Oracle Add 1 hour in SQL - sql

I am just trying to add 1 hour to a value, it is kind of complicated on where and why i am doing this but basically i just need to query something like this
select DATE_ADD(hh,1,'2014-10-15 03:30:00 pm') from dual
I keep reading old articles that say to use dateAdd or date_add but I keep getting invalid identifier errors.

select sysdate + 1/24 from dual;
sysdate is a function without arguments which returns DATE type
+ 1/24 adds 1 hour to a date
select to_char(to_date('2014-10-15 03:30:00 pm', 'YYYY-MM-DD HH:MI:SS pm') + 1/24, 'YYYY-MM-DD HH:MI:SS pm') from dual;

Use an interval:
select some_date_column + interval '1' hour
from your_table;

You can use INTERVAL type or just add calculated number value - "1" is equal "1 day".
first way:
select date_column + INTERVAL '0 01:00:00' DAY TO SECOND from dual;
second way:
select date_column + 1/24 from dual;
First way is more convenient when you need to add a complicated value - for example, "1 day 3 hours 25 minutes 49 seconds".
See also: http://www.oracle-base.com/articles/misc/oracle-dates-timestamps-and-intervals.php
Also you have to remember that oracle have two interval types - DAY TO SECOND and YEAR TO MONTH.
As for me, one interval type would be better, but I hope people in oracle knows, what they do ;)

Old way:
SELECT DATE_COLUMN + 1 is adding a day
SELECT DATE_COLUMN + N /24 to add hour(s) - N being number of hours
SELECT DATE_COLUMN + N /1440 to add minute(s) - N being number of minutes
SELECT DATE_COLUMN + N /86400 to add second(s) - N being number of seconds
Using INTERVAL:
SELECT DATE_COLUMN + INTERVAL 'N' HOUR or MINUTE or SECOND - N being a number of hours or minutes or seconds.

To add/subtract from a DATE, you have 2 options :
Method #1 :
The easiest way is to use + and - to add/subtract days, hours, minutes, seconds, etc.. from a DATE, and ADD_MONTHS() function to add/subtract months and years from a DATE. Why ? That's because from days, you can get hours and any smaller unit (1 hour = 1/24 days), (1 minute = 1/1440 days), etc... But you cannot get months and years, as that depends on the month and year themselves, hence ADD_MONTHS() and no add_years(), because from months, you can get years (1 year = 12 months).
Let's try them :
SELECT TO_CHAR(SYSDATE, 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints current date: 19-OCT-2019 20:42:02
SELECT TO_CHAR((SYSDATE + 1/24), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date + 1 hour: 19-OCT-2019 21:42:02
SELECT TO_CHAR((SYSDATE + 1/1440), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date + 1 minute: 19-OCT-2019 20:43:02
SELECT TO_CHAR((SYSDATE + 1/86400), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date + 1 second: 19-OCT-2019 20:42:03
-- Same goes for subtraction.
SELECT SYSDATE FROM dual; -- prints current date: 19-OCT-19
SELECT ADD_MONTHS(SYSDATE, 1) FROM dual; -- prints date + 1 month: 19-NOV-19
SELECT ADD_MONTHS(SYSDATE, 12) FROM dual; -- prints date + 1 year: 19-OCT-20
SELECT ADD_MONTHS(SYSDATE, -3) FROM dual; -- prints date - 3 months: 19-JUL-19
Method #2 : Using INTERVALs, you can or subtract an interval (duration) from a date easily. More than that, you can combine to add or subtract multiple units at once (e.g 5 hours and 6 minutes, etc..)
Examples :
SELECT TO_CHAR(SYSDATE, 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints current date: 19-OCT-2019 21:34:15
SELECT TO_CHAR((SYSDATE + INTERVAL '1' HOUR), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date + 1 hour: 19-OCT-2019 22:34:15
SELECT TO_CHAR((SYSDATE + INTERVAL '1' MINUTE), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date + 1 minute: 19-OCT-2019 21:35:15
SELECT TO_CHAR((SYSDATE + INTERVAL '1' SECOND), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date + 1 second: 19-OCT-2019 21:34:16
SELECT TO_CHAR((SYSDATE + INTERVAL '01:05:00' HOUR TO SECOND), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date + 1 hour and 5 minutes: 19-OCT-2019 22:39:15
SELECT TO_CHAR((SYSDATE + INTERVAL '3 01' DAY TO HOUR), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date + 3 days and 1 hour: 22-OCT-2019 22:34:15
SELECT TO_CHAR((SYSDATE - INTERVAL '10-3' YEAR TO MONTH), 'DD-MON-YYYY HH24:MI:SS') FROM dual; -- prints date - 10 years and 3 months: 19-JUL-2009 21:34:15

The calculation is simple
if you want to add 1 hour in the date .
every day have 24 hour , you can add .
select sysdate + 1/24 from dual;
if you want 1 day to add
select sysdate + 24/24 from dual;
or
select sysdate + 1 from dual;
same as for 2, 3 , 4 day
For static date you have the answer below.

Related

Column based on time range in Oracle

I have a sales table with created datetime, my business hours are from 9 AM to 2 AM in the night on the following day. I am trying to convert the dates into my business date.
01/08/22 09:39:12.000000000 AM +04:00
Lets say I have a sale at 1 AM, this sale has to be considered in the previous day.
Any function that can help me solve this issue would be appreciated
It might be a bit of an overkill, but you could just use EXTRACT:
WITH dat AS
(
SELECT to_date('01/08/22 09:39:12','DD/MM/YY HH24:MI:SS') AS t_stmp FROM dual UNION ALL
SELECT to_date('02/08/22 01:03:15','DD/MM/YY HH24:MI:SS') FROM dual UNION ALL
SELECT to_date('02/08/22 08:27:33','DD/MM/YY HH24:MI:SS') FROM dual UNION ALL
SELECT to_date('02/08/22 14:11:51','DD/MM/YY HH24:MI:SS') FROM dual UNION ALL
SELECT to_date('02/08/22 02:01:15','DD/MM/YY HH24:MI:SS') FROM dual
)
SELECT CASE WHEN EXTRACT(HOUR FROM CAST(t_stmp AS TIMESTAMP)) BETWEEN 2 AND 8 THEN -1
ELSE 0
END + TRUNC(t_stmp,'DD') AS business_date
FROM dat;
business_date
01.08.2022
02.08.2022
01.08.2022
02.08.2022
01.08.2022
It looks like you just need to make a 2 hour shift to get your sales in the right date. You can add or substract hours from DATE/DATETIME/TIMESTAMP data type. If your column is TIMESTAMP then it would be like this:
-- when selecting data for date of sales
SELECT TRUNC(your_column_name - INTERVAL '2' HOUR, 'dd') "SALE_DATE"
-- And/Or
WHERE TRUNC(your_column_name - INTERVAL '2' HOUR, 'dd') = :DATE_OF_SALES
-- TRUNC function always returns DATE datatype
--
-- The opposite conversion would be
CAST(your_datetime_column + INTERVAL '2' HOUR as TIMESTAMP) ...
Here is the small sample with result:
SELECT
to_char(SYSDATE, 'dd.mm.yyyy hh24:mi:ss') "DATETIME",
to_char(SYSDATE - INTERVAL '2' HOUR, 'dd.mm.yyyy hh24:mi:ss') "DATETIME_MINUS_2H",
to_char(SYSDATE + INTERVAL '2' HOUR, 'dd.mm.yyyy hh24:mi:ss') "DATETIME_PLUS_2H",
to_char(SYSDATE - INTERVAL '10' HOUR, 'dd.mm.yyyy hh24:mi:ss') "DATETIME_MINUS_10H"
FROM
DUAL
--
-- R e s u l t
--
-- DATETIME DATETIME_MINUS_2H DATETIME_PLUS_2H DATETIME_MINUS_10H
-- ------------------- ------------------- ------------------- -------------------
-- 07.08.2022 09:58:38 07.08.2022 07:58:38 07.08.2022 11:58:38 06.08.2022 23:58:38
The last column now has the date from day before.

last day/time of month

How do I get the last day of the month to be 11:59:59 pm? Right now it is showing 05/31/2019 12:00 AM. If I just do sysdate it just shows the time I ran it. Thanks.
LAST_DAY(TRUNC(SYSDATE,'MM'))
Try adding one day to the last day of the current month, then subtracting one minute:
SELECT
LAST_DAY(TRUNC(SYSDATE,'MM')) + INTERVAL '1' DAY - INTERVAL '1' SECOND
FROM dual;
Or, if you want to also see the time component, use CURRENT_TIMESTAMP in place of SYSDATE:
SELECT
LAST_DAY(TRUNC(CURRENT_TIMESTAMP,'MM')) + INTERVAL '1' DAY - INTERVAL '1' SECOND
FROM dual;
This outputted:
30.06.2019 23:59:59
The oldfashioned way would be to subtract a second (that's what 1 / (24*60*60) represents as there are 24 hours in a day, 60 minutes in an hour and 60 seconds in a minute) from the first day of the next month (and that's what trunc(add_months(sysdate, 1)) does):
SQL> select sysdate today,
2 trunc(add_months(sysdate, 1)) - 1/(24*60*60) result
3 from dual;
TODAY RESULT
---------------------- ----------------------
06/01/2019 07:52:40 AM 06/30/2019 11:59:59 PM
SQL>
I would use:
SELECT TRUNC(SYSDATE, 'MM') + INTERVAL '1' MONTH - INTERVAL '1' SECOND
FROM dual;
(This has one less step than Tim's solution.)
Or in the older-fashioned method:
SELECT ADD_MONTHS(TRUNC(SYSDATE, 'MM'), 1) - 1/(24*60*60)
If you convert this to a timestamp, you will see the time value:
SELECT CAST(TRUNC(sysdate, 'MM') + INTERVAL '1' MONTH - INTERVAL '1' SECOND as TIMESTAMP)
FROM dual;

How to show interval between two timestamps in hh24:mi notation? [duplicate]

This question already has answers here:
Oracle SQL Hours Difference between dates in HH:MM:SS
(5 answers)
Closed 4 years ago.
How can I show interval between 2 timestamps in 'hh24:mi' format?
I have a Table with two Datestamps StartTime and EndTime between which I would like to see the difference in Hours and Minutes as HH24:mi
The difference is never above 24 hours but StartTime can be on day 1 and EndTime can be on day .
Example :
StartTime = 19/02/2019 22:52:42
EndTime = 20/02/2019 02:56:42
Result wanted = 04:04
So far the best results I managed to have are :
4,8 : ROUND ((EndTime - StartTime) * 24,2) INTERVAL
4:4 : EXTRACT (hour from numtodsinterval (EndTime - add_months (StartTime, floor (months_between (EndTime,StartTime))), 'day')) || ':'
|| EXTRACT (minute from numtodsinterval (EndTime - add_months (StartTime, floor (months_between (EndTime, StartTime))), 'day'))
Hours & Minutes in a seperate field : Hrs : 4 | Min : 4
trunc(((86400*(EndTime-StartTime))/60)/60)-24*(trunc((((86400*(EndTime-StartTime))/60)/60)/24)) "Hrs"
trunc((86400*(EndTime-StartTime))/60)-60*(trunc(((86400*(EndTime-StartTime))/60)/60)) "Min"
I wrote an anonymous block using your values. but if you are querying from a table it is not required. you can apply this logic in select query itself to get output.
declare
start_time date;
end_time date;
output number;
f_out varchar2(10);
begin
start_time :=to_Date('19/02/2019 22:52:42','dd/mm/yyyy hh24:mi:ss');
end_time:= to_Date('20/02/2019 02:56:42','dd/mm/yyyy hh24:mi:ss');
select (end_time-start_time) into output from dual;
--output := trunc(output*86400/3600);
f_out:=lpad(trunc(output*86400/3600),2,'0')||':'||lpad(mod(output*86400,3600)/60,2,'0');
dbms_output.put_line(f_out);
end;
You can do this by processing dates:
select to_char(date '2000-01-01' + (end_time - start_time), 'hh24:mi')
from (select to_Date('2019-02-19 22:52:42', 'yyyy-mm-dd hh24:mi:ss') as start_time,
to_Date('2019-02-19 02:56:42', 'yyyy-mm-dd hh24:mi:ss') as end_time
from dual
) t
Here is a db<>fiddle.

Oracle - fetch first 3 days record of the current month

I need to fetch first 3 days record of the current month from Oracle database. Something like below,
Select * from test.purchase where create_ts=( first 3 days of the current month)
Select *
from test.purchase
where create_ts between trunc(sysdate,'mm') and trunc(sysdate,'mm') + 3
You can get the first day of the current month with the trunc(date) function, using the MM date format element.
select to_char(trunc(sysdate, 'MM'), 'YYYY-MM-DD Hh24:MI:SS') from dual;
TO_CHAR(TRUNC(SYSDA
-------------------
2017-06-01 00:00:00
You can then use date arithmetic to either add a number of days or an interval representing that number to get the fourth day of the month:
select to_char(trunc(sysdate, 'MM') + 3, 'YYYY-MM-DD Hh24:MI:SS') from dual;
TO_CHAR(TRUNC(SYSDA
-------------------
2017-06-04 00:00:00
If you want data up to the start of that fourth day, i.e. up to 23:59:59 on the 3rd, you can look for values less than midnight on the 4th:
select * from test.purchase
where create_ts >= trunc(sysdate, 'MM')
and create_ts < trunc(sysdate, 'MM') + 3;
You could potentially use between, but as that is inclusive you would need to specify the absolute latest time on the 3rd - checking whether the column is a date or a timestamp, which might change, and can be a little confusing. If you used between trunc(sysdate, 'MM') and trunc(sysdate, 'MM') + 3 then you would include any records at exactly midnight on the 4th, which isn't what you want. I find using >= and < clearer and less ambiguous, even if it is a little more typing.
If the column is actually a timestamp then you can cast the calculated dates to timestamp too, and/or an use interval for the upper bound:
select * from test.purchase
where create_ts >= cast(trunc(sysdate, 'MM') as timestamp)
and create_ts < cast(trunc(sysdate, 'MM') + 3 as timestamp);
... or:
...
and create_ts < cast(trunc(sysdate, 'MM') as timestamp) + interval '3' day;

Increment Oracle time in varchar field by a certain amount?

We have times stored in an Oracle varchar(5) field.
Times stored using HH24:MI (ie: 22:30, 10:15).
How can we run a query that will increase or decrease these times by a certain amount? Ie: increase by one hour or decrease by 45 minutes.
you could use the built-in date (and interval -- thanks Alex for the link) calculation:
to_char(to_date(:x, 'hh24:mi') + INTERVAL :y MINUTE,'hh24:mi')
for instance:
SQL> WITH my_data AS (
2 SELECT '12:15' t FROM dual
3 UNION ALL SELECT '10:30' FROM dual
4 )
5 SELECT t,
6 to_char(to_date(t, 'hh24:mi') + INTERVAL '15' MINUTE,'hh24:mi')"t+15"
7 FROM my_data;
T t+15
----- -----
12:15 12:30
10:30 10:45
The functions to convert from/to date and char are TO_DATE() and TO_CHAR():
SELECT
TO_DATE('31/12/2001 23:55:00', 'DD/MM/YYYY HH24:MI:SS') AS "DATE",
TO_CHAR(CURRENT_TIMESTAMP, 'DD/MM/YYYY HH24:MI:SS') AS "CHAR"
FROM DUAL
So you can do:
SELECT
TO_DATE('23:45', 'HH24:MI'),
TO_DATE('23:45', 'HH24:MI') + INTERVAL '45' MINUTE,
TO_CHAR(TO_DATE('23:45', 'HH24:MI') + interval '45' MINUTE, 'HH24:MI')
FROM DUAL
There're possibly other better ways to do it (I'm not sure this will work as expected if ran when DST is about to start) but I'm still learning :)