>>How can I take a number like 1.345 and round it to a specific precision like .2?
>>
>>1.3456 precision .2 = 1.4
>>1.6345 precision .5 = 1.5
>>1.3456 precision .005 = 1.345
>>
>>I think that I need to figure out how many decmil places and ROUND() the number then somehow round to the precision
>>
>>
>>Thanks
>
>
>John, I just cooked this up and it should handle all your needs... 3 simple parameters...
>
>

>? 1.3456, PreciseRound( 1.3456, .2, 1 )
>? 1.6345, PreciseRound( 1.6345, .5, 1 )
>? 1.3456, PreciseRound( 1.3456, .005, 3 )
>
>*/ function to round precision
>*/ Ex: PreciseRound( 123.3321, .4, 1 )
>function PreciseRound( lnValue, lnPrecVal, lnDec )
>   local lnNewVal, lnRem
>	
>   */ Put into whole number and do primary rounding
>   */ before determining precision up/down...
>   lnNewVal = round( lnValue, lnDec +1 )
>	
>   */ Now, based on precision what remainder would
>   */ be left in case not exactly divisible
>   lnRem    = mod( lnNewVal, lnPrecVal )
>	
>   */ if over the halfway precision mark, round up
>   if lnRem / lnPrecVal >= .5
>      lnNewVal = lnNewVal + lnPrecVal - lnRem
>   else
>      lnNewVal = lnNewVal - lnRem
>   endif
>return( round( lnNewVal, lnDec ) )
>

Try this out for 2 decimal places

lnVar = 1.545

lnVar=(int(lnVar*100)/100)
?lnVar
the answer should be 1.54

if you want to round it up to 1.6

lnVar = (int(lnVar*10)+.5)/10

HTH



HTH