DateDiff in hours , minutes and seconds between two dates should include only working hours Crystal Report
Hi,
I am working on a crystal report to show the Responded and Completed time of work orders based on work order created date and Responded Date. Created date and completed date
I want to display the difference of dates in hours minutes and seconds which should only include working hours
WOrking hours window - Mon - Fri 7:30 am - 5:00 pm
Thanks !
Crystal ReportsMicrosoft SQL Server
Last Comment
Mike McCracken
8/22/2022 - Mon
arnold
IN crystall you would need to perform the calculations
The logic shoukd exclude the first day of the work order and the last when the work order completed the says in between multiplied by 9.5 hours.
Then adding the difference in time between work order entry and end of day. Plus the time diff from start of day till when the work order was completed.
The combination of getting number of days in the wororder....
I.e. Processing wise...........
Nitin Sontakke
Efficient and maintainable would be a sql server scalar udf receiving two dates and returning a value, i suppose!
PortletPaul
There are some well known Crystal formula by Ken Hamady (here) for calculating working hours.
Good luck with that. The only way you're going to be able to come up with a rock-solid answer is to build a SQL Server Calendar Table , but build it hourly (half hour?) instead of daily like in the article, with a column 'open for business' that identifies if the place is open or not.
Otherwise you could come up with a function that handles the Mon - Fri 7:30 am - 5:00 pm, but it will not handle any exception to that such as staying open and closing early.
Recommend doing a Google Search for 'steve wake sql', as there's a SQL expert named Steve Wake who worked at Chipotle and did a presentation at PASS Summit 2015 on how to pull off a calendar table by hour. Maybe he published some of this presentation for download.
Mike McCracken
The formulas from Ken Hamady do work. I have used them in the past. They can be a bit of a pain to get working.
I have the below logic from other reports, it is calculating the hours by excluding the week end (Saturday & sunday)
but not calculating for working hours. Mon - Fri (7:30 am - 5:00 pm)
How can i add the logic to it to calculate time only with working hours
Function CF_getDuration (StartingDate As DateTime, EndingDate As DateTime, Sought As String) As String
DIM SECONDS_PER_DAY:SECONDS_PER_DAY=24*60*60
DIM SECONDS_PER_HOUR: SECONDS_PER_HOUR=60*60
DIM SECONDS_PER_MINUTE: SECONDS_PER_MINUTE=60
DIM Days as number,Hours as number, Minutes as number, Seconds as number
DIM TotalSeconds as number
DIM WeekendDayCount as number
DIM NewEndingDate as DateTime
' Get the number of business days between the 2 dates
WeekendDayCount = DateDiff("ww", StartingDate, EndingDate, crSaturday) + DateDiff ("ww", StartingDate, EndingDate, crSunday)
' Set a new end date that removes the business day count
NewEndingDate = EndingDate - WeekendDayCount
'Determine the total number of seconds
Seconds=datediff("s",StartingDate,NewEndingDate)
TotalSeconds=Seconds
'calc days
Days=Seconds \ SECONDS_PER_DAY 'integer division
Seconds=Seconds-Days*SECONDS_PER_DAY 'remove days
Double check where you're using the start and end dates. The obvious guess would be that you have them backwards. If you're not sure, post your formulas, or the actual report (the .RPT file), so that we can take a look at them.
Having said that, if you're looking for the difference down to the second, Ken's formulas give you the difference in hours, so that's probably only a partial solution (Unless the result includes fractions of hours. I haven't checked). I didn't look at the report that mlmcc posted, so I don't know if he calculated down to the second.
James
Mike McCracken
Can you post your report?
If not can you copy the formulas here
mlmcc
ram27
ASKER
Below is the formula I am using from the above attached TimeToComplete.rpt
I need the result in d: h:m which I had in my code before I sent.. If it is only 5 min , formula should return 0:0:05 ,
If it is 65 min then 0:1:05 and if it is more than one day and 1 hour 45 min then 1:1:45
I need other formula to return only hours for the bar graph
Thanks !
Mike McCracken
My correction should display it as x days hh:mm:ss
What other formula are you referring to?
The chart will only show the count of the records. The hour values won't enter into it at all.
What are you trying to show with the chart?
DateVar Array Holidays := [
Date (2017,1,1),
Date (2017,2,20),
Date (2017, 5, 29)
];
Local DateVar Start := Date({TASKS.OPENDATE});
Local DateVar End := Date({TASKS.COMPLETED});
Local NumberVar Weeks;
Local NumberVar Days;
Local Numbervar Hol;
DateVar Array Holidays;
Days := DayOfWeek(End) - DayOfWeek(Start) + 1;
Days := Days + (if DayOfWeek(Start) = 1 then 1 else 0);
Days := Days + (if DayOfWeek(End) = 7 then 1 else 0);
Local NumberVar i;
For i:= 1 to Count (Holidays) do
(
if DayOfWeek ( Holidays(i) ) in 2 to 6 and
Holidays(i) in start to end then
Hol:=Hol+1
);
Days := Weeks + Days - Hol;
TimeVar SetStart := TimeValue("7:30");
TimeVar SetEnd := TimeValue("17:00");
TimeVar StartTime := TimeValue({TASKS.OPENDATE});
TimeVar EndTime := TimeValue({TASKS.COMPLETED});
Local NumberVar NumDays;
Local NumberVar NumHours;
Local NumberVar NumMinutes;
Local NumberVar NumSeconds;
The logic shoukd exclude the first day of the work order and the last when the work order completed the says in between multiplied by 9.5 hours.
Then adding the difference in time between work order entry and end of day. Plus the time diff from start of day till when the work order was completed.
The combination of getting number of days in the wororder....
I.e. Processing wise...........