Using awk to search for lines that have a number of digits

Question

Here is file1

200
201
202
203
204
205
2001
2002
2003
2004
2005

Is there an awk oneliner that finds only lines with three digits in the the first field?

Answer 1

awk '$1 ~ /^[0-9][0-9][0-9]$/' file1

This will match the first field ($1) against three digits only (note the forced start and stop range denoted by ^ and $). It then prints the entire line ($0). You don't need a {print $0} after the regex match, because the default action is to print the line anyway.

If you want to use the interval expression operator {} in your regex then you will need to use gawk and the --posix switch:

gawk --posix '$1 ~ /^[0-9]{3}$/' file1