07-31-2007, 09:18 PM
Hi,
I have a variable that I need to truncate to 5 numbers plus a decimal point from a larger number in a string. The 6th digit of the should be rounded.
The problem is that the decimal point is not always in the same place so I can't easily use the
command
Some number examples are as follows:
310.417389
484.000000
180.000000
50.027470
6.547363
My strategy so far has been to extract the first 7 digits (6 digits plus the decimal point).
I would like to round up the 7th digit(it won't ever be the decimal point)
and then get the first 6 digits (which will be what I need 5 digits + a decimal point)
But I don't know how to just round up a number (regardless of its decimal location)
I thought about reducing the number to an integer using Variant.fix
and then getting the length of the integer and then depending on that making individual regex's to get out the right number
but that seems too cumbersome, Especially if there is a way to round up last digit of any number regardless of its decimal location.
Thanks for any advice.
Stuart
I have a variable that I need to truncate to 5 numbers plus a decimal point from a larger number in a string. The 6th digit of the should be rounded.
The problem is that the decimal point is not always in the same place so I can't easily use the
command
Some number examples are as follows:
310.417389
484.000000
180.000000
50.027470
6.547363
My strategy so far has been to extract the first 7 digits (6 digits plus the decimal point).
I would like to round up the 7th digit(it won't ever be the decimal point)
and then get the first 6 digits (which will be what I need 5 digits + a decimal point)
But I don't know how to just round up a number (regardless of its decimal location)
I thought about reducing the number to an integer using Variant.fix
str first7.left(sample number 7)
double first7=val(first7 2);; converts first 6 characters into a double variable
Variant v.fix(first7)
and then getting the length of the integer and then depending on that making individual regex's to get out the right number
but that seems too cumbersome, Especially if there is a way to round up last digit of any number regardless of its decimal location.
Thanks for any advice.
Stuart