Paul Salyers wrote:
How is this statement wrote in Rev,
len(right(number, 3)
this in VB will take the number ex:
number = "12345678"
and cut it down to
number = "678"
starting at the right and give you the first 3 numbers.
I need to do this in Rev.
I searched "len" and got
put char (length(it) - 3) to (length(it)) of it into myExtension
but don't understand what it means.
Paul,
The thing you need to know about Run Rev is that it treats a string of characters like words or numbers depending on the context. For example
If you write:
Put 123 & 456 into theResult
Then theResult is 123456
If you write:
put 123 * 456 into theResult
Then the result is 56088
If you write:
Put char 1 to 4 of 123*456
Then theReuslt is 56088
Which is confusing. But if you
put char 1 to 4 of (123*456) into theResult
Then theResult is 5608
Which is what you might hope for.
Moral: Don't abuse flexibility.
I think the function you want is:
function rightCharacters tNum,howMany put the number of chars in tNum into tNumChars return char -(tNumChars-howMany) to -1 of tNum end rightCharacters
Notice: You count characters from the right as -1, -2, -3 etc.
Jim _______________________________________________ use-revolution mailing list use-revolution@lists.runrev.com http://lists.runrev.com/mailman/listinfo/use-revolution
