I need to calculate hours between datetime fields and I can achieve it by simply doing
select date1,date2,(date1-date2) from table; --This gives answer in DD:HH:MM:SS format
select date1,date2,(trunc(date1)-trunc(date2))*24 --This doesn't take into account the time, it only gives hours between two dates.
Is there a way I can find the difference between date times that gives the output in Hours as a number?
The 'format' comment on your first query suggests your columns are timestamps, despite the dummy column names, as the result of subtracting two timestamps is an interval. Your second query is implicitly converting both timestamps to dates before subtracting them to get an answer as a number of days - which would be fractional if you weren't truncating them and thus losing the time portion.
You can extract the number of hours from the interval difference, and also 24 * the number of days if you expect it to exceed a day:
extract(day from (date1 - date2)) * 24 + extract(hour from (date1 - date2))
If you want to include fractional hours then you can extract and manipulate the minutes and seconds too.
You can also explicitly convert to dates, and truncate or floor after manipulation:
floor((cast(date1 as date) - cast(date2 as date)) * 24)
db<>fiddle demo
Use the DATEDIFF function in sql.
Example:
SELECT DATEDIFF(HOUR, '2021-09-05 12:00:00', GETDATE());
You can find it using the differnece of dates and multiplying with 24
select date1
,date2
,(date1-date2)*24 as diff_in_hrs
from table
Related
I have a dataset with 2 columns of datetime datatype as shown here:
I want to take the difference between the two dates and I try it with this code:
Select
*,
original_due_date - due_date as difference
from
Table
However I'm not sure if the same would suffice as this is a datetime and not just date.
Any inputs would be much appreciated.
Desired output
The question was originally tagged Postgres, so this answers the original question.
Presumably, you are storing the values as timestamps. If you just want the results in days, then convert to dates and take the difference:
Select t.*,
(t.original_due_date::date - t.due_date::date) AS difference
from Table t;
If you want fractional days, then a pretty simple method is to extract the "epoch", which is measured in seconds, and use arithmetic:
Select t.*,
( extract(epoch from t.original_due_date -
extract(epoch from t.due_date
) / (24.0 * 60 * 60) AS decimal_days
from Table t;
transform timestamps to seconds (unix_timestamp), calculate difference and divide by (60*60*24) to get days
select (unix_timestamp(original_due_date, 'MM-dd-yyyy HH:mm')-unix_timestamp(due_date, 'MM-dd-yyyy HH:mm'))/(60*60*24) as difference_days
from (select '07-01-2021 00:00' as due_date, '02-10-2020 00:00' as original_due_date) t
Result:
-507
I have two timestamp columns: arrTime and depTime.
I need to find the number of munites the bus is late.
I tried the following:
SELECT RouteDate, round((arrTime-depTime)*1440,2) time_difference
FROM ...
I get the following error: inconsistent datatype . expected number but got interval day to second
How can i parse the nuber of minutes?
If i simply subtract: SELECT RouteDate, arrTime-depTime)*1440 time_difference
The result is correct but not well formatted:
time_difference
+00000000 00:01:00 0000000
The result of timestamp arithmetic is an INTERVAL datatype. You have an INTERVAL DAY TO SECOND there...
If you want the number of minutes one way would be to use EXTRACT(), for instance:
select extract( minute from interval_difference )
+ extract( hour from interval_difference ) * 60
+ extract( day from interval_difference ) * 60 * 24
from ( select systimestamp - (systimestamp - 1) as interval_difference
from dual )
Alternatively you can use a trick with dates:
select sysdate + (interval_difference * 1440) - sysdate
from (select systimestamp - (systimestamp - 1) as interval_difference
from dual )
The "trick" version works because of the operator order of precedence and the differences between date and timestamp arithmetic.
Initially the operation looks like this:
date + ( interval * number ) - date
As mentioned in the documentation:
Oracle evaluates expressions inside parentheses before evaluating those outside.
So, the first operation performed it to multiply the interval by 1,440. An interval, i.e. a discrete period of time, multiplied by a number is another discrete period of time, see the documentation on datetime and interval arithmetic. So, the result of this operation is an interval, leaving us with:
date + interval - date
The plus operator takes precedence over the minus here. The reason for this could be that an interval minus a date is an invalid operation, but the documentation also implies that this is the case (doesn't come out and say it). So, the first operation performed is date + interval. A date plus an interval is a date. Leaving just
date - date
As per the documentation, this results in an integer representing the number of days. However, you multiplied the original interval by 1,440, so this now represented 1,440 times the amount of days it otherwise would have. You're then left with the number of seconds.
It's worth noting that:
When interval calculations return a datetime value, the result must be an actual datetime value or the database returns an error. For example, the next two statements return errors:
The "trick" method will fail, rarely but it will still fail. As ever it's best to do it properly.
SELECT (arrTime - depTime) * 1440 time_difference
FROM Schedule
WHERE ...
That will get you the time difference in minutes. Of course, you can do any rounding that you might need to to get whole minutes....
Casting to DATE first returns the difference as a number, at least with the version of Oracle I tried.
round((cast(arrTime as date) - cast(depTime as date))*1440)
You could use TO_CHAR then convert back to a number. I have never tested the performance compared to EXTRACT, but the statement works with two dates instead of an interval which fit my needs.
Seconds:
(to_char(arrTime,'J')-to_char(depTime,'J'))*86400+(to_char(arrTime,'SSSSS')-to_char(depTime,'SSSSS'))
Minutes:
round((to_char(arrTime,'J')-to_char(depTime,'J'))*1440+(to_char(arrTime,'SSSSS')-to_char(depTime,'SSSSS'))/60)
J is julian day and SSSSS is seconds in day. Together they give an absolute time in seconds.
Hello everyOne I need to get the AVERAGE of difference of two dates (timestamp)
I tried this
select AVG((sva.endTime - sva.startTime)) as seconds from SVATable sva;
but I got an error
93/5000
ORA-00932: Inconsistent data types; expected: NUMBER; got: INTERVAL DAY TO SECOND
You may use EXTRACT to get AVG seconds.
SELECT AVG (EXTRACT (SECOND FROM (sva.endTime - sva.startTime)))
AS avg_seconds
FROM SVATable sva;
This is an insidious problem in Oracle. Your calculation would work with the date data type, but it does not work with timestamps.
One solution is to extract the days, hours, minutes, and seconds from the interval. Another is to use date arithmetic. You can get fractions of a day by using:
select (date '2000-01-01' + (sva.endTime - sva.startTime)) - date '2000-01-01'
You can use the average and convert to seconds:
select avg( (date '2000-01-01' + (sva.endTime - sva.startTime)) - date '2000-01-01') * (60*60*24)
Suppose I have given two dates, if difference is one month then it should be display as 30 days.Even months also need to convert into days
I have tried with age( date,now()::timestamp without time zone) but it is giving months-dates-years format. ex: 10 mons 24 days 13:57:40.5265.But I need following way.
For example if two months difference is there then I need output as 60 days.
Don't use the age() function for date/time arithmetic. It only returns "symbolic" results (which are good enough for human representation, but almost meaningless for date/time calculations; compared to the standard difference).
The standard difference operator (-) returns day-based results for both date, timestamp and timestamp with time zone (the former returns days as int, the latter two return day-based intervals):
From the day-based intervals you can extract days with the extract() function:
select current_date - '2017-01-01',
extract(day from now()::timestamp - '2017-01-01 00:00:00'),
extract(day from now() - '2017-01-01 00:00:00Z');
http://rextester.com/RBTO71933
Both of those will give that result.
select make_interval(days => ((extract(epoch from '2016-04-10'::date) - extract(epoch from '2016-02-10'::date))/(60*60*24))::integer)::text;
select format('%1$s days', ((extract(epoch from '2016-04-10'::date) - extract(epoch from '2016-02-10'::date))/(60*60*24))::integer);
I need to substract 2 timestamps in the given format:
16/01/17 07:01:06,165000000
16/01/17 07:01:06,244000000
I want to express the result with 2 decimal values but somewhere in the CAST process I am loosing precision. My atempt by now goes this way:
select
id,
trunc((CAST(MAX(T.TIMESTAMP) AS DATE) - CAST(MIN(T.TIMESTAMP) AS DATE))*24*60*60,2) as result
from table T
group by id;
But I get id_1 '0' as a result for the two timestamps above even after I set the truncate decimals at 2.
Is there a way that I can obtain the 0.XX aa a result of the substraction?
It's because you are casting the timestamp to date.
Use to_timestamp to convert your string into timestamp.
Try this:
with your_table(tstamp) as (
select '16/01/17 07:01:06,165000000' from dual union all
select '16/01/17 07:01:06,244000000' from dual
),
your_table_casted as (
select to_timestamp(tstamp,'dd/mm/yy hh24:mi:ss,ff') tstamp from your_table
)
select trunc(sysdate + (max(tstamp) - min(tstamp)) * 86400 - sysdate, 2) diff
from your_table_casted;
The difference between two timestamps is INTERVAL DAY TO SECOND.
To convert it into seconds, use the above trick.
DATE—This datatype stores a date and a time, resolved to the second. It does not include the time zone. DATE is the oldest and most commonly used datatype for working with dates in Oracle applications.
TIMESTAMP—Time stamps are similar to dates, but with these two key distinctions: you can store and manipulate times resolved to the nearest billionth of a second (9 decimal places of precision), and you can associate a time zone with a time stamp, and Oracle Database will take that time zone into account when manipulating the time stamp.
The result of a substraction of two timestamps is an INTERVAL:
INTERVAL—Whereas DATE and TIMESTAMP record a specific point in time, INTERVAL records and computes a time duration. You can specify an interval in terms of years and months, or days and seconds.
You can find more information here