How can I ensure a Bash string is alphanumeric, without an underscore?

Question

I'm adding a feature to an existing script that will allow the user to configure the hostname of a Linux system. The rules I'm enforcing are as follows:

• Must be between 2 and 63 characters long
• Must not start or end with a hyphen
• Can only contain alphanumeric characters, and hyphens; all other characters are not allowed (including an underscore, which means I can't use the \W regex symbol)

I've solved the first two in the list, but I'm having trouble figuring out how to check if a bash string contains only letters, digits, and hyphens. I think I can do this with a regex, but I'm having trouble figuring out how (I've spent the past hour searching the web and reading man pages).

I'm open to using sed, grep, or any of the other standard tools, but not Perl or Python.

Answer 1

Seems like this ought to do it:

^[a-zA-Z0-9][-a-zA-Z0-9]{0,61}[a-zA-Z0-9]$

Match any one alphanumeric character, then match up to 61 alphanumeric characters (including hyphens), then match any one alphanumeric character. The minimum string length is 2, the maximum is 63. It doesn't work with Unicode. If you need it to work with Unicode you'll need to add different character classes in place of a-zA-Z0-9 but the principle will be the same.

I believe the correct grep expression that will work with Unicode is:

^[[:alnum:]][-[:alnum:]]{0,61}[[:alnum:]]$

Example Usage:

---

echo 123-abc-098-xyz | grep -E '^[[:alnum:]][-[:alnum:]]{0,61}[[:alnum:]]$'

result=$(grep -E '^[[:alnum:]][-[:alnum:]]{0,61}[[:alnum:]]$' <<< "this-will-work"); echo $result;

echo "***_this_will_not_match_***" | grep -E '^[[:alnum:]][-[:alnum:]]{0,61}[[:alnum:]]$'

Answer 2

This is a bash script testing the first parameter whether it contains only alphanumerics or hyphens. It "pipes" the contents of $1 into grep:

#!/bin/bash
if grep '^[-0-9a-zA-Z]*$' <<<$1 ;
  then echo ok;
  else echo ko;
fi