When we use this query
select '2930:41:51.974223'::interval(0);
Output
interval
------------
2930:41:52
Desired Output
2930:42
I tried
select to_char('2930:41:51.974223'::interval(0),'HH24:MI');
Results:
to_char
---------
2930:41
It won't round 41 to 42 by adding the seconds)
Simply add 30 seconds to the interval:
select to_char('2930:41:51.974223'::interval(0)+ '30s','HH24:MI');
to_char
---------
2930:42
(1 row)
select to_char('2930:41:51.974223' + interval '30','HH24:MI');
or
select to_char('2930:41:51.974223' + interval '30 seconds','HH24:MI');
Related
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;
I would like to convert in PL/SQL miliseconds to Time(hh:mm:ss)
21649000 to 06:00:49
or
83293000 to 23:08:13
Use NUMTODSINTERVAL with SECOND option :
select NUMTODSINTERVAL( 83293000 / 1000, 'SECOND' ) "Time" from dual t;
Time
-------------------
+000000000 23:08:13
This is a quite self-explanatory way:
select val,
floor( val / 1000 / 60 / 60 ) as hours,
floor( mod(val / 1000 / 60 , 60) ) as minutes,
floor( mod(val / 1000 , 60) ) as seconds
from (
select 21649000 val from dual union
select 83293000 val from dual
)
VAL HOURS MINUTES SECONDS
---------- ---------- ---------- ----------
21649000 6 0 49
83293000 23 8 13
This does not handle days, so the number of milliseconds must be less than 24 hours.
This gives numbers, you can edit it the way you need to get your desired output format/type.
If you only want second precision you could divide the number by 1000 to get seconds, and by 86400 to get a fraction of a day, then add that to midnight on any nominal date - and convert the result to a string:
select to_char(date '1970-01-01' + (21649000/86400000), 'HH24:MI:SS') as time
from dual;
TIME
--------
06:00:49
select to_char(date '1970-01-01' + (83293000/86400000), 'HH24:MI:SS') as time
from dual;
TIME
--------
23:08:13
This only works properly for values less than a day, i.e. where your original number is less than 86400000; higher than that and you only see the leftover in the second day.
The below query worked for me to convert to the Milliseconds to the HH:MM:SS
SELECT
TO_CHAR(TRUNC([ColumnName]/3600000),'FM9900') || ':' ||
TO_CHAR(TRUNC(MOD([ColumnName],3600000)/60000),'FM00') || ':' ||
TO_CHAR( trunc(MOD([ColumnName],60000)/1000),'FM00') FROM [TableName]`
I have a table (incident) that has column Create_date(DataType=Date).
I want to get difference in Days OR Hours from Previous Record. Like the screenshot below.
From Second Record Create_Date I want to minus First Create_Date and from Third Create Date to Second and so on. I'm using LAG function in Oracle, but not sure how its calculating there. Could any one please help me regarding that issue.
incident.create_date - lag(incident.create_date,1) OVER (ORDER BY incident.create_date) AS CREATEDATE_DIFF,
RN 1 We have Create_date (05/01/017 10:40:17 AM
Date differences in Oracle are calculated in numbers of days. If the difference is less than a day, you're going to get a value of less than 1 returned.
If you want to convert that into hours, you'll have to multiply the result by 24, for minutes multiply by 24*60 and for seconds it's 24*60*60.
e.g.:
select sysdate - trunc(sysdate) diff_in_days,
(sysdate - trunc(sysdate))*24 diff_in_hours,
(sysdate - trunc(sysdate))*24*60 diff_in_mins,
(sysdate - trunc(sysdate))*24*60*60 diff_in_secs
from dual;
DIFF_IN_DAYS DIFF_IN_HOURS DIFF_IN_MINS DIFF_IN_SECS
------------ ------------- ------------ ------------
0.4342245370 10.4213888888 625.28333333 37517
You may then wish to apply ROUND (or maybe TRUNC/CEIL) depending on how you want the output to look like (e.g. to 2 d.p., to nearest minute, etc).
If you subtract one date from another you will get the difference in days (or fractions thereof) as a number.
You can get the days/hours/minutes/seconds of this using an interval:
SELECT EXTRACT( DAY FROM createdate_diff ) AS days,
EXTRACT( HOUR FROM createdate_diff ) AS hours,
EXTRACT( MINUTE FROM createdate_diff ) AS minutes,
EXTRACT( SECOND FROM createdate_diff ) AS seconds,
createdate_diff
FROM (
SELECT NUMTODSINTERVAL(
create_date - lag(create_date) OVER (ORDER BY create_date),
'DAY'
) AS CREATEDATE_DIFF
FROM incident
);
Or you can perform the same calculations manually:
SELECT TRUNC( createdate_diff ) AS days,
TRUNC( MOD( createdate_diff * 24, 24 ) ) AS hours,
TRUNC( MOD( createdate_diff * 24 * 60, 60 ) ) AS minutes,
MOD( createdate_diff * 24 * 60 * 60, 60 ) AS seconds,
createdate_diff
FROM (
SELECT create_date - lag(create_date) OVER (ORDER BY create_date)
AS CREATEDATE_DIFF
FROM incident
);
Use
......
(incident.create_date -
lag(incident.create_date,1) OVER (ORDER BY incident.create_date))
*24*60
AS CREATEDATE_DIFF_IN_MINS,.....
to get the output in minutes, which seams suitable for your sample data. Or multiply further by 60 to get output in seconds.
I've got 2 columns in a database of type TIMESTAMP(6) WITH TIME ZONE. I've subtracted one from the other to get the time between the two timestamps.
select lastprocesseddate-importeddate
from feedqueueitems
where eventid = 2213283
order by written desc;
How can I get an average of the list of time differences I have?
Here are a small sample of time differences:
+00 00:00:00.488871
+00 00:00:00.464286
+00 00:00:00.477107
+00 00:00:00.507042
+00 00:00:00.369144
+00 00:00:00.488918
+00 00:00:00.354797
+00 00:00:00.378801
+00 00:00:00.320040
+00 00:00:00.361242
+00 00:00:00.302327
+00 00:00:00.331441
+00 00:00:00.324065
EDIT: I also should have noted - I've tried the AVG function, and it just returns
ORA-00932: inconsistent datatypes: expected NUMBER got INTERVAL DAY TO SECOND
00932. 00000 - "inconsistent datatypes: expected %s got %s"
*Cause:
*Action:
Error at Line: 3 Column: 29
EDIT2: Just to clarify the above snippet. Line 3 is my SQL query all on one line in the following format:
select AVG(lastprocesseddate-importeddate) from feedqueueitems where eventid = 2213283;
EDIT3: Massive thanks to Matt and Alex Poole. You've both helped massively and I appreciate you taking the time to help with this and to both consistently return with updated help in response to the feedback/further problems! Thanks guys!
Use the AVG function
SELECT avg(cast(lastprocesseddate as date)-cast(importeddate as date))
FROM feedqueueitems
WHERE eventid = 2213283
ORDER BY written DESC;
On the Database with the +1 timezone for importeddate and lastprocesseddate is UTC
SELECT avg(cast(cast(lastprocesseddate as timestamp with time zone) at time zone '+01:00' as date)-cast(importeddate as date))
FROM feedqueueitems
WHERE eventid = 2213283
ORDER BY written DESC;
You could extract the time components from each gap value, which is an interval data type, so you end up with a figure in seconds (including the fractional part), and then average those:
select avg(extract(second from gap)
+ extract(minute from gap) * 60
+ extract(hour from gap) * 60 * 60
+ extract(day from gap) * 60 * 60 * 24) as avg_gap
from (
select lastprocesseddate-importeddate as gap
from feedqueueitems
where eventid = 2213283
);
A demo using a CTE to provide the interval values you showed:
with cte as (
select interval '+00 00:00:00.488871' day to second as gap from dual
union all select interval '+00 00:00:00.464286' day to second from dual
union all select interval '+00 00:00:00.477107' day to second from dual
union all select interval '+00 00:00:00.507042' day to second from dual
union all select interval '+00 00:00:00.369144' day to second from dual
union all select interval '+00 00:00:00.488918' day to second from dual
union all select interval '+00 00:00:00.354797' day to second from dual
union all select interval '+00 00:00:00.378801' day to second from dual
union all select interval '+00 00:00:00.320040' day to second from dual
union all select interval '+00 00:00:00.361242' day to second from dual
union all select interval '+00 00:00:00.302327' day to second from dual
union all select interval '+00 00:00:00.331441' day to second from dual
union all select interval '+00 00:00:00.324065' day to second from dual
)
select avg(extract(second from gap)
+ extract(minute from gap) * 60
+ extract(hour from gap) * 60 * 60
+ extract(day from gap) * 60 * 60 * 24) as avg_gap
from cte;
AVG_GAP
----------
.397544692
Or if you wanted it as an interval:
select numtodsinterval(avg(extract(second from gap)
+ extract(minute from gap) * 60
+ extract(hour from gap) * 60 * 60
+ extract(day from gap) * 60 * 60 * 24), 'SECOND') as avg_gap
...
which gives
AVG_GAP
--------------------
0 0:0:0.397544692
SQL Fiddle with answer in seconds. (It doesn't seem to like displaying intervals at the moment, so can't demo that).
This query should solve the issue.
WITH t AS
(SELECT
TIMESTAMP '2015-04-23 12:00:00.5 +02:00' AS lastprocesseddate,
TIMESTAMP '2015-04-23 12:05:10.21 UTC' AS importeddate
FROM dual)
SELECT
AVG(
EXTRACT(SECOND FROM SYS_EXTRACT_UTC(lastprocesseddate) - SYS_EXTRACT_UTC(importeddate))
+ EXTRACT(MINUTE FROM SYS_EXTRACT_UTC(lastprocesseddate) - SYS_EXTRACT_UTC(importeddate)) * 60
+ EXTRACT(HOUR FROM SYS_EXTRACT_UTC(lastprocesseddate) - SYS_EXTRACT_UTC(importeddate)) * 60 * 60
+ EXTRACT(DAY FROM SYS_EXTRACT_UTC(lastprocesseddate) - SYS_EXTRACT_UTC(importeddate)) * 60 * 60 * 24
) AS average_gap
FROM t;
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.