Reports Greater Than 30 Days

Question

I need to display the number of problem tickets that are more than 30 days open based on the date created. and calculate the number of open problem tickets more than 30 days. So far the formula I have is:
if {table.datecreated} > Date (30) Then
Sum ({table.status}, {table.itemnumber})
Am I on the right track? Any ideas? I am having trouble with the date functions in the formula workshop. The reading materials have all lead me to a dead end.
Thanks,
ITSteveSE

Answer 1

Hello,
please also see below some explanation plus samples from the CR Help (DateDiff) :
Basic and Crystal syntax.
Overloads
DateDiff (intervalType, startDateTime, endDateTime)
DateDiff (intervalType, startDateTime, endDateTime, firstDayOfWeek)
Arguments
intervalType is a String expression that is the interval of time you use to calculate the difference between startDateTime and endDateTime. Possible values can be:
Interval type value Description
yyyy
Year
q
Quarter
m
Month
y
Day of year
d
Day (both "y" and "d" find the difference in days)
w
Number of weeks between startDateTime and endDateTime
ww
Number of firstDayOfWeek's between startDateTime and endDateTime
h
Hour
n
Minute
s
Second
startDateTime is the first DateTime value used in calculating the difference.
endDateTime is the second DateTime value used in calculating the difference.
firstDayOfWeek is an optional constant specifying the first day of the week. If not specified, crSunday is assumed. See Day of Week constants.
Returns
A Number value.
Action
DateDiff returns a number of time intervals between two specified dates.
Examples
The following examples are applicable to both Basic and Crystal syntax:
Use DateDiff with the "d" or "y" interval type parameter to find the number of days between two dates:
DateDiff ("d", #10/7/1999#, #10/10/1999#)
Returns 3.
Use DateDiff with the "yyyy" interval type parameter to find the number of years difference between two dates. This use of DateDiff is the same as finding the difference between the year of endDateTime and the year of startDateTime.
DateDiff ("yyyy", #10/7/1999#, #2/10/2005#)
Returns 6.
DateDiff ("yyyy", #12/31/1999#, #1/1/2000#)
Returns 1 (a 1 year difference), even though there is only a 1 day difference between the dates.
DateDiff ("yyyy", #1/1/1999#, #12/31/1999#)
Returns 0 (a 0 year difference), even though there is a 364 day difference.
Suppose that for the above examples, the first date is the date that you bought a mutual fund, and the second date is the date you sold it. The mutual fund company must send you an annual report for every year in which you owned units in the fund. Thus, you would get 7, 2 and 1 annual reports respectively, in the above cases.
Use DateDiff with the "q" parameter to find the number of quarters (3 month periods) difference between two dates.
DateDiff ("q", #10/6/1999#, #5/20/2003#)
Returns 14.
DateDiff ("q", #3/31/1999#, #4/1/1999#)
Returns 1. The two dates are in adjacent quarters.
DateDiff ("q", #1/1/1999#, #3/31/1999#)
Returns 0. The two dates are in the same quarter.
Suppose the mutual fund company in the "yyyy" example mailed out quarterly reports. Then it would need to mail out 15, 2 and 1 quarterly reports respectively, in the above cases.
Use DateDiff with the "m" parameter to find the number of months difference between two dates.
DateDiff ("m", #3/15/1999#, #7/13/1999#)
Returns 4.
Use DateDiff with the "w" parameter to calculate the number of weeks between two dates. For example, if startDateTime is on a Tuesday, DateDiff counts the number of Tuesdays between startDateTime and endDateTime not including the initial Tuesday of startDateTime. Note however that it counts endDateTime if endDateTime is on a Tuesday.
DateDiff ("w", #10/19/1999#, #10/25/1999#)
Returns 0.
DateDiff ("w", #10/19/1999#, #10/26/1999#)
Returns 1.
Use DateDiff with the "ww" parameter to calculate the number of firstDayOfWeek's occurring between two dates. For the DateDiff function, the "ww" parameter is the only one that makes use of the firstDayOfWeek argument. It is ignored for all the other interval type parameters. For example, if firstDayOfWeek is crWednesday, it counts the number of Wednesday's between startDateTime and endDateTime. It does not count startDateTime even if startDateTime falls on a Wednesday, but it does count endDateTime if endDateTime falls on a Wednesday. For the examples, note that October 6, 13, 20 and 27 all fall on a Wednesday in 1999.
DateDiff ("ww", #10/5/1999#, #10/29/1999#, crWednesday)
Returns 4.
DateDiff ("ww", #10/6/1999#, #10/29/1999#, crWednesday)
Returns 3.
DateDiff ("ww", #10/5/1999#, #10/27/1999#, crWednesday)
Returns 4.
For example, suppose you want to calculate the number of days between the order date and ship date, excluding Saturdays and Sundays:
Rem Basic syntax
Dim d1, d2
d1 = {Orders.Order Date}
d2 = {Orders.Ship Date}
formula = DateDiff("d", d1, d2) - _
            DateDiff("ww", d1, d2, crSaturday) - _
            DateDiff("ww", d1, d2, crSunday)
//Crystal syntax
Local DateTimeVar d1 := {Orders.Order Date};
Local DateTimeVar d2 := {Orders.Ship Date};
DateDiff ("d", d1, d2) -
     DateDiff ("ww", d1, d2, crSaturday) -
     DateDiff ("ww", d1, d2, crSunday)