Performing IF Statement in awk

Question

I have an array of users and I need to see if the owner of a file (logfile.log) exists within the array. Using awk I am able to pull the owner ($3) but when I use try to see if $3 is in user I get a syntax error at the beginning of my if statement. My limited understanding is that awk is not liking the syntax.

user=('michael' 'mark' 'luke' 'john' 'phil' 'sam' 'kevin'); 
ls -ldL logfile.log 2>/dev/null | 
/bin/awk '{ 
    Result = $NF ":\tPermissions=" $1; 
    if ([[ "${user[*]}" =~ (^|[^[:alpha:]])$3([^[:alpha:]]|$) ]]) { 
        Result = Result "\tOwner=SUPPORT"; 
    } 
    else { 
        Result = Result "\tOwner=" $3; 
    } 
    print Result;
}'

Answer 1

Don't parse ls ( http://mywiki.wooledge.org/ParsingLs ). Use stat to get the owner (check your stat man page, there are different implementations of different OS's)

# give your arrays a plural variable name
users=('michael' 'mark' 'luke' 'john' 'phil' 'sam' 'kevin')
owner=$(stat -c '%U' logfile.log)

if [[ " ${users[*]} " == *" $owner "* ]]; then    # spaces are deliberate
    echo logfile.log has a valid owner: $owner
else
    echo logfile.log is not owned by a valid user: $owner
fi

The other approach is to iterate over the array and look for an exact match:

valid=false
for user in "${users[@]}"; do
    if [[ $user == $owner ]]; then
        valid=true
        break
    fi
done
if $valid; then
    echo file has a valid owner
fi

The main problem in your code is that you expect awk to understand bash syntax. It doesn't.