Hey grimm...
here's the one I use.
on removeDups li
newLi = li.duplicate()
newLi.sort()
newLiCnt = newLi.count
finalLi = []
repeat with i = 1 to newLiCnt
if finalLi.getOne(newLi[i]) = 0 then
finalLi.append(newLi[i])
end if
end repeat
return finalLi
end removeDups
...Neil
At 10:24 PM 6/6/2002 -0500, you wrote:
>Prob messy but it works:
>
>on checkList whichList
> tList = whichList
> repeat with tEntry = tList.count down to 1
> this = tList[tEntry]
> prev = false
> repeat with i = 1 to tList.count
> if tList[i] = this and i <> tEntry then
> prev = true
> end if
> end repeat
> if prev then tList.deleteAt(tEntry)
> end repeat
> return(tList)
>end
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