I want SQL query to fetch/select records which are taken only from previous half an hour only. For example if my scheduler ran at 2 pm, and then again in 2:30, during the 2:30 run it should only pick rows from between 2pm and 2:30pm and not earlier, using the column created_timestamp which stores the time as unix epoch format eg:
|created_timestamp|
|1497355750350 |
|1497506182344 |
We can do arithmetic with Oracle dates. Subtracting one date from another gives the interval as a fractional number. Multiplying by 86400 gives us the number of seconds. So this is the current unix expoch:
(sysdate - date '1970-01-01') * 86400
This means your query will be something like
select * from your_table
where created_timestamp >= (:last_run_time - date '1970-01-01') * 86400
The trick is that your scheduler needs to pass in the time of the previous run - last_run_time - to pick up all the records which have been added since then.
You can do Flashback query
SELECT * FROM TABLE
AS OF TIMESTAMP (SYSTIMESTAMP - INTERVAL '30' MINUTE);
Related
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.
I have two fields dateTS and closingTime.
dateTS is a normal timestamp field (e.g. 2019-07-13 22:31:10.000000)
closingTime is a HH:MM format of when a store closes (e.g. 23:00)
I need a PostgreSQL query to subtract the two field and get the number of minutes difference between them.
Using the examples given above the difference between the two fields would be 28 minutes
So far I've tried different variations of the datediff function, but it won't work.
My guess is I either have to
a. generate a fake timestamp for closingTime which is the same day as the dateTs field and subtract the 2 timestamps.
or
b. convert the hour/minutes of both field to a float and subtract the two values to get the hours difference and convert that to minutes
You can just subtract them by converting the timestamp to a time:
select closingtime - datets::time
from your_table;
That will give you an interval as the result.
To convert that to minutes you can get the number of seconds and divide it by 60:
select (extract epoch from closingtime - datets::time) / 60
from your_table;
Cast your closing time to an interval and the timestamp to time and then subtract the two. By casting the timestamp to time you are effectively discarding the date part. You can the subtract one from the other to generate the difference as an interval.
select closingTime::interval - dateTS::time...
e.g.:
# select '23:00'::interval - now()::time;
?column?
-----------------
05:31:00.031141
(1 row)
If needed you can then convert the interval to minutes:
# select extract(epoch from ('23:00'::interval - now()::time)) / 60;
?column?
------------------
327.435313083333
(1 row)
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 wish to convert a single Oracle datetime value to minutes.
As an example, I need to convert the current date/time alone into minutes, i.e.:
select (sysdate)*24*60 from dual
but this is giving me an error.
I basically need to perform a check to see that a certain operation cannot be performed until 30 minutes before a particular date/start time, that exists in the database.
So if the Start Time in the DB is:
24/04/2014 22:00:00 and the current date/time (SYSDATE) is 24/04/2014 21:29:59,
then operation CANNOT be performed but if the current date/time (SYSDATE) is:
24/04/2014 21:30:00,
then operation CAN be performed.
You probably want something like
startTime - interval '30' minute >= sysdate
or
startTime >= sysdate + interval '30' minute
You could also subtract the two date values which gives you the number of days between them and multiply
(startTime - sysdate)*24*60 >= 30
but I generally find the interval notation clearer and easier to read. It's also easier to structure in a way that allows you to use indexes on columns like startTime.
select (sysdate - trunc(sysdate)) *24 *60 from dual
You seem to want to know if the seconds component of sysdate is 0. So, test for that:
where extract(second from sysdate) = 0
Oops, I misread the question. You just need a difference of 30 minutes. That is also easy:
where starttime <= sysdate + 30/(24*60)
When you add an integer to a datetime, it is interpreted as a number of days. The expression 30/(24*60) is an expression for half an hour measured in days.
I thought it was really simple but it isn't.
SELECT TO_TIMESTAMP('10/08/2012','DD/MM/YYYY')
- 1/(24*50*60*1000) data
FROM dual;
It simply doesn't work.
Other details:
SELECT TO_TIMESTAMP('10/08/2012','DD/MM/YYYY') -
NUMTODSINTERVAL(1/(24*50*60*1000),'HOUR') data
FROM dual;
doesn't work..
The right seems to be
SELECT TO_TIMESTAMP('10/08/2012','DD/MM/YYYY') -
NUMTODSINTERVAL(1/(24*25*60*1000),'HOUR') data
FROM dual;
Why? How does it work?
For adding or subtracting an amount of time expressed as a literal you can use INTERVAL.
SELECT TO_TIMESTAMP('10/08/2012','DD/MM/YYYY')
- INTERVAL '0.001' SECOND
FROM dual;
As well there are now standard ways to express date and time literals and avoid the use of various database specific conversion functions.
SELECT TIMESTAMP '2012-10-08 00:00:00'
- INTERVAL '0.001' SECOND DATA
FROM dual;
For your original question the time part of a day is stored in fractional days. So one second is:
1 / (hours in day * minutes in hour * seconds in a minute)
Divide by 1000 to get milliseconds.
1 / (24 * 60 * 60 * 1000)
SELECT TO_TIMESTAMP('10/08/2012','DD/MM/YYYY') - NUMTODSINTERVAL(1/(24*50*60*1000),'HOUR') data
FROM dual;
OUTPUT
DATA
---------------------------------
09/AUG/12 11:59:59.999950000 PM
1 row selected.
The answer posted above subtracts a tenth of a millisecond from the date. I think what you want is the following:
SELECT TO_TIMESTAMP('10/08/2012','DD/MM/YYYY')-NUMTODSINTERVAL(1/1000,'SECOND')
FROM dual;
Output:
DATA
---------------------------------------------------------------------------
09-AUG-12 11.59.59.999000000 PM
^^^
|||
tenths|thousandths
|
hundredths
The following NUMTODSINTERVAL(1/(24*25*60*1000),'HOUR') seems to work only because 24*25 = 600. But that number is wrong because 1/(600*60*1000) of an hour is a tenth of a millisecond, not a millisecond. If you want to use 'HOUR' in NUMTODSINTERVAL() you should use 1/(60*60*1000) (sixty minutes in an hour, sixty seconds in a minute, 1000 ms in a second).
This is correct (with a millisecond being 1000th of a second):-
SELECT TO_TIMESTAMP('10/08/2012','DD/MM/YYYY') - NUMTODSINTERVAL(1/1000,'SECOND') data FROM dual;
DATA
-----------------------------
09-AUG-12 23.59.59.999000000
As to why the other code isn't working it's because you aren't calculating a millisecond correctly. An hour must be divided by 60 to give minutes and again by 60 to given seconds then by 1000 to give a millisecond, thus if you must use HOUR as the interval then it is:-
SELECT TO_TIMESTAMP('10/08/2012','DD/MM/YYYY') - NUMTODSINTERVAL(1/(60*60*1000),'HOUR') as data FROM dual;
DATA
---------------------------------------------------------------------------
09-AUG-12 23.59.59.999000000
select TO_CHAR(TO_TIMESTAMP('10.05.2012', 'DD.MM.YYYY') -
NUMTODSINTERVAL(1/1000, 'SECOND'), 'DD.MM.YYYY HH24:MI:SS:FF3') Res
from dual;
RES
-----------------------------
09.05.2012 23:59:59.999