This query: select EXTRACT(HOUR FROM 1435047532) as hour from TABLENAME;
Returns this error: "invalid extract field for extract source".
I'm trying to extract the hour from a given timestamp. Maybe the problem is the format of timestamp field is NUMBER and not TIMESTAMP?
You can convert your numeric seconds-since-epoch time to a TIMESTAMP type using:
TIMESTAMP '1970-01-01 00:00:00' + NUMTODSINTERVAL( your_time_since_epoch, 'SECOND' )
So to get the hours:
SELECT EXTRACT( HOUR FROM TIMESTAMP '1970-01-01 00:00:00'
+ NUMTODSINTERVAL( 1435047532, 'SECOND' ) )
FROM DUAL;
If you need to handle leap seconds then you will need to create a function/package to handle this.
try this
select extract(hour from (select to_date('19700101', 'YYYYMMDD')
+ ( 1 / 24 / 60 / 60 ) * 1435047532 from dual)) from dual
Related
I'm trying to get the minute value from a Timestamp after a subtraction
First I made a subtraction:
Current Time - Target Time(INSP_FIN_DT)
(
TO_TIMESTAMP(TO_CHAR(SYSDATE,'YYYY-MM-DD HH24:MI:SS'),'YYYY-MM-DD HH24:MI:SS') -
TO_TIMESTAMP(INSP_FIN_DT,'YYYY-MM-DD HH24:MI:SS')
) AS REST_TIME
the output:
+00 00:09:44.000000
What I need:
09
I already tried
SELECT SUBSTR(REST_TIME, 7,2)
But the return is ever 00, even If I convert it to Char like:
SELECT SUBSTR(TO_CHAR(REST_TIME),7,2)
How can I solve it by query?
Information here https://dba.stackexchange.com/questions/53924/how-do-i-get-the-difference-in-minutes-from-2-timestamp-columns
select
round(
(SYSDATE - cast(<other_timestamp> as date))
* 24 * 60
) as diff_minutes
from <some_table>;
For my current implementation, I'll never have more than a hour, in this case I used the bellow solution:
SYSTIMESTAMP - TO_TIMESTAMP(INSP_FIN_DT,'YYYY-MM-DD HH24:MI:SS') AS REST_TIME
Select..
EXTRACT(MINUTE FROM REST_TIME) AS REST_MINUTES
In case of more than 60 minutes, I should use:
(EXTRACT(HOUR FROM REST_TIME) * 60)
+ EXTRACT(MINUTE FROM REST_TIME) AS REST_MINUTES
I have a table with two columns, Start_time and SLA.
Start time updates for each day and is in a date format e.g., 01-Jun-2021 19:15:38
SLA column is having fixed HH24MI as 2010
I want 1915 - 2010 to be -00:55 (as in HH24MI format)
SELECT TO_CHAR((TO_CHAR(START_TIME,'HH24')||TO_CHAR(START_TIME,'MI'))-2010,'0000')
FROM DUAL;
Above will give the result as -0095 but I want it to be -00:55
Store the start_time as a DATE data type and the sla as an INTERVAL DAY TO SECOND data type:
CREATE TABLE table_name (
start_time DATE,
sla INTERVAL DAY TO SECOND
);
Then your data would be:
INSERT INTO table_name ( start_time, sla ) VALUES (
TO_DATE('01-Jun-2021 19:15:38', 'DD-MON-YYYY HH24:MI:SS', 'NLS_DATE_LANGUAGE=American'),
INTERVAL '20:10:00' HOUR TO SECOND
);
And, to find the difference, you can use:
SELECT start_time,
sla,
(start_time - TRUNC(start_time)) DAY TO SECOND - sla AS difference
FROM table_name
Which outputs:
START_TIME
SLA
DIFFERENCE
2021-06-01 19:15:38
+00 20:10:00.000000
-000000000 00:54:22.000000000
If you want the output as a formatted string, rather than as an interval, then:
SELECT start_time,
sla,
CASE WHEN difference < INTERVAL '0' HOUR THEN '-' END
|| TO_CHAR( ABS( EXTRACT( HOUR FROM difference ) ), 'FM00' )
|| TO_CHAR( ABS( EXTRACT( MINUTE FROM difference ) ), 'FM00' )
AS difference
FROM (
SELECT start_time,
sla,
(start_time - TRUNC(start_time)) DAY TO SECOND - sla AS difference
FROM table_name
)
Which outputs:
START_TIME
SLA
DIFFERENCE
2021-06-01 19:15:38
+00 20:10:00.000000
-0054
db<>fiddle here
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.
How to round up time interval to next day in Oracle SQL?
select apppackage
, numtodsinterval(
sum( trunc(extract (day from (periods)) * 86400
+ extract (hour from (periods)) *3600
+ extract (minute from (periods))*60
+ extract (second from (periods)))
), 'SECOND') as retention_period
, count(apppackage) as users
from retentions
where apppackage = 'com.Freesoul.Rotter'
group by apppackage;
The output of this is
'com.Freesoul.Rotter' '+2969 04:32:47.000000' '3'
and desired output is
'com.Freesoul.Rotter' '2970' '3'
but if the output of query is
'com.Freesoul.Rotter' '+2969 00:00:00.000000' '3'
then desired output is
'com.Freesoul.Rotter' '2969' '3'
column period is of INTERVAL DAY(9) TO SECOND(6) type
and i won't mind if the retention_period is changed to number datatype.
I'll be grateful if anyone can suggest change in my query to attain the desired output.
The result of your sum is in seconds, so you don't really need to convert it to an interval at all. Just divide by 60*60*24 to get the answer in days, and round it up with ceil():
select apppackage
, ceil(sum( trunc(extract (day from (periods)) * 86400
+ extract (hour from (periods)) *3600
+ extract (minute from (periods))*60
+ extract (second from (periods)))
) / 86400) as retention_period
, count(apppackage) as users
from retentions
where apppackage = 'com.Freesoul.Rotter'
group by apppackage;
Demo with artificial data in a CTE just to mimic your expected results, for both scenarios:
-- CTE for sample data
with retentions (apppackage, periods) as (
select 'com.Freesoul.Rotter', interval '+2967 04:32:47.000000' day(9) to second(6) from dual
union all
select 'com.Freesoul.Rotter', interval '1' day from dual
union all
select 'com.Freesoul.Rotter', interval '1' day from dual
union all
select 'com.Freesoul.XYZ', interval '+2967 00:00:00.000000' day(9) to second(6) from dual
union all
select 'com.Freesoul.XYZ', interval '1' day from dual
union all
select 'com.Freesoul.XYZ', interval '1' day from dual
)
-- actual query
select apppackage
, ceil(sum( trunc(extract (day from (periods)) * 86400
+ extract (hour from (periods)) *3600
+ extract (minute from (periods))*60
+ extract (second from (periods)))
) / 86400) as retention_period
, count(apppackage) as users
from retentions
where apppackage = 'com.Freesoul.Rotter'
-- extra clause for dummy data
or apppackage = 'com.Freesoul.XYZ'
group by apppackage;
APPPACKAGE RETENTION_PERIOD USERS
------------------- ---------------- ----------
com.Freesoul.XYZ 2969 3
com.Freesoul.Rotter 2970 3
Your expected output shows a plain number. If you actually want it as an interval, but as the whole number of days, just pass ceil'd number into numtodsinterval or more simply (and usually faster for some reason) multiply by interval '1' day.
With the same dummy data:
select apppackage
, ceil(sum( trunc(extract (day from (periods)) * 86400
+ extract (hour from (periods)) *3600
+ extract (minute from (periods))*60
+ extract (second from (periods)))
) / 86400) * interval '1' day as retention_period
, count(apppackage) as users
...
APPPACKAGE RETENTION_PERIOD USERS
------------------- --------------------- ----------
com.Freesoul.XYZ +2969 00:00:00.000000 3
com.Freesoul.Rotter +2970 00:00:00.000000 3
As #mathguy pointed out, you probably don't need or want the trunc() call in there; that is removing the fractional seconds from each period before they are summed, which sounds insignificant but could easily affect the result you get.
How about this?
CASE WHEN numtodsinterval(extract (day from periods), 'DAY') = periods THEN
extract (day from periods)
ELSE
extract (day from periods) + 1
END
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.