Listing all digits behind the comma with grep

Shell Programming and Scripting
1

Hello, Unix-Forums!

1.23456789

This an example number. It can be any number. I want grep to only find the digits behind the "."
That means

23456789

should be the output in this case.

How would I do that?

2

Try:

nr=1.23456789
echo "${nr#*.}"
3

So is this in a stream or already in a shell variable? Which shell?

For example if it is a string you can use your shells parameter expansion (bash/ksh):

var='1.23456789'
echo "${var#*.}"

a stream:

$ echo 1.23456789 | awk -F. '{print $2}'
23456789

$ echo 1.23456789 | cut -d. -f2
23456789