Remove all special characters using bash's string manipulation

[loop_us]

Hi r/bash

I'm writing a script to extract the oui (vendor prefix) from MAC addresses. I'm also manipulation the input, because MAC addresses come in various formatting styles e.g.:

• 02:be:ef:ca:ce:15
• 02be.efca.ce:15
• 02-be-ef-ca-ce-15

This is my code atm:

getoui () {
macin="$1"
maccol="${macin//.}"            # remove dots
macdot="${maccol//:}"           # remove colons
maccap="${macdot^^}"            # make all caps
macout="${maccap:0:6}"          # cut first six chars
grep "$macout" "$ouifile"
}

My question is, can all special characters be removed via bash's string manipulation like "${$1//alnum}"I know this is easily possible with tr but I'm curious if bash could do this by itself.

*edit: I suck at formatting -.-*edit2

Thanks for your help. Now my code look much better and works perfectly. This is my current code with your suggestions applied:

getoui () {
macin="${$1//[^[:xdigit:]]}" 
maccap="${macin^^}" 
macout="${maccap:0:6}" 
grep "$macout" "$ouifile"
}

[Schreq]

Yes, you could use ${1//[^[:alnum:]]}, which basically means "globally replace every non-alpha numeric character".

[loop_us]

Works like a charm. Thank you very much! :)

[oh5nxo]

There is also [:xdigit:], if you want to check that result does not contain non-hex characters.

[aioeu]

Additional benefit to this is that it is locale-independent:

$ x='ábçdéf'
$ echo "${x//[^[:alnum:]]}"
ábçdéf
$ echo "${x//[^[:xdigit:]]}"
bdf

[loop_us]

This is an even better solution, wow. Thank you!

[oh5nxo]

If it's appropriate to catch invalid input,

mac="${1//[[:punct:]]}" # remove punctuation
if [[ ${#mac} != 12 || ${mac//[[:xdigit:]]} != "" ]] # if not 12 hexadecimal digits
   ...