SELECT MIN (snap_id) AS FIRST_SNAP,
MAX (snap_id) AS LAST_SNAP,
MIN (BEGIN_INTERVAL_TIME) AS FIRST_QUERY,
MAX (END_INTERVAL_TIME) AS LAST_QUERY,
max(end_interval_time) - min(begin_interval_time) as "TIME_ELAPSED"
FROM dba_hist_snapshot
ORDER BY snap_id;
2931 3103 5/28/2012 6:00:11.065 AM 6/4/2012 11:00:40.967 AM +07 05:00:29.902000
I would like the last columns output to be 7 (for the days). I have tried trunc and extract like some other posts mentioned but can't seem to get the syntax right. Any ideas?
Judging from your comment, you're using timestamp columns, not datetime. You could use extract to retrieve the hour difference, and then trunc(.../24) to get the whole number of days:
trunc(extract(hour from max(end_interval_time) - min(begin_interval_time))/24)
Or you could cast the timestamp to a date:
trunc(cast(max(end_interval_time) as date) -
cast(min(begin_interval_time) as date))
Related
Following is a query in oracle.
SELECT start_date - TO_DATE('1900-01-01','YYYY-MM-DD') FROM start_table
In oracle it gives the output 44680.3646, where start_date is 01-MAY-22.
what query would require to form to get the same output in EDB and postgresql
If you want to get the fractional part of a day, then you need to convert each value to number of seconds using EXTRACT(EPOCH FROM ...) and divide by 86400(number of seconds in 1 day) and then find the difference of the results.
SELECT extract(epoch from '2022-05-01 11:44:16'::timestamp - '1900-05-02'::timestamp) / 86400 as date
Result: 44559.489074074074
Demo in DBfiddle
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
How do I get the days interval for prestodb? I can convert to milliseconds and convert these to number of days but I am looking if there is any shorter way to do this.
Example: I want to see how many days has it been since the first row inserted in a table.
SELECT
to_milliseconds(date(current_date) - min(created)) / (1000*60*60*24) as days_since_first_row
FROM
some_table
What I am hoping to see: (Either 1 of below)
SELECT
to_days(date(current_date) - min(created)) / (1000*60*60*24) as days_since_first_row
,cast(date(current_date) - min(created)) as days) as days_since_first_row2
FROM
some_table
Unfortunately, daylight savings breaks the solution from the accepted answer. DAY(DATE '2020-09-6' - DATE '2020-03-09') and DAY(DATE '2020-09-6' - DATE '2020-03-08') are both equal to 181 due to daylight savings time and DAY acting as a floor function on timestamps.
Instead, use DATE_DIFF:
DATE_DIFF('day', DATE '2020-09-6', DATE '2020-03-09')
Use subtraction to obtain an interval and then use day on the interval to get number of days elapsed.
presto:default> select day(current_date - date '2018-07-01');
_col0
-------
86
The documentation for this is at https://trino.io/docs/current/functions/datetime.html
I am trying to use Datediff function while subtracting two dates. One date is with Date and time stamp and the other with date only. How to get the difference of dates?
Here Column1 is 7/11/2017 4:24:38 PM and Column2 is 15/12/2017 where there is no timestamp.
DateDiff("d",[Column1],[Column2])
Convert the date column into datetime using Format function. See example below.
EDIT: since you want the difference in days and decimal point, I get the difference in hours then divide by 24. You can be as accurate if you want by getting the difference in minutes or seconds but using a different divisor.
SELECT DateDiff("h",
Now(),
Format('04/05/2018','mm/dd/yyyy hh:nn:ss'))/24 AS Expr1;
result: 1.125 days
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)