I hope any one will help me with this.
I have two datetime start and end time . I get the time difference between two.
eg.
Start Time 18/07/2011 08:49:48
End time 18/07/2011 08:49:52
Diff +000000000 00:00:04.000000000 ( 04 seconds)
i need to add this time difference to the start time like this
18/07/2011 08:49:48
,18/07/2011 08:49:49
,18/07/2011 08:49:50
,18/07/2011 08:49:51
,18/07/2011 08:49:52
In simple words , Need to add the time difference to the start time one by one. With that i need to do some other calculation.
Thanks in advance.
SELECT CAST('18/07/2011 08:49:48' AS DATE) + (level - 1) / 86400
FROM dual
CONNECT BY
level <= (CAST('18/07/2011 08:49:52' AS DATE) - CAST('18/07/2011 08:49:48' AS DATE)) * 86400 + 1
Related
my company has numbers of shops around all the locations. They raised a request for delivering the item to their shop which they can sell . We wanted to understand how much time the company takes to deliver the item in minutes.However, we don't want to add the time in our elapsed time when the shop is closed i.e.
lets consider shop opening and closing time are
now elapsed time
When I deduct complain time and resolution time then I get calculatable elasped time in minutes but I need Required elapsed time in minutes so in the first case out of 2090 minutes those minutes are deducated when shop was closed. I need to write an oracle query to calcualted the required elapsed time in minutes which is in green.
help what query we can write.
One formula to get the net time is as follows:
For every day involved add up the opening times. For your first example this is two days 2021-01-11 and 2021-01-12 with 13 daily opening hours (09:00 - 22:00). That makes 26 hours.
If the first day starts after the store opens, subtract the difference. 10:12 - 09:00 = 1:12 = 72 minutes.
If the last day ends before the store closes, subtract the difference. 22:00 - 21:02 = 0:58 = 58 minutes.
Oracle doesn't have a TIME datatype, so I assume you are using Oracle's datetime data type they call DATE to store the opening and closing time and we must ignore the date part. And you are probably using the DATE type for the complain_time and the resolution_time, too.
In below query I convert the time parts to minutes right away, so the calculations get a tad more readable later.
with s as
(
select
shop,
extract(hour from opening_time) * 60 + extract(minute from opening_time) as opening_minute,
extract(hour from closing_time) * 60 + extract(minute from closing_time) as closing_minute
from shops
)
, r as
(
select
request, shop, complain_time, resolution_time,
trunc(complain_time) as complain_day,
trunc(resolution_time) as resolution_day,
extract(hour from complain_time) * 60 + extract(minute from complain_time) as complain_minute,
extract(hour from resolution_time) * 60 + extract(minute from resolution_time) as resolution_minute
from requests
)
select
r.request, r.shop, r.complain_time, r.resolution_time,
(r.resolution_day - r.complain_day + 1) * 60
- case when r.complain_minute > s.opening_minute) then r.complain_minute - s.opening_minute else 0 end
- case when r.resolution_minute < s.opening_minute) then s.closing_minute - r.resolution_minute else 0 end
as net_duration_in_minutes
from r
join s on s.shop = r.shop
order by r.request;
I have a list of time in a decimal format of seconds, and I know what time the series started. I would like to convert it to a time of day with the offset of the start time applied. There must be a simple way to do this that I am really missing!
Sample source data:
\Name of source file : 260521-11_58
\Recording from 26.05.2021 11:58
\Channels : 1
\Scan rate : 101 ms = 0.101 sec
\Variable 1: n1(rpm)
\Internal identifier: 63
\Information1:
\Information2:
\Information3:
\Information4:
0.00000 3722.35645
0.10100 3751.06445
0.20200 1868.33350
0.30300 1868.36487
0.40400 3722.39355
0.50500 3722.51831
0.60600 3722.50464
0.70700 3722.32446
0.80800 3722.34277
0.90900 3722.47729
1.01000 3722.74048
1.11100 3722.66650
1.21200 3722.39355
1.31300 3751.02710
1.41400 1868.27539
1.51500 3722.49097
1.61600 3750.93286
1.71700 1868.30334
1.81800 3722.29224
The Start time & date is 26.05.2021 11:58, and the LH column is elapsed time in seconds with the column name [Time] . So I just want to convert the decimal / real to a time or timespan and add the start time to it.
I have tried lots of ways that are really hacky, and ultimately flawed - the below works, but just ignores the milliseconds.
TimeSpan(0,0,0,Integer(Floor([Time])),[Time] - Integer(Floor([Time])))
The last part works to just get milli / micro seconds on its own, but not as part of the above.
Your formula isn't really ignoring the milliseconds, you are using the decimal part of your time (in seconds) as milliseconds, so the value being returned is smaller than the format mask.
You need to convert the seconds to milliseconds, so something like this should work
TimeSpan(0,0,0,Integer(Floor([Time])),([Time] - Integer(Floor([Time]))) * 1000)
To add it to the time, this would work
DateAdd(Date("26-May-2021"),TimeSpan(0,0,0,Integer([Time]),([Time] - Integer([Time])) * 1000))
You will need to set the column format to
dd-MMM-yyyy HH:mm:ss:fff
I want to get the current system time in milliseconds in Fortran. I can't use system_clock, because I can't figure out how to get the current time from it.
This illustrates how to get the time since midnight in milliseconds using date_and_time:
program time
integer :: values(8)
real :: rTime
! Get the values
call date_and_time(values=values)
! From https://gcc.gnu.org/onlinedocs/gfortran/DATE_005fAND_005fTIME.html
! values(5) ... The hour of the day
! values(6) ... The minutes of the hour
! values(7) ... The seconds of the minute
! values(8) ... The milliseconds of the second
! Calculate time since midnight
rTime = ( values(5) )*60. ! Hours to minutes
rTime = ( rTime + values(6) )*60. ! Minutes to seconds
rTime = ( rTime + values(7) )*1e3 ! Seconds to milliseconds
rTime = rTime + values(8) ! Add milliseconds
! Time in seconds
print *, 'Time (ms) since midnight', rTime
end program
I think your question is: "how can I call date_and_time subroutin and access it to calculate ms?" Am I right?
Alexander's answer was true.also you can use this code:
program time
integer :: values(8)
call date_and_time(values=values)
print *, values(5),":",values(6),":",values(7),":",values(8)
end program time
You can use ITIME(), a function embedded in Fortran. It returns a real number in units of milliseconds. You just need to call it twice and subtract the two values in order to calculate the time interval.
There is the following task: I need to get minutes between one time and another one: for example, between "8:15" and "7:45". I have the following code:
(Time.parse("8:15") - Time.parse("7:45")).minute
But I get result as "108000.0 seconds".
How can I fix it?
The result you get back is a float of the number of seconds not a Time object. So to get the number of minutes and seconds between the two times:
require 'time'
t1 = Time.parse("8:15")
t2 = Time.parse("7:45")
total_seconds = (t1 - t2) # => 1800.0
minutes = (total_seconds / 60).floor # => 30
seconds = total_seconds.to_i % 60 # => 0
puts "difference is #{minutes} minute(s) and #{seconds} second(s)"
Using floor and modulus (%) allows you to split up the minutes and seconds so it's more human readable, rather than having '6.57 minutes'
You can avoid weird time parsing gotchas (Daylight Saving, running the code around midnight) by simply doing some math on the hours and minutes instead of parsing them into Time objects. Something along these lines (I'd verify the math with tests):
one = "8:15"
two = "7:45"
h1, m1 = one.split(":").map(&:to_i)
h2, m2 = two.split(":").map(&:to_i)
puts (h1 - h2) * 60 + m1 - m2
If you do want to take Daylight Saving into account (e.g. you sometimes want an extra hour added or subtracted depending on today's date) then you will need to involve Time, of course.
Time subtraction returns the value in seconds. So divide by 60 to get the answer in minutes:
=> (Time.parse("8:15") - Time.parse("7:45")) / 60
#> 30.0
I want to trigger a notification for all my users at a specific time in their time zone. I want to compute the delay the server should wait before firing the notification. I can compute the time at the users Time Zone using Time.now.in_time_zone(person.time_zone)
I can strip out the hours, minutes and seconds from that time and find out the seconds remaining to the specific time. However, I was wondering if there's a more elegant method where I could set 9:00 AM on today and tomorrow in a timezone and compare it with Time.now.in_time_zone(person.time_zone) and just find out the number of seconds using arithmetic operations in the ruby Time Class.
Or in short my question is: (was: before the downvote!)
How do I compute the number of seconds to the next 9:00 AM in New York?
What about this
next9am = Time.now.in_time_zone(person.time_zone).seconds_until_end_of_day + 3600 * 9
next9am -= 24 * 60 * 60 if Time.now.in_time_zone(person.time_zone).hour < 9
NOTIFICATION_HOUR = 9
local_time = Time.now.in_time_zone(person.time_zone)
desired_time = local_time.hour >= NOTIFICATION_HOUR ? local_time + 1.day : local_time
desired_time = Time.new(desired_time.year, desired_time.month, desired_time.day, NOTIFICATION_HOUR, 0, 0, desired_time.utc_offset)
return desired_time - local_time