Bash: Change String variable to the first word that starts with - string

Question

I am trying to isolate the first interface that starts with enp
Example:
foo = "bar0s0 foo0s8 enp0s0"
startword = "enp"
result:
$foo = enp0s0

Answer 1

WORDS="bar0s0 foo0s8 enp0s0"
START="enp"
for w in $WORDS; do
case $w in
${START}*)
echo $w
break # break out of the for loop for first instance
;;
esac
done

Answer 2

foo=$(echo $foo|sed -e "s/^.*\(${startword}[^ ]*\).*$/\1/")

Answer 3

Here's a solution using awk, modify to suit your needs:
echo "bar0s0 foo0s8 enp888 enp0s0" | awk ' { for(i = 1; i <= NF; i++) if($i ~ "enp") { print $i; break;} } '

Answer 4

You tagged your question as bash, so I'll provide a simple, portable solution.
Since you provide no context to your question it's hard to determine what the best approach might be but in most cases, this should do the trick.
bash>bash --version
GNU bash, version 3.2.52(2)-release (i586-suse-linux-gnu)
Copyright (C) 2007 Free Software Foundation, Inc.
bash>foo="bar0s0 foo0s8 enp0s0 random word123"
bash>startword="enp"
bash>index=$(expr index "$foo" $startword)
bash>echo ${foo:$(($index -1)):6}
enp0s0
A couple of caveats:
This assumes you are in fact working with bash and your tools are somewhat limited.
This also assumes that interface names are always 6 characters.
This approach uses sub-string extraction, if you want to read up on it take a look at http://www.tldp.org/LDP/abs/html/string-manipulation.html