Re: Number of decimal places
Re: Number of decimal places
- Subject: Re: Number of decimal places
- From: email@hidden
- Date: Mon, 22 Apr 2002 17:03:37 -0600
What a fun challenge! I am sure someone will come up with a mathematical
algorithm to do the job in one line, but the following is how I would do
it. But I had to cheat. Notice that I removed the coercion of the string
value to a real out of the first line. First, to keep this coercion, you
would have to trap to make sure the user enters a string that can be
coerced to a number. It is much simpler to leave the value as a string and
do string manipulation. My solution does not test whether the value is
coerceable to a number, it just returns the number of places the decimal is
from the right of the string.
Here is my solution:
set decimalPlaces to 0
set x to (the text returned of (display dialog "Number:" default answer
""))
if x contains "." then set decimalPlaces to (offset of "." in ((reverse of
(characters of x)) as string)) - 1
display dialog "The number of decimal places of " & x & " is " &
decimalPlaces
The AppleScript Text Item Delimiter is a good solution but as it stands
does not take into consideration the possibility of a whole number. Looking
forward to seeing other solutions.
Shirley Hopkins
FaceSpan Support
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