In oracle I have function which take number of days after 1.1.1900 (Java) and convert it to timestamp it looks like:
TO_TIMESTAMP( day + 2447893, 'J');
so for example when day = 8843 it should return 19.3.2014
what is the equivalent this function in informix ?
You can wrote stored function that uses INTERVAL with days of precision 5:
... DATETIME(1990-01-01) YEAR TO DAY + INTERVAL(8843) DAY(5) TO DAY ...
Related
select (current_date - TO_DATE('20210817124015','YYYYMMDDHH24MISS')) from dual;
Outputs:
0.1229282407407407407407407407407407407407
I want to know how oracle internally achieves this value.
ps: the current_date and the hardcoded date are same, only time is the difference.
CURRENT_DATE returns the current date and time in the user's session time zone.
TO_DATE('20210817124015','YYYYMMDDHH24MISS') returns the date 2021-08-17T12:40:15.
Note: A DATE data type always has year, month, day, hour, minute and second components. However, the user interface you are using may chose not to show all the components.
Subtracting one date from another returns the number of days between the two values.
0.1229282407407407407407407407407407407407 days is:
2.950277778 hours; or
177.016666667 minutes; or
10621 seconds; or
2 hours 57 minutes and 1 second.
So your current date was 2021-08-17T12:40:15 + 10621 seconds or 2021-08-17T15:37:16.
For example:
ALTER SESSION SET NLS_DATE_FORMAT = 'YYYY-MM-DD"T"HH24:MI:SS';
ALTER SESSION SET TIME_ZONE = 'Asia/Samarkand';
SELECT CURRENT_DATE,
TO_DATE('20210817124015','YYYYMMDDHH24MISS') As other_date,
CURRENT_DATE - TO_DATE('20210817124015','YYYYMMDDHH24MISS') as difference,
(CURRENT_DATE - TO_DATE('20210817124015','YYYYMMDDHH24MISS')) DAY TO SECOND
as interval_difference
FROM DUAL;
Outputs:
CURRENT_DATE
OTHER_DATE
DIFFERENCE
INTERVAL_DIFFERENCE
2021-08-17T15:40:01
2021-08-17T12:40:15
.124837962962962962962962962962962962963
+00 02:59:46.000000
db<>fiddle here
Subtracting two dates returns a difference in days.
0.1229282407407407407407407407407407407407 days is
2.9502777777768 hours
177.016666666608 minutes
10621 seconds
Or, put another way, current_date is returning a date value that is 2 hours 57 minutes and 1 second after the hard-coded date. Since the hard-coded date has a time of 12:40:51, that means that current_date has a time of 15:37:52.
I have a column with interval year to month datatype. With Extract function I get only month value from interval. But how to get full month (year * 12 + month)?
I can use Extract for year, multiply it by 12 and add it to the extracted month. But how to do it simply, using 1 function?
I am trying to add a day with a specific date using add_months in oracle database.
I wrote this line:
SELECT ADD_MONTHS('01-JAN-2018', MONTHS_BETWEEN('02-JAN-2018', '01-JAN-2018')) FROM DUAL;
this returns:
01-JAN-18
Why doesn't it return 02-JAN-18?? Can I add one day to the date using this function?
Why doesn't it return 02-JAN-18??
According to MONTHS_BETWEEN documentation,
The MONTHS_BETWEEN function calculates the number of months between
two dates. When the two dates have the same day component or are both
the last day of the month, then the return value is a whole number.
Otherwise, the return value includes a fraction that considers the
difference in the days based on a 31-day month
So,
select MONTHS_BETWEEN('02-JAN-2018', '01-JAN-2018') FROM DUAL ;
yields
.0322580645161290322580645161290322580645
ADD_MONTHS returns the date date plus integer months.
So, .0322.. is considered as integer 0 and your query is equivalent to
SELECT ADD_MONTHS('01-JAN-2018', 0) FROM DUAL;
In order to add 1 months, simply take the difference of two dates.
SELECT ADD_MONTHS(DATE '2018-01-01', DATE '2018-01-02' - DATE '2018-01-01') FROM DUAL;
Or better, add an INTERVAL of 1 month
SELECT DATE '2018-01-01' + INTERVAL '1' MONTH FROM DUAL;
To answer your question, add 1 day, simply use
SELECT DATE '2018-01-01' + 1 FROM DUAL;
I'm trying to get a new date from the product of 'date' + 'time interval'.
Something like this.
'15/02/2016 18:00:00' + '+00 02:00:00.000000'
Expected result:
'15/02/2016 20:00:00'
But using the columns in database.
CREATE TABLE timerest
(
DATE_ASIGN DATE,
TIME_ASIGN INTERVAL DAY(2) TO SECOND(0)
);
Thanks for your help.
You can just add them together:
insert into timerest (date_asign, time_asign)
values (to_date('15/02/2016 18:00:00', 'DD/MM/YYYY HH24:MI:SS'),
to_dsinterval('+00 02:00:00.000000'));
alter session set NLS_DATE_FORMAT = 'DD/MM/YYYY HH24:MI:SS';
select date_asign + time_asign from timerest;
DATE_ASIGN+TIME_ASIGN
---------------------
15/02/2016 20:00:00
This follows the rules for datetime/interval arithmetic: date + interval = date.
If you have a date in DATE format you can simply add the a numeric interval that represents days (for example 1.5 is 1 day and a half)
You can extract from the time interval days and hours and then add them to to you date because, if I remember correctly, you can't add directly them to a date type (maybe to a timestamp type you can)
To extract the days you can use the extract function:
(
extract(second from TIME_ASIGN)/3600)+(extract(hour from TIME_ASIGN)/24)+(extract(day from TIME_ASIGN)/24)
then you add the number to your DATE_ASIGN
There are many question posted about getting the difference between two dates in Oracle. My question is requires the query to do a couple more things.
Here's how far I have got at the moment
select m_bug_t.date_submitted, m_bug_history_t.date_modified
from m_bug_t, m_bug_history_t
where m_bug_t.id = m_bug_history_t.bug_id
and field_name = 'status'
and new_value = '100'
So far I get a set of date pairs returned like this
date_submitted | date_modified
1314894774 | 1315906468
...
...
I want to convert these numbers to dates, find the difference between them and then get the minimum of all the results. I want the difference to be represented as days.
Any ideas how you do this?
Thanks very much :).
Well, Unix timestamps are expressed as a number of seconds since 01 Jan 1970, so if you subtract one from the other you get the difference in seconds. The difference in days is then simply a matter of dividing by the number of seconds in a day:
(date_modified - date_submitted) / (24*60*60)
or
(date_modified - date_submitted) / 86400
To convert UNIX time to a date you can use:
DATE '1970-01-01' + numtodsinterval(:unix_time_stamp, 'second')
In SQL when you substract two dates you will get the difference in days so you could write:
SELECT MIN(dt_mod - dt_sub)
FROM (SELECT DATE '1970-01-01'
+ numtodsinterval(m_bug_t.date_submitted, 'second') dt_sub,
DATE '1970-01-01'
+ numtodsinterval(m_bug_history_t.date_modified, 'second') dt_mod
FROM m_bug_t, m_bug_history_t
WHERE m_bug_t.id = m_bug_history_t.bug_id
AND field_name = 'status'
AND new_value = '100')
Of course as others have suggested you don't really need to do this DATE conversion, you could just substract your 2 timestamps (difference in seconds) and convert the result in days.