[Solved] randomized Date Function needed

[everseeker]

I am testing a couple forms at work. To do it, I need to generate a couple thousand date related fields:

I got 90% of my work done, but I am stumped on 1 thing that I THOUGHT was going to be simple:

Given a min and max age, return a random DOB in "01/01/1999" format

Any thoughts?

Edited July 2, 2010 by everseeker

[Melba23]

everseeker,

This gets you dates in YYYY/MM/DD - I leave the reformatting as an exercise for the student (as my old maths master used to say! ):

#Include <Date.au3>

$sYoung = "1990/01/01"

$sOld = "1950/01/01"

$iMaxDaysDiff = _DateDiff("D", $sOld, $sYoung)

For $i = 1 To 10

    $iRandom = Random(0, $iMaxDaysDiff, 1)

    $sRandomDate = _DateAdd("D", $iRandom, $sOld)

    ConsoleWrite($sRandomDate & @CRLF)

Next

I hope it does what you want.

M23

[everseeker]

Hmmm... I was working along the lines of

Func CalculateDOB($MinAge, $MaxAge)
;Takes Min and Max age into account and calculates a random age
Return ("10/10/"&Random (_DateAdd( 'Y',-$MinAge, _NowCalcDate()), _DateAdd( 'Y',-$MaxAge, _NowCalcDate()),1))
EndFunc   ;==>CalculateDOB

But was failing... will look at your solution...

(Thing is, I have a min and max AGE, not Date.... )

[everseeker]

Silly me... Looking at yours , I realized my order was off... Here's the correct one, if anyone else browsing needs it:

Func CalculateDOB($MinAge, $MaxAge)
;Takes Min and Max age into account and calculates a random age
$Mon = Random(1, 12, 1)
If StringLen($Mon) = 1 Then $Mon = "0" & $Mon

$day = Random(1, 28, 1)
If StringLen($day) = 1 Then $day = "0" & $day
Return ($Mon & "/" & $day & "/" & Random(_DateAdd('Y', -$MaxAge, _NowCalcDate()), _DateAdd('Y', -$MinAge, _NowCalcDate()), 1))
EndFunc   ;==>CalculateDOB