I am trying to get days from given data like this:
In this data suppose ID B start date is 4/10/2019 and end date is 10/25/2019. Here there is 7 months: April to October, so for the first month start date is 4/10/2019 and end date is 4/30/2019 so this means he only avail 10 days from this month and remaining days is 21.. same for here end date is 10/25/2019 so if we look calendar end date 10/31/2019 we only avail 6 days so in data I want to get above data which is mentioned in image .. where as I am trying this formula in qlikview:
=sum(
If(
MonthName(CalendarMonthEnd) = MonthName([End Date]),
([End Date]-CalendarMonthStart+1),
(RangeMin([End Date],CalendarMonthEnd)-RangeMax([Start Date],CalendarMonthStart))
)
)
and through this formula I get this data which is remaining days where i want to get days which is availed...
this is link of folder please download and check ..
https://www.dropbox.com/s/v48373io1bv9qqj/file_qlik.rar?dl=0
in this folder the excel file "output.. " in this excel file the first table output which i need
Just add another if
=Sum(
If(CalendarMonthStart >= [Start Date] and CalendarMonthEnd <= [End Date],
CalendarMonthEnd-CalendarMonthStart,
If([Start Date]>CalendarMonthStart,
[Start Date]-CalendarMonthStart+1,
CalendarMonthEnd-[End Date])
)
)
)
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 need to test if in specified day has any events in the built-in Apple Calendar app. Here is my current idea as pseudocode:
// input variables
day_to_test = 23
month_to_test = 8
year_to_test = 2016
testing_date = day_to_test + month_to_test + year_to_test as date
// actual "code"
if testing_date has any event in Apple calendar Then
display dialog "There is any event on that day!"
else
display dialog "Free of events!"
end if
Try:
tell application "Calendar"
tell calendar "Main" -- national holiday calendar in my iCloud
set today to current date
set today's year to 2013
set today's month to 3
set today's day to 5
set today's time to 0
copy (today + days) to tomorrow
set currentEvents to events whose start date ≥ today and end date ≤ tomorrow
end tell
end tell
Is there anything like the Oracle "NEXT_DAY" function available in the syntax that Crystal Reports uses?
I'm trying to write a formula to output the following Monday # 9:00am if the datetime tested falls between Friday # 9:00pm and Monday # 9:00am.
So far I have
IF DAYOFWEEK ({DATETIMEFROMMYDB}) IN [7,1]
OR (DAYOFWEEK({DATETIMEFROMMYDB}) = 6 AND TIME({DATETIMEFROMMYDB}) in time(21,00,00) to time(23,59,59))
OR (DAYOFWEEK({DATETIMEFROMMYDB}) = 2 AND TIME({DATETIMEFROMMYDB}) in time(00,00,00) to time(08,59,59))
THEN ...
I know I can write seperate IF statements to do a different amount of DateAdd for each of Fri, Sat, Sun, Mon, but if I can keep it concise by lumping all of these into one I would much prefer it. I'm already going to be adding additional rules for if the datetime falls outside of business hours on the other weekdays so I want to do as much as possible to prevent this from becoming a very overgrown and ugly formula.
Since there is no CR equivalent that I know of, you can just cheat and borrow the NEXT_DAY() function from the Oracle database. You can do this by creating a SQL Expression and then entering something like:
-- SQL Expression {%NextDay}
(SELECT NEXT_DAY("MYTABLE"."MYDATETIME", 'MONDAY')
FROM DUAL)
then you could either use that directly in your formula:
IF DAYOFWEEK ({MYTABLE.MYDATETIME}) IN [7,1]
OR (DAYOFWEEK({MYTABLE.MYDATETIME}) = 6 AND TIME({MYTABLE.MYDATETIME}) in time(21,00,00) to time(23,59,59))
OR (DAYOFWEEK({MYTABLE.MYDATETIME}) = 2 AND TIME({MYTABLE.MYDATETIME) in time(00,00,00) to time(08,59,59))
THEN DateTime(date({%NextDay}),time(09,00,00))
Or, the even better way would be to just stuff ALL of the logic into the SQL Expression and do away with the formula altogether.
Considering Sunday is 1
And the first 7 is the week we want to back
7 = 1 week
14 = 2 weeks
The last Number (1) is 1 for Sunday, 2 for Monday, 3 for Tuestday
Last Sunday 1 week ago
Today - 7 + ( 7 - WEEKDAY(TODAY) )+1
Last Monday 2 weeks ago
Today - 14 + ( 7 - WEEKDAY(TODAY) )+2
So this 2 formulas give me MONDAY LAST WEEK and SUNDAY LAST WEEK.
EvaluateAfter({DATETIMEFROMMYDB}) ;
If DayOfWeek ({DATETIMEFROMMYDB}) In [crFriday,crSaturday,crSunday,crMonday]
then
IF DayOfWeek ({DATETIMEFROMMYDB}) In [crFriday]
AND TIME({DATETIMEFROMMYDB}) >= time(21,00,00)
then //your code here
Else if Not(DayOfWeek ({DATETIMEFROMMYDB}) In [crFriday] )
AND (TIME({DATETIMEFROMMYDB}) >= time(00,00,00) AND TIME({DATETIMEFROMMYDB}) <= time(23,59,59))
then //your code here
Else if DayOfWeek ({DATETIMEFROMMYDB})In [crMonday]
AND TIME({DATETIMEFROMMYDB}) < time(09,00,00)
then //your code here
I often need to convert (long) character strings into the date class in R. I notice that this step seems quite slow.
Example:
date <- c("5/31/2013 23:30", "5/31/2013 23:35", "5/31/2013 23:40", "5/31/2013 23:45", "5/31/2013 23:50", "5/31/2013 23:55")
Date <- as.POSIXct(date, format="%m/%d/%Y %H:%M")
This isn't a huge problem, but I wonder if I'm overlooking an easy route to increased efficiency. Any tips for speeding this up? Thanks.
Since I wrote this before it was pointed out this is a duplicate, I'll add it as an answer anyway. Basically package fasttime can help you IF you have dates AFTER 1970-01-01 00:00:00 AND they are GMT AND they are of the format year, month, day, hour, minute, second. If you can rewrite your dates to this format then fastPOSIXct will be quick:
# data
date <- c( "2013/5/31 23:30" , "2013/5/31 23:35" , "2013/5/31 23:40" , "2013/5/31 23:45" )
require(fasttime)
# fasttime function
dates.ft <- fastPOSIXct( date , tz = "GMT" )
# base function
dates <- as.POSIXct( date , format= "%Y/%m/%d %H:%M")
# rough comparison
require(microbenchmark)
microbenchmark( fastPOSIXct( date , tz = "GMT" ) , as.POSIXct( date , format= "%Y/%m/%d %H:%M") , times = 100L )
#Unit: microseconds
# expr min lq median uq max neval
# fastPOSIXct(date, tz = "GMT") 19.598 21.699 24.148 25.5485 215.927 100
# as.POSIXct(date, format = "%Y/%m/%d %H:%M") 160.633 163.433 168.332 181.9800 278.220 100
But the question would be, is it quicker to transform your dates to a format fasttime can accept or just use as.POSIXct or buy a faster computer?!
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