How one can convert integer to week interval?
CREATE TABLE integers( i integer);
INSERT INTO integers VALUES ('10');
Output would be table with one column indicating 10 weeks interval.
http://sqlfiddle.com/#!17/4b404/5/0
One take would be to create constant interval of 1 week and multiply it by integer.
I would prefer function to do it directly, but I am not aware of it.
SELECT interval '1 week' * i AS weeks_interval FROM integers;
Your solution is well accepted.
If you don't want to keep the "1" in the string you could write this instead
SELECT (i || 'week')::interval FROM intervals
demo: db<>fiddle
Related
I am looking to subtract 2 dates to get the number of days in between. However, the columns are defined as "timestamp without time zone". I'm not sure how to get a whole integer.
I have a stored procedure with this code:
v_days_between_changes := DATE_PATH('day',CURRENT_DATE - v_prev_rec.date_updated)::integer;
But I get this error:
HINT: No function matches the given name and argument types. You might need to add explicit type casts.
QUERY: SELECT DATE_PATH('day',CURRENT_DATE - v_prev_rec.date_updated)::integer
Any help would be great.
You can compute the difference between the dates, which returns an interval. Then, extract the number of days from this interval.
WITH src(dt1,dt2) AS (select '1/1/2019'::timestamp without time zone, CURRENT_DATE )
select EXTRACT(day FROM dt2-dt1) nbdays
from src;
nbdays
-----------
98
I'm looking to cast/convert a decimal data into date data. I've looked online and am still struggling. Can you help? "value" is a double precision, and here I convert it to varchar. From that point on, I've tried using convert but I'm still unable to get a date value.
Thanks in advance!
select cast(value as varchar(8) )date_value, value,
For example: The value 43099 should read 12/30/2017.
Guessing from your meager example, you might want this:
SELECT date '1899-12-30' + 43099; -- returns date '2017-12-30'
You need to cast your value column to integer for this:
SELECT date '1899-12-30' + value::int;
Meaning, the number might represent the count of days since 1900 (with an off-by-2 error I can't explain).
Adding an integer to a date means adding the number of days.
If your value can have fractional digits, you can compute a timestamp in similar fashion:
SELECT timestamp '1899-12-30' + interval '1 day' * value;
Can anyone suggest me, the easiest way to find summation of time field in POSTGRESQL. i just find a solution for MYSQL but i need the POSTGRESQL version.
MYSQL: https://stackoverflow.com/questions/3054943/calculate-sum-time-with-mysql
SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(timespent))) FROM myTable;
Demo Data
id time
1 1:23:23
2 4:00:23
3 9:23:23
Desired Output
14:47:09
What you want, is not possible. But you probably misunderstood the time type: it represents a precise time-point in a day. It doesn't make much sense, to add two (or more) times. f.ex. '14:00' + '14:00' = '28:00' (but there are no 28th hour in a day).
What you probably want, is interval (which represents time intervals; hours, minutes, or even years). sum() supports interval arguments.
If you use intervals, it's just that simple:
SELECT sum(interval_col) FROM my_table;
Although, if you stick to the time type (but you have no reason to do that), you can cast it to interval to calculate with it:
SELECT sum(time_col::interval) FROM my_table;
But again, the result will be interval, because time values cannot exceed the 24th hour in a day.
Note: PostgreSQL will even do the cast for you, so sum(time_col) should work too, but the result is interval in this case too.
I tried this solution on sql fieddle:
link
Table creation:
CREATE TABLE time_table (
id integer, time time
);
Insert data:
INSERT INTO time_table (id,time) VALUES
(1,'1:23:23'),
(2,'4:00:23'),
(3,'9:23:23')
query the data:
SELECT
sum(s.time)
FROM
time_table s;
If you need to calculate sum of some field, according another field, you can do this:
select
keyfield,
sum(time_col::interval) totaltime
FROM myTable
GROUP by keyfield
Output example:
keyfield; totaltime
"Gabriel"; "10:00:00"
"John"; "36:00:00"
"Joseph"; "180:00:00"
Data type of totaltime is interval.
I'm trying to get difference in days, casting result to decimal:
SELECT
CAST( TO_DATE('2999-01-01','yyyy-mm-dd') - TO_DATE('2909-01-01','yyyy-mm-dd') AS DECIMAL )
;
Now if I add 1 month to the 2nd date:
SELECT
CAST( TO_DATE('2999-01-01','yyyy-mm-dd') - (TO_DATE('2909-01-01','yyyy-mm-dd') + INTERVAL '1 MONTH' * (1) ) AS DECIMAL )
;
I'm getting an error:
ERROR: cannot cast type interval to numeric
OK, I can cast to char to get result:
SELECT
CAST( TO_CHAR( TO_DATE('2909-02-10','yyyy-mm-dd') - (TO_DATE('2909-01-01','yyyy-mm-dd') + INTERVAL '1 MONTH' * (1) ), 'DD') AS DECIMAL )
;
But in this case the 1st query modified with TO_CHAR casting stop working:
SELECT
CAST( TO_CHAR(TO_DATE('2999-01-01','yyyy-mm-dd') - TO_DATE('2909-01-01','yyyy-mm-dd'), 'DD') AS DECIMAL )
;
I'm getting ERROR: multiple decimal points.
So, my question is, how can I get days using the same sql statement? For both sql queries.
Look at your first two examples again. If you remove the outer CAST ... AS DECIMAL you get
?column?
----------
32872
?column?
------------
32841 days
Clearly the difference is in the "days". The second is an interval value rather than a simple number. You only want the number (because you always just want days) so you need to extract that part. Then you can cast to whatever precision you like:
SELECT extract(days FROM '32841 days'::interval)::numeric(9,2);
date_part
-----------
32841.00
Edit responding to Alexandr's follow-up:
Your first example fails with a fairly specific error:
SELECT extract(days FROM (TO_DATE('2999-01-01','yyyy-mm-dd') - TO_DATE('2909-01-01','yyyy-mm-dd'))::interval)::numeric(9,2);
ERROR: cannot cast type integer to interval
LINE 1: ...yyyy-mm-dd') - TO_DATE('2909-01-01','yyyy-mm-dd'))::interval...
Here you've got an integer (which is what you originally wanted) and try to cast it to an interval (for reasons I don't understand). It's complaining it doesn't know what units you want. You want 32872 what in your interval - seconds, hours, weeks, centuries?
The second example is complaining because you are trying to extract the "day" part from a simple integer, and of course there's no extract() function in the system to do that.
I think you probably need to take a step back and just take the time to understand the values your various expressions return.
Subtracting one date from another gives the number of days separating them - as an integer. There is no other sensible measure, really.
Adding (or subtracting) an interval to a date gives you a timestamp (without time zone) since the interval may contain whole days, days and hours, seconds etc.
Subtracting a timestamp from a date will give you an interval since the result may contain days, hours, seconds etc.
If you have an interval and you just want the days part then you use extract() on it and you will get an integer number of days back.
You will need an integer (or floating-point) number of days if you want to cast to numeric, not an interval because casting an interval to an scalar number makes no sense without units.
So - either stick to dates and date arithmetic (easy), or realise you are using timestamps (flexible) but understand which it is.
To get an illustration of what's happening you can do something like this (in psql):
CREATE TEMP TABLE tt AS SELECT
('2909-01-02'::date - '2909-01-01'::date) AS a,
('2909-01-02'::date - '2909-01-02 00:00:00'::timestamp) AS b;
\x
SELECT * FROM tt;
\d tt
That will show you the values and types you are dealing with. Repeat for as many columns as you find useful.
HTH
If you're doing interval arithmetic with dates, you should generally be using timestamps instead, as mentioned in the docs.
# SELECT extract(days FROM TO_TIMESTAMP('2999-01-01','yyyy-mm-dd') - TO_TIMESTAMP('2909-01-01','yyyy-mm-dd'))
date_part
-----------
32872
# SELECT extract(days FROM TO_TIMESTAMP('2999-01-01','yyyy-mm-dd') - (TO_TIMESTAMP('2909-01-01','yyyy-mm-dd') + '1 month'::interval) );
date_part
-----------
32841
The result of adding an interval to a date is actually a timestamp, not another date (the interval might have contained time portions), so you have to cast the result of the addition back down to date first:
SELECT
CAST( TO_DATE('2999-01-01','yyyy-mm-dd')
- CAST( (TO_DATE('2909-01-01','yyyy-mm-dd') + INTERVAL '1 MONTH' * (1) ) AS DATE)
AS DECIMAL )
With a date field I can do this:
ORDER BY ABS(expiry - CURRENT_DATE)
With a timestamp field I get the following error:
function abs(interval) does not exist
Use now() or CURRENT_TIMESTAMP for the purpose.
The reason for the different outcome of your queries is this:
When you subtract two values of type date, the result is an integer and abs() is applicable.
When you subtract two values of type timestamp (or just one is a timestamp), the result is an interval, and abs() is not applicable. You could substitute with a CASE expression:
ORDER BY CASE WHEN expiry > now() THEN expiry - now() ELSE now() - expiry END
Or you can extract() the unix epoch from the resulting interval like #Craig already demonstrated. I quote: "for interval values, the total number of seconds in the interval". Then you can use abs() again:
ORDER BY abs(extract(epoch from (expiry - now())));
age() would just add a more human readable representation to the interval by summing up days into months and years for for bigger intervals. But that's beside the point: the value is only used for sorting.
As your column is of type timestamp, you should use CURRENT_TIMESTAMP (or now()) instead of CURRENT_DATE, or you will get inaccurate results (or even incorrect for "today").
Compare with current_timestamp
SELECT the_timestamp > current_timestamp;
The age function is probably what you want when comparing them:
SELECT age(the_timestamp);
eg:
regress=# SELECT age(TIMESTAMP '2012-01-01 00:00:00');
age
----------------
8 mons 17 days
(1 row)
If you want an absolute distance, use:
SELECT abs( extract(epoch from age(the_timestamp)) );
This works (and gives the correct sorting):
ABS(EXTRACT(DAY FROM expiry - CURRENT_TIMESTAMP))
Unfortunately, as Erwin Brandstetter pointed out, it reduces the granularity of the sorting to a full day.