Jeff,
I think parseDecimal will do the trick with the addition of 1 parameter. You can add a format parameter to this function call. In your case, it would look something like this parseDecimal(‘1,470.00’,’###,###.##’) Note that you can make the number of digits as large as you think will be needed (i.e. millions, billions, etc.). # means that digit is option. if you replaced # with 0, that would make that digit required.
The documentation for this is not very thorough. I’ve submitted some suggested changes to the documentation.
Hope that helps,
Mike