Shell how to test every input for file

Question

so i am doing a shell script and am trying to figure out how to do this safe guard function:

if ! [[ -e "$@" ]]; then
    echo"Not a file!"
    exit 0
fi

my ideia is to check every input for file. so when i call the script: ./scriptname file1.txt file2.tar NOTfile4 it will output NOT a file

Use a loop. (This suggestion brought to you by [shellcheck](https://shellcheck.net)) — that other guy, Mar 23 '18 at 18:47
+1. see https://stackoverflow.com/questions/255898/how-to-iterate-over-arguments-in-a-bash-script for how to use a loop. — tk421, Mar 23 '18 at 18:53

score 1 · Accepted Answer · answered Mar 23 '18 at 19:09

1

thanks guys! did the loop and it worked. heres what i did if anyone who has the same problem:

for input in "$@"; do
if ! [ -e "$input" ]; then
    echo "Not a file!"
    exit 0
fi
done

answered Mar 23 '18 at 19:09

2

That should probably be `exit 1`; Non-zero exit statuses indicate errors. If every file exists, your script will exit 0 anyway. – chepner Mar 23 '18 at 19:26

1 Answers1