How can i convert the result of select statement of time interval field in respective time in Am/Pm format.
My Field is:
Interval Day(2) To Second(6)
I tried this:
select To_Char(Att_EntranceTime , 'HH:MI AM') From EMPLOYEEATTENDENCETABLE;
however this does not help me, i have also tried to add the basetime from systime to my interval field but that did not help.. can someone suggest me what to do?
Intervals can't be directly formatted, as you've discovered. You can add your interval to any date which has its time set to midnight, and then format the resulting date to show the time in your desired format. For example you could add it to today's date using trunc(sysdate):
to_char(trunc(sysdate) + my_interval, 'HH:MI AM')
You need to truncate it to set the time to midnight; otherwise the result will be your interval plus the current system time.
Or you can use any fixed date; here's an example with some dummy data set-up:
create table my_table (my_interval interval day(2) to second(6));
insert into my_table (my_interval) values (interval '0 12:34:56.78' day to second);
insert into my_table (my_interval) values (interval '99 01:02:03.456' day to second);
select my_interval, to_char(date '1970-01-01' + my_interval, 'HH:MI AM') as formatted
from my_table;
MY_INTERVAL FORMATTED
-------------------- ---------
+00 12:34:56.780000 12:34 PM
+99 01:02:03.456000 01:02 AM
The second value shows a potential problem. Your interval is defined to allow a two-digit day number, which means the interval can span anything less than 100 days. If you only extract the time portion you lose that information about the number of days. That may be what you want to happen though. If the interval is supposed to be representing a time of day, which wanting to show AM/PM implies - and it's unusual to store an actual time separate from its date - then having or allowing a number of days seems strange.
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 want to SELECT a TIMESTAMP(6) with milliseconds, but at the same time I need to subtract 3 hours (0.125 of a day) from that TIMESTAMP to convert it to my timezone. So I tried:
SELECT To_Char(UTCSCANTIME-0.125,'YYYY-MM-DD HH24:MI:SS') AS LOCALSCANTIME
Outcome: 2018-08-01 19:22:39
If I append "FF" to show milliseconds:
SELECT To_Char(UTCSCANTIME-0.125,'YYYY-MM-DD HH24:MI:SS.FF') AS LOCALSCANTIME
Outcome: ORA-01821: date format not recognized
However if I keep the "FF" but I don't subtract 0.125:
SELECT To_Char(UTCSCANTIME,'YYYY-MM-DD HH24:MI:SS.FF') AS LOCALSCANTIME
Outcome: 2018-08-01 22:22:39.259000
How can I achieve both things?
Thanks in advance!
Subtract a 3 hour INTERVAL instead of 0.125.
SELECT To_Char(UTCSCANTIME-INTERVAL '3' HOUR,'YYYY-MM-DD HH24:MI:SS.FF') AS LOCALSCANTIME
Subtracting a number like 0.125 implicitly converts the result to a DATE, losing your fractional seconds.
Also, note, there are better ways in Oracle to convert time zones than to add and subtract intervals.
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
I need help to write query do the following:
subtract between two columns (start date and end date), and please note that the type for the columns are char not date, this is the exact format: 10-MAR-12 11.11.40.288389000 AM), then get the average for the result.
I assume this is homework, so some hints...
First don't ever store dates as varchar, it will cause you and the optimiser all sorts of problems.
Second, Oracle's date datatype can only store to second precision, and your string has fractions of a second, so you are looking at timestamp rather than date. You can convert your string to a timestamp with the to_timestamp() function, passing a suitable format mask. Oh OK, I'm feeling generous:
select to_timestamp(start_date, 'DD-Mon-RR HH.MI.SS.FF9 AM') from your_table;
Third, subtracting two timestamps will give you an interval data type, from which you will need to extract the information you want in a readable format. Search this site or elsewhere for timestamp subtraction, but I'll point you at this recent one as a sample.
The average is a bit trickier, so you may want to convert your intervals to numbers for that; again search for previous questions, such as this one. The size of the intervals, the precision you actually care about, and the way you want the output formatted, etc. will have some bearing on the approach you want to take.
If you need an approximate result then #Joachim Isaksson's answer will give you that - 'approximate' because of rounding; a duration of less than a second will show up as zero, for example. The same effect can be seen with timestamps cast to dates, which also loses the fractional seconds:
select 24*60*60*avg(
cast(to_timestamp(step_ending_time, 'DD-Mon-RR HH.MI.SS.FF9 AM') as date)
- cast(to_timestamp(step_starting_time, 'DD-Mon-RR HH.MI.SS.FF9 AM') as date)
) as avg_duration
from process_audit;
A more accurate answer can be found by extracting the various components of the timestamps, as in a question I linked to earlier. You may not need them all if you know that your durations are always less then an hour, say, but if you need more than one (i.e. if a duration could be more than a minute) then using an intermediate common table expression simplifies things a bit:
with cte as (
select to_timestamp(step_ending_time, 'DD-Mon-RR HH.MI.SS.FF9 AM')
- to_timestamp(step_starting_time, 'DD-Mon-RR HH.MI.SS.FF9 AM') as duration
from process_audit
)
select avg(extract(second from duration)
+ extract(minute from duration) * 60
+ extract(hour from duration) * 60 * 60
+ extract(day from duration) * 60 * 60 * 24) as avg_duration
from cte;
With two sample rows, one with a gap of exactly a second and one with exactly 1.5 seconds, this gives the result 1.25.
Comments about storing times in VARCHAR aside; Oracle's to_date to the rescue; this should work for you to show the average number of seconds between the times. Since you're a bit low on details on precision, I didn't bother about the "sub seconds";
SELECT 24*3600*AVG(
to_date(enddate, 'DD-Mon-YY HH.Mi.SS.????????? AM') -
to_date(startdate, 'DD-Mon-YY HH.Mi.SS.????????? AM')) avg_seconds
FROM TableA;
Demo here.
I need to add 30 minutes to values in a Oracle date column. I do this in my SELECT statement by specifying
to_char(date_and_time + (.000694 * 31)
which works fine most of the time. But not when the time is on the AM/PM border. For example, adding 30 minutes to 12:30 [which is PM] returns 1:00 which is AM. The answer I expect is 13:00. What's the correct way to do this?
In addition to being able to add a number of days to a date, you can use interval data types assuming you are on Oracle 9i or later, which can be somewhat easier to read,
SQL> ed
Wrote file afiedt.buf
SELECT sysdate, sysdate + interval '30' minute FROM dual
SQL> /
SYSDATE SYSDATE+INTERVAL'30'
-------------------- --------------------
02-NOV-2008 16:21:40 02-NOV-2008 16:51:40
All of the other answers are basically right but I don't think anyone's directly answered your original question.
Assuming that "date_and_time" in your example is a column with type DATE or TIMESTAMP, I think you just need to change this:
to_char(date_and_time + (.000694 * 31))
to this:
to_char(date_and_time + (.000694 * 31), 'DD-MON-YYYY HH24:MI')
It sounds like your default date format uses the "HH" code for the hour, not "HH24".
Also, I think your constant term is both confusing and imprecise. I guess what you did is calculate that (.000694) is about the value of a minute, and you are multiplying it by the number of minutes you want to add (31 in the example, although you said 30 in the text).
I would also start with a day and divide it into the units you want within your code. In this case, (1/48) would be 30 minutes; or if you wanted to break it up for clarity, you could write ( (1/24) * (1/2) ).
This would avoid rounding errors (except for those inherent in floating point which should be meaningless here) and is clearer, at least to me.
UPDATE "TABLE"
SET DATE_FIELD = CURRENT_TIMESTAMP + interval '48' minute
WHERE (...)
Where interval is one of
YEAR
MONTH
DAY
HOUR
MINUTE
SECOND
from http://www.orafaq.com/faq/how_does_one_add_a_day_hour_minute_second_to_a_date_value
The SYSDATE pseudo-column shows the current system date and time. Adding 1 to SYSDATE will advance the date by 1 day. Use fractions to add hours, minutes or seconds to the date
SQL> select sysdate, sysdate+1/24, sysdate +1/1440, sysdate + 1/86400 from dual;
SYSDATE SYSDATE+1/24 SYSDATE+1/1440 SYSDATE+1/86400
-------------------- -------------------- -------------------- --------------------
03-Jul-2002 08:32:12 03-Jul-2002 09:32:12 03-Jul-2002 08:33:12 03-Jul-2002 08:32:13
I prefer using an interval literal for this, because interval '30' minute or interval '5' second is a lot easier to read then 30 / (24 * 60) or 5 / (24 * 60 * 69)
e.g.
some_date + interval '2' hour
some_date + interval '30' minute
some_date + interval '5' second
some_date + interval '2' day
You can also combine several units into one expression:
some_date + interval '2 3:06' day to minute
Adds 2 days, 3 hours and 6 minutes to the date value
The above is also standard SQL and also works in several other DBMS.
More details in the manual: https://docs.oracle.com/database/121/SQLRF/sql_elements003.htm#SQLRF00221
If the data type of the field is date or timestamp, Oracle should always give the correct result if you add the correct number given in number of days (or a the correct fraction of a day in your case). So if you are trying to bump the value in 30 minutes, you should use :
select field + 0.5/24 from table;
Based on the information you provided, I believe this is what you tried to do and I am quite sure it works.
Can we not use this
SELECT date_and_time + INTERVAL '20:00' MINUTE TO SECOND FROM dual;
I am new to this domain.
like that very easily
i added 10 minutes to system date and always in preference use the Db server functions not custom one .
select to_char(sysdate + NUMTODSINTERVAL(10,'MINUTE'),'DD/MM/YYYY HH24:MI:SS') from dual;
Be sure that Oracle understands that the starting time is PM, and to specify the HH24 format mask for the final output.
SELECT to_char((to_date('12:40 PM', 'HH:MI AM') + (1/24/60) * 30), 'HH24:MI') as time
FROM dual
TIME
---------
13:10
Note: the 'AM' in the HH:MI is just the placeholder for the AM/PM meridian indicator. Could be also 'PM'
Oracle now has new built in functions to do this:
select systimestamp START_TIME, systimestamp + NUMTODSINTERVAL(30, 'minute') end_time from dual
Based on what you're asking for, you want the HH24:MI format for to_char.
To edit Date in oracle you can try
select to_char(<columnName> + 5 / 24 + 30 / (24 * 60),
'DD/MM/RRRR hh:mi AM') AS <logicalName> from <tableName>
SELECT to_char(sysdate + (1/24/60) * 30, 'dd/mm/yy HH24:MI am') from dual;
simply you can use this with various date format....