date and to_timestamp". Well frankly I wish it wasn't ignored :) This means this code will return a result I don't want:
select to_date('2014q3', 'YYYY\qQ');
**result**
2014-01-01
**desired result**
2014-07-01
How could I parse this string to the proper date?
Use string manipulation functions format(), left() and right():
with quarters(q) as (
values ('2014q1'), ('2015q2'), ('2016q3'), ('2017q4')
)
select format('%s-%s-1', left(q, 4), right(q, 1)::int* 3- 2)::date
from quarters;
format
------------
2014-01-01
2015-04-01
2016-07-01
2017-10-01
(4 rows)
Create a function for convenience:
create or replace function quarter_to_date(text)
returns date language sql as $$
select format('%s-%s-1', left($1, 4), right($1, 1)::int* 3- 2)::date
$$;
with quarters(q) as (
values ('2014q1'), ('2015q2'), ('2016q3'), ('2017q4')
)
select quarter_to_date(q)
from quarters;
quarter_to_date
-----------------
2014-01-01
2015-04-01
2016-07-01
2017-10-01
(4 rows)
In PostgreSQL 9.4 and greater there is a function that will create a date.
make_date(year, month, day)
You can use it as follows:
make_date(year, quarter * 3 -2 , 1)::date