Counting Fridays in date range (for loop) 3

[putts]

I need to figure out how many Fridays fall in a certain date range.

I had a good algorithm for it, but I dont know how to go about making a for loop.

Here's what I had

if {?postDate} = "" then
for i=0 to ((CurrentDate-{?preDate}) mod 7)
if DayOfWeek (DateAdd ("d",i,{?preDate})) = "Friday" then
truncate((CurrentDate-{?preDate})/7,0) + 1
next

truncate((CurrentDate-{?preDate})/7,0)
else
for i=0 to (({?postDate}-{?preDate}) mod 7)
if DayOfWeek (DateAdd ("d",i,{?preDate})) = "Friday" then
truncate(({?postDate}-{?preDate})/7,0) + 1
next

truncate(({?postDate}-{?preDate})/7,0)

So the question is, is there to run a for loop (which I'm guessing there's gotta be) and how do I do it.

TIA

 

[synapsevampire]

You might simplify this, what you're really doing is getting the number of weeks, adjusting for any missing Fridays.

Datediff("w",currentdate,{?preDate}) will return the number of weeks.

Now allow for the start and end dates possibly missing a Friday.

Sorry, I don't have Cryswtal installed right now or I'd work out the formula, but you get the idea.

-k

 

[putts]

Well it's the second part that I'm having a problem with.

The basis of my algorithm is this:
1. determine the number of full weeks in between my two dates
2. I figure out how many days are leftover after calculating the number of weeks [(difference in dates) mod 7] and figure out if a Friday falls into that leftover time.


For Example,
if the preDate is 04/01/03 and the end date is 04/17/03. The difference between the dates is 16 days which means that there's at least two fridays in there.
The number of days leftover when I subtract the 14 days for the two weeks from the original 16 leaves with 2 days leftover and now I have to figure out if there's a Friday in there.
This is what I have the problem with because this number could be in the range of 1 - 6 days. (thus my for loop)

So, that's where I'm stuck with this.

 

[JoeCrystal]

numberVar n;
numberVar i;

n:= DateDiff("d",{?sDate},{?eDate})\7;

for i:= 0 to (DateDiff("d",{?sDate},{?eDate}) Mod 7) do
(
If DayOfWeek (DateAdd("d",-i,{?eDate})) = 6 Then
(
n:= n + 1;
exit for);
);
n

 

[putts]

Thanks for that loop - I need to do some reading on syntaxes.

I found a way to do it with a downloadable Seagate component:

if (isDate({?preDate}) and isDate({?postDate})) then
cstr(wdaysclass1wdnumworkdays(CDate({?preDate}),CDate({?postDate}),"-----6-&quot)
else
"Error in date parameters"

Which, as you can see, is very compact and modular, so I'm gonna try to use this.

Thanks for the responses.

 

[dgillz]

Here is my formula which works:

datevar firstday:=Date(2002,8,1);
datevar lastday:=Date(2002,8,31);
numbervar loop;
numbervar Days;
numbervar span:=datediff("d",Firstday,LastDay);

For loop:= 0 to span do(
if dayofweek(firstday+loop) = 6 then Days:=Days+1 else Days:=Days);

Days


Software Sales, Training, Implementation and Support for Exact Macola, eSynergy, and Crystal Reports

http://www.trianglepartners.com

dgilsdorf@trianglepartners.com

 

[putts]

Thanks dGillz.

That was perfect.

I'm gonna modify it a bit to calculate some time logs out of a table but I needed that for loop with Fridays to get it going.

Thanks again to everyone for responding.

 

[Reebo99]

I've used the following in the past to calculate the number of Fridays between 2 dates:

DateDiff ("ww", Date(2003,04,01), Date(2003,04,17), 6)

Just to break it down, "ww" is the number of first day of weeks between the 2 dates, and the 6 at the end makes Friday the first day of the week, hence giving the number of Fridays between 2 dates.


Reebo
Scotland (Sunny with a Smile)

 

[dgillz]

Reebo,

A star for you. I did not know that datediff() had a fourth argument.

Software Sales, Training, Implementation and Support for Exact Macola, eSynergy, and Crystal Reports

http://www.trianglepartners.com

dgilsdorf@trianglepartners.com

 

[Reebo99]

Thank you.....I'd like to thank my family and friends who have stood by me all these years...This ones for you Mum!!

Reebo
Scotland (Sunny with a Smile)

 

[Naith]

Reebo, you are a crazy fool. But I like your style.

 

[Reebo99]

Takes one to know one Naith

Reebo
Scotland (Sunny with a Smile)