Let's suppose I have 2 intervals:
INTERVAL '0 00:30:00' DAY TO SECOND
INTERVAL '0 04:00:00' DAY TO SECOND
What is the most elegant way to get amount of minutes in each interval. 30 and 240 accordingly.
Yes, I know I can perform EXTRACT(HOUR FROM interval) * 60 + EXTRACT(MINUTE FROM interval), but this looks terrible to me.
Any better solutions?
What looks terrible to you, looks perfectly acceptable to me. If you look at the documentation at the arithmetic you can perform on INTERVALs:
http://download.oracle.com/docs/cd/E11882_01/server.112/e17118/sql_elements001.htm#sthref175
then you see you can multiply them with numerics. So if you multiply your intervals to 24 and 60, you can get the number of minutes by extracting the number of days. It's more compact, but I doubt if it's more elegant in your view.
SQL> create table t (my_interval interval day to second)
2 /
Table created.
SQL> insert into t
2 select numtodsinterval(30,'minute') from dual union all
3 select numtodsinterval(4,'hour') from dual
4 /
2 rows created.
SQL> select my_interval
2 , 60 * extract(hour from my_interval)
3 + extract(minute from my_interval) minutes_terrible_way
4 , extract(day from 24*60*my_interval) minutes_other_way
5 from t
6 /
MY_INTERVAL MINUTES_TERRIBLE_WAY MINUTES_OTHER_WAY
------------------------------ -------------------- -----------------
+00 00:30:00.000000 30 30
+00 04:00:00.000000 240 240
2 rows selected.
Regards,
Rob.
(extract(day from my_interval) * 24 * 60 + extract(hour from my_interval)) * 60 + extract(minute from my_interval) my_way
Don't forget about days.
Reusing Rob's example above,
select (sysdate-(sysdate-to_dsinterval(my_interval)))*24*60 from t;
Related
can you help me please.
I'm use:
round((24 * 60 * (to_date(EndPick, 'HH24:MI') - to_date(StartPick, 'HH24:MI'))), 2) as WorkTime
All time EndPick bigger then StartPick. But we have a chance, what Start bigger where
estimate time interval if there is a crossing over 00:00 into the next date?
In relult for example:
3
(minutes)
But we have a chance, what Start bigger where estimate time interval if there is a crossing over 00:00 into the next date?
Use a CASE expression and add a day if startpick > endpick:
SELECT round(
(24 * 60 * (to_date(EndPick, 'HH24:MI')
+ CASE
WHEN startpick > endpick
THEN INTERVAL '1' DAY
ELSE INTERVAL '0' DAY
END
- to_date(StartPick, 'HH24:MI')
)),
2
) as WorkTime
FROM table_name;
Which, for the sample data:
CREATE TABLE table_name (startpick, endpick) AS
SELECT '00:00', '01:00' FROM DUAL UNION ALL
SELECT '23:00', '00:00' FROM DUAL;
Outputs:
WORKTIME
60
60
fiddle
I was curious to see how in Oracle 12c you can take a timestamp datatype and convert the records into EPOCH time to make them a number and then use that number to find any records within that date column that are within 1 minute of each other (assuming the same day if needed, or simply any calculations within 1 minute).
I tried the following but got an ORA-01873: the leading precision of the interval is too small error.
select (sold_date - to_date('1970-01-01 00:00:00','YYYY-MM-DD HH24:MI:SS'))*86400 as epoch_sold_date from test1;
What is SOLD_DATE? For e.g. SYSDATE (function that returns DATE datatype), your code works OK.
SQL> select (sysdate
2 - to_date('1970-01-01 00:00:00','YYYY-MM-DD HH24:MI:SS')
3 ) * 86400 as epoch_sold_date
4 from dual;
EPOCH_SOLD_DATE
---------------
1600807918
SQL>
As SOLD_DATE is a timestamp, but - it appears that fractions of a second aren't or special interest to you, cast it to DATE:
select (cast (systimestamp as date) --> this
- to_date('1970-01-01 00:00:00','YYYY-MM-DD HH24:MI:SS')
) * 86400 as epoch_sold_date
from dual;
Saying that you get the same result for all rows: well, I don't, and you shouldn't either if SOLD_DATE differs.
SQL> with test (sold_date) as
2 (select timestamp '2020-09-22 00:00:00.000000' from dual union all
3 select timestamp '2015-03-18 00:00:00.000000' from dual
4 )
5 select sold_date,
6 (cast (sold_date as date)
7 - to_date('1970-01-01 00:00:00','YYYY-MM-DD HH24:MI:SS')
8 ) * 86400 as epoch_sold_date
9 from test;
SOLD_DATE EPOCH_SOLD_DATE
------------------------------ ---------------
22.09.20 00:00:00,000000000 1600732800
18.03.15 00:00:00,000000000 1426636800
SQL>
One more edit: when you subtract two timestamps, result is interval day to second. If you extract minutes from it, you get what you wanted:
SQL> with test (sold_date) as
2 (select timestamp '2020-09-22 10:15:00.000000' from dual union all
3 select timestamp '2015-03-18 08:05:00.000000' from dual
4 )
5 select sold_date,
6 lead(sold_date) over (order by sold_date) next_sold_date,
7 --
8 lead(sold_date) over (order by sold_date) - sold_date diff,
9 --
10 extract(day from lead(sold_date) over (order by sold_date) - sold_date) diff_mins
11 from test
12 order by sold_date;
SOLD_DATE NEXT_SOLD_DATE DIFF DIFF_MINS
------------------------------ ------------------------------ ------------------------------ ----------
18.03.15 08:05:00,000000000 22.09.20 10:15:00,000000000 +000002015 02:10:00.000000000 2015
22.09.20 10:15:00,000000000
SQL>
In your case, you'd check whether extracted minutes value is larger than 1 (minute).
If you just want to see how many minutes are there between two timestamps, then
cast them to dates
subtract those dates (and you'll get number of days)
multiply it by 24 (as there are 24 hours in a day) and by 60 (as there are 60 minutes in an hour)
Something like this:
SQL> with test (date_1, date_2) as
2 (select timestamp '2020-09-22 10:15:00.000000',
3 timestamp '2020-09-22 08:05:00.000000' from dual
4 )
5 select (cast(date_1 as date) - cast(date_2 as date)) * 24 * 60 diff_minutes
6 from test;
DIFF_MINUTES
------------
130
SQL>
If you are just looking to compare dates and find rows that are within one minute of each other, you do not need to use epoch time. There are several solutions to this problem on this thread.
I would like to compare two time values. The first time value is a custom time which reprsents the start time, for example the column name is Business_Start_time and set to 6:00:00 am. I would also like to extract the time only from a column in Oracle which is a date field that looks like '5/1/2019 12:57:19 PM' and is called 'Completed_Date_Time'. The purpose of this is to compare the businses start date to the time a file was completed. I've tried to convert the 'Completed_Date_Time' field to 'HH24:MI:SS' format which seems to change the datatype to a char(8) value which does not allow me to compare two timestamps.
CAST(TO_CHAR(Completed_Date_Time, 'HH:MI:SS AM') AS CHAR(8))
Convert the values to TIMESTAMP and then you can subtract the values from the values truncated to the start of the day to get an INTERVAL containing the time since midnight and to get the difference you can subtract.
Oracle Setup:
CREATE TABLE table_name ( Business_Start_time, Completed_Date_Time ) AS
SELECT '6:00:00 AM',
TO_DATE( '5/1/2019 12:57:19 PM', 'DD/MM/YYYY HH12:MI:SS AM' )
FROM DUAL
Query:
SELECT ( completed_time - TRUNC( completed_time ) ) -
( start_time - TRUNC( start_time ) ) AS time_difference
FROM (
SELECT TO_TIMESTAMP( business_start_time, 'HH12:MI:SS AM' ) AS start_time,
CAST( Completed_Date_Time AS TIMESTAMP ) AS completed_time
FROM table_name
)
Output:
| TIME_DIFFERENCE |
| :---------------------------- |
| +000000000 06:57:19.000000000 |
db<>fiddle here
Although you wrote both the question and a comment, I'm still not sure what you have and what you want to get. Sample case would help (create table & insert into).
Meanwhile, a few words about it: when subtracting two DATE datatype values, the result is number of days, which means that - if you want to display it in a format which is easier to read & understand - you have to do some calculations (a day has 24 hours; an hour has 60 mintues; and so forth).
Here's an example:
SQL> create table test
2 (business_Start_time date,
3 completed_date_Time date
4 );
Table created.
SQL> insert into test (business_start_time, completed_date_time) values
2 (to_date('05.01.2019 12:57:19', 'dd.mm.yyyy hh24:mi:ss'),
3 to_date('05.01.2019 18:58:20', 'dd.mm.yyyy hh24:mi:ss'));
1 row created.
Simply subtracted, you'd get
SQL> select completed_date_time - business_start_time result from test;
RESULT
----------
,250706019
SQL>
Here's a function which presents such a value in another format, dd:hh:mi (days:hours:minutes) (you can omit days by setting the second parameter to 0):
SQL> create or replace
2 function f_days2ddhhmi (par_broj_dana in number, par_cb_dd in number)
3 return varchar2
4 is
5 /* Converting number of days into dd:hh:mi format
6
7 Date from Date to Diff (days) Retval
8 -------------------- -------------------- -------------- ----------------------------------
9 20.11.2018. 07:00:00 - 20.11.2018. 13:45:00 0,28125 0:06:45 (6 hours 45 minutes)
10 23.10.2018. 07:00:00 - 25.10.2018. 22:12:00 2,63333 2:15:12 (2 daysa 15 hours 12 minutes)
11
12 PAR_BROJ_DANA: 0.28125
13 PAR_CB_DD : display number of days or not? 1 - yes --> 0:06:45
14 0 - no --> 06:45
15 */
16 l_broj_dana number := round (par_broj_dana, 15); -- to avoid 1.99999999999999 days = 1 day 24 hours
17 retval varchar2 (20);
18 begin
19 with podaci
20 as (select trunc (l_broj_dana) broj_dana,
21 round (mod (l_broj_dana * 24, 24), 2) broj_sati
22 from dual)
23 select decode (par_cb_dd,
24 1, lpad (p.broj_dana, 2, '0') || ':',
25 0, null)
26 || lpad (trunc (p.broj_sati), 2, '0')
27 || ':'
28 || lpad (round ( (p.broj_sati - trunc (p.broj_sati)) * 60),
29 2,
30 '0')
31 into retval
32 from podaci p;
33
34 return retval;
35 end f_days2ddhhmi;
36 /
Function created.
Applied to the test table, you'd get
SQL> select f_days2ddhhmi(completed_date_time - business_start_time, 0) result
2 from test;
RESULT
--------------------------------------------------------------------------------
06:01
which means that the difference is 6 hours and 1 minute.
If that's what you asked, see whether you can use it. Feel free to enhance it to seconds etc. if necessary.
I want to subtract "X" days and "X" minutes from sysdate, where the days and minutes are an integer as input parameter. For instance, 10 days and 5 minutes.
I found many examples to subtract either minutes or hours but not a combination of days and minutes.
select sysdate - 5 / 24 / 60 from dual -- will retrieve sysdate minus 5 minutes.
--What about the days?
Thank you!
Use an interval literal:
SELECT SYSDATE - INTERVAL '10 00:05' DAY(2) TO MINUTE
FROM DUAL
Or:
SELECT SYSDATE - INTERVAL '10' DAY - INTERVAL '5' MINUTE
FROM DUAL
Or just use arithmetic:
SELECT SYSDATE - 10 /* Days */ - 5 / 24 /60 /* Minutes */
FROM DUAL
Or use NUMTODSINTERVAL:
SELECT SYSDATE - NUMTODSINTERVAL( 10, 'DAY' ) - NUMTODSINTERVAL( 5, 'MINUTE' )
FROM DUAL
You can use Interval day to minute - http://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements003.htm#i38598
select sysdate - interval '1 00:05' day to minute from dual
I have a requirement to take every 3 seconds data within the specific time interval in SQL. I am new to SQL so can anyone help me on the scenario
This is my select query which returns all the values but i need data for every 3 seconds only
SELECT ton_nbr
FROM
icr_file_interface
WHERE
(
reading_dttm BETWEEN
TO_DATE(concat('2016-10-19',to_char(0930)),'yyyy-mm-dd HH24MISS')
AND TO_DATE(concat('2016-10-19',to_char('0945')),'yyyy-mm-dd HH24MISS')
)
AND
(
ton_nbr BETWEEN
(SELECT value FROM text_para WHERE para_cd='ICR_ST_RNG')
AND (SELECT value FROM text_para WHERE para_cd='ICR_ED_RNG')
)
If you only need to subtract 3 seconds from a date, you can use the following:
SQL> select to_char(sysdate, 'yyyy-mm-dd hh24:mi:ss'),
2 to_char(sysdate - 3*1/24/60/60, 'yyyy-mm-dd hh24:mi:ss')
3 from dual;
TO_CHAR(SYSDATE,'YY TO_CHAR(SYSDATE-3*1
------------------- -------------------
2016-10-19 09:38:17 2016-10-19 09:38:14
Given that sysdate -1 means "subtract one day to sysdate", you can derive the number of seconds you need with a bit af arithmetic
This selects you data between last 3 seconds. Hope you got the idea.
select ton_nbr
from icr_file_interface
where reading_dttm between dateadd(ss, -3, getdate()) and getdate() ;