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)
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 need to know a difference between start time and end time. Both are DATETIME fields, I tried to use "-" and DATADIFF.
I already tried using DATADIFF and simple subtraction converting the field to just time.
(to_date(Fim_Hora,'HH24:MI') - to_date(Inicio_Hora,'HH24:MI')) AS Diferenca
DATADIFF(MIN,Fim_Hora,Inicio_Hora)
I need to know the time in minutes for use as parameters.
Oracle does not have a time data type. Usually, subtraction works well enough:
select (end_time - start_time) as diff
You may need to convert to a string if you want it formatted in a particular way.
In Oracle, you can directly substract dates, it returns the difference between the dates in days. To get the difference in minutes, you can multiply the result by 24 (hours per days) and 60 (minutes per hour):
(Fim_Hora - Inicio_Hora) * 24 * 60 diff_minutes
This assumes that both Fim_Hora and Inicio_Hora are of datatype DATE.
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 have a table that contains an epoch timestamp column(transaction_date), if I have two rows and I want to determine if they are on the same day; without converting them to date using to_date. how can I do that and what is the best way that takes less operations while determining it.
An epoch "timestamp" is a number which should be generated by converting the date to UTC (if it has a timezone) and subtracting the epoch to give a number of seconds (or milliseconds) since the epoch.
Without converting to a date, you can just divide by the number of seconds (or milli-seconds) in a day and truncate to get the number of days since the epoch and then compare those values:
SELECT TRUNC( transaction_date / ( 24 * 60 * 60 ) ) FROM your_table
If the values given by two rows have an identical number of days since the epoch then they are on the same day.
Using this transformation from timestamp to day, you can compare
SELECT
* (all but transaction_date),
(EXTRACT (DAY FROM (transaction_date))*24*60*60) transaction_day
FROM yourtable;
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