Given the following sqlfiddle: http://www.sqlfiddle.com/#!17/f483a/2/0
create table test (
start_date date
);
insert into test values ('2019/01/01');
select
start_date,
age(now()::date,start_date) as date_diff
from test;
Which generates the following output:
date_diff | 0 years 7 mons 27 days 0 hours 0 mins 0.00 secs
How could I instead generate the correct number of calendar days
239 days
without using a custom function?
Don't use the age function. Subtracting a date from a date yields an integer. now() returns a timestamp so you need to use current_date instead.
select start_date,
current_date - start_date as date_diff
from test;
Related
I am trying to get week numbers in a Year starting from a certain day
I've checked the stack but quite confused.
SELECT EXTRACT(WEEK FROM TIMESTAMP '2021-01-01'),
extract('year' from TIMESTAMP '2021-01-01')
The output is 53|2021
I want it to be 01|2021
I understand the principle of the isoweek but I want the year to start in 01-01-2021
The aim is to use intervals from this day to determine week numbers
Week N0| End Date
1 | 01-01-2021
2 | 01-08-2021
5 | 01-29-2021
...
This is really strange way to determine the week number, but in the end it's a simple math operation: the number of days since January first divided by 7.
You can create a function for this:
create function custom_week(p_input date)
returns int
as
$$
select (p_input - date_trunc('year', p_input)::date) / 7 + 1;
$$
language sql
immutable;
So this:
select date, custom_week(date)
from (
values
(date '2021-01-01'),
(date '2021-01-08'),
(date '2021-01-29')
) as v(date)
yields
date | custom_week
-----------+------------
2021-01-01 | 1
2021-01-08 | 2
2021-01-29 | 5
i have following query in postgresql for dates between 2 ranges.
select generate_series('2019-04-01'::timestamp, '2020-03-31', '1 month')
as g_date
I need to generate specific date in every month .i.e 15 th of every month. Following is my query to generate series
DO $$
DECLARE
compdate date = '2019-04-15';
BEGIN
CREATE TEMP TABLE tmp_table ON COMMIT DROP AS
select *,
case
when extract('day' from d) <> extract('day' from compdate) then 0
when ( extract('month' from d)::int - extract('month' from compdate)::int ) % 1 = 0 then 1
else 0
end as c
from generate_series('2019-04-01'::timestamp, '2020-03-31', '1 day') d;
END $$;
SELECT * FROM tmp_table
where c=1;
;
But every thing is perfect if input date between (1..29)-04-2019 ..
2019-04-25
2019-05-25
2019-06-25
2019-07-25
2019-08-25
2019-09-25
2019-10-25
2019-11-25
2019-12-25
2020-01-25
2020-02-25
2020-03-25
but if i give compdate: 31-04-2019 or 30-04-2019 giving out put:
2019-05-31
2019-07-31
2019-08-31
2019-10-31
2019-12-31
2020-01-31
2020-03-31
Expected Output:
date flag
2019-04-01 0 ----start_date
2019-04-30 1
2019-05-31 1
2019-06-30 1
2019-07-31 1
2019-08-31 1
2019-09-30 1
2019-10-31 1
2019-11-30 1
2019-12-31 1
2020-01-31 1
2020-02-29 1
2020-03-31 0 ---end_date
If matched day not found in the result it should take last day of that month..i.e if 31 not found in month of feb it
should take 29-02-2019 and also in april month instead of 31 it should take 2019-04-30.
Please suggest.
to generate the last days of the month, just generate first days & subtract a 1 day interval
example: the following generates all last day of month in the year 2010
SELECT x - interval '1 day' FROM
GENERATE_SERIES('2010-02-01', '2011-01-01', interval '1 month') x
You cannot accomplish what you want with generate_series. This results due to that process applying a fixed increment from the previous generated value. Your case 1 month. Now Postgres will successfully compute correct end-of-month date from 1 month to the next. So for example 1month from 31-Jan yields 28-Feb (or 29), because 31-Feb would be an invalid date, Postgres handles it. However, that same interval from 28-Feb gives the valid date 28-Mar so no end-of-month adjustment is needed. Generate_Series will return 28th of the month from then on. The same applies to 30 vs. 31 day months.
But you can achieve what your after with a recursive CTE by employing a varying interval to the same initial start date. If the resulting date is invalid for date the necessary end-of-month adjustment will be made. The following does that:
create or replace function constant_monthly_date
( start_date timestamp
, end_date timestamp
)
returns setof date
language sql strict
as $$
with recursive date_set as
(select start_date ds, start_date sd, end_date ed, 1 cnt
union all
select (sd + cnt*interval '1 month') ds, sd, ed, cnt+1
from date_set
where ds<end_date
)
select ds::date from date_set;
$$;
-- test
select * from constant_monthly_date(date '2020-01-15', date '2020-12-15' );
select * from constant_monthly_date(date '2020-01-31', date '2020-12-31' );
Use the least function to get the least one between the computed day and end of month.
create or replace function test1(day int) returns table (t timestamptz) as $$
select least(date_trunc('day', t) + make_interval(days => day-1), date_trunc('day', t) + interval '1 month' - interval '1 day') from generate_series('2019-04-01', '2020-03-31', interval '1 month') t
$$ language sql;
select test1(31);
Hi I am having a Postgresql query like below to calculate DateTime difference for {1} and {2} in minutes.
CAST(ROUND(EXTRACT(EPOCH from (({2}::timestamp) - ({1}::timestamp)))/60) AS INT)
I want to calculate the difference in hours, minutes and seconds displayed like:
3 hrs 31 minutes 42 secs
What manipulation do I need for displaying like above?
SELECT to_char((col1 - col0), 'HH24 hrs MI "minutes" SS "seconds"') FROM T1;
Here is a sqlfiddle : link
The to_char function takes an interval (an interval is the time span between two timestamps, and subtracting timestamps gives you an interval). It then takes a formatting, and you can apply pretty much what you want.
Formatting functions in PostgreSQL
Try use this sql:
SELECT to_char(column2 - column1, 'DD" days "HH24" hours "MI" minutes "SS" seconds"');
The subtraction of two timestamp or timestamptz values produces an interval. (While subtracting two date values produces an integer!)
Details about date/time types in the manual.
The default text representation of an interval may be sufficient:
SELECT timestamp '2017-1-6 12:34:56' - timestamp '2017-1-1 0:0';
Result is an interval, displayed as:
5 days 12:34:56
If you need the format in the question, precisely, you need to specify how to deal with intervals >= 24 hours. Add 'days'? Or just increase hours accordingly?
#Nobody provided how to use to_char(). But add days one way or the other:
SELECT to_char(ts_col2 - ts_col1, 'DD" days "HH24" hours "MI" minutes "SS" seconds"');
Result:
05 days 12 hours 34 minutes 56 seconds
'days' covers the rest. There are no greater time units in the result by default.
Simple
SELECT
EXTRACT(year FROM LOCALTIMESTAMP(0) - yourFieldTime)||' year '||
EXTRACT(month FROM LOCALTIMESTAMP(0) - yourFieldTime)||' month '||
EXTRACT(day FROM LOCALTIMESTAMP(0) - yourFieldTime)||' day '||
EXTRACT(hour FROM LOCALTIMESTAMP(0) - yourFieldTime)||' hour '||
EXTRACT(minute FROM LOCALTIMESTAMP(0) - yourFieldTime)||' minute '||
EXTRACT(second FROM LOCALTIMESTAMP(0) - yourFieldTime)||' second '
AS full_time_as_you_wish FROM your_table;
Result
full_time_as_you_wish
---------------------------------
0 year 0 month 0 day 0 hour 0 minute 0 second
How can I combine these two columns and add integer of 7 to it? In postgres.
Both columns are in the table a_event.
start_time is a timestamp without time zone data type.
start_date is a date data type.
so far I have:
SELECT start_time || start_date AS start_date_time
FROM a_event;
This works but it shows:
2015-03-06 21:17:162015-02-06
How can I get it to simply show:
2015-03-06 13:17:16 with an added 7 days so: 2015-03-13 in the desired result.
The || operator concatenates two strings, so the 'start_time' and 'start_date' columns are converted to text and then concatenated to give the result you see.
You want to add 7 days to the starting date. That is simply:
SELECT start_date + 7 AS start_date_time
FROM a_event;
On a date data type the + operator adds a number of days.
If you want to keep the starting time information, the solution is slightly more complex:
SELECT start_time + interval '7 days' AS start_date_time
FROM a_event;
The timestamp data type records microseconds since 01-01-1970, so you have to explicitly indicate that you want to add an interval of 7 days (and not 7 microseconds).
How do I add a dynamic (column based) number of days to NOW?
SELECT NOW() + INTERVAL a.number_of_days "DAYS" AS "The Future Date"
FROM a;
Where a.number_of_days is an integer?
I usually multiply the number by interval '1 day' or similar, e.g.:
select now() + interval '1 day' * a.number_of_days from a;
I know this is a year old, but if you need to use a column to specify the actual interval (e.g. 'days', 'months', then it is worth knowing that you can also CAST your string to an Interval, giving:
SELECT now()+ CAST(the_duration||' '||the_interval AS Interval)
So the the original question would become:
SELECT now() + CAST(a.number_of_days||" DAYS" AS Interval) as "The Future Date" FROM a;
I prefer this way. I think its pretty easy and clean.
In Postgres you need interval to use + operator with timestamp
select (3||' seconds')::interval;
select now()+ (10||' seconds')::interval,now();
where you can use seconds, minutes, days, months...
and you can replace the numbers to your column.
select now()+ (column_name||' seconds')::interval,now()
from your_table;
Use make_interval()
SELECT NOW() + make_interval(days => a.number_of_days) AS "The Future Date"
FROM a;
But in general it might be a better idea to use a column defined as interval, then you can use any unit you want when you store a value in there.
To creating intervals those based on column values, I recommend to add two columns in your table. For example, column "period_value"::INT4 and column "period_name"::VARCHAR.
Column "period_name" can store the following values:
microsecond
milliseconds
second
minute
hour
day
week
month
quarter
year
decade
century
millennium
+--------------+-------------+
| period_value | period_name |
+--------------+-------------+
| 2 | minute |
+--------------+-------------+
Now you can write:
SELECT NOW() - (period_value::TEXT || ' ' || period_name::TEXT)::INTERVAL FROM table;
If we have field with interval string value such as '41 years 11 mons 4 days' and want to convert it to date of birth use this query :
UPDATE "february14" set dob = date '2014/02/01' - (patient_age::INTERVAL)
dob is date field to convert '41 years 11 mons 4 days' to '1972/10/14' for example
patient_age is varchar field that have string like '41 years 11 mons 4 days'
And this is query to convert age back to date of birth
SELECT now() - INTERVAL '41 years 10 mons 10 days';
Updating based on a column ID was a useful way to create some randomised test data for me.
update study_histories set last_seen_at = now() - interval '3 minutes' * id;