Unzipping file with exclamation mark from command line in bash script

Question

I'm trying to do a simple script that unzips a zip file which has an exclamation mark in its name:

#!/bin/bash
UNZIP='/usr/bin/unzip'
CUT="/usr/bin/cut"
GREP="/usr/bin/grep"
FILENAME="test"

FILE="/usr/local/var/www/htdocs/"$FILENAME"\!3.zip"
UNZIPPEDFOLDER=$($UNZIP ${FILE} | $GREP -m1 'creating:' | $CUT -d' ' -f5-)
echo $UNZIPPEDFOLDER

but when script is being executed, unzip returns:

unzip:  cannot find or open /usr/local/var/www/htdocs/test\!3.zip, /usr/local/var/www/htdocs/test\!3.zip.zip or /usr/local/var/www/htdocs/test\!3.zip.ZIP.

It works fine when there is no "!" sign in filename, but to make some of my work automated, I need to stay with original file names.

Answer 1

! is the command history expansion character so needs to be quoted on the command-line when history expansion is enabled. It does not need to be quoted inside a script.

From the man page:

When the command history expansion facilities are being used (see HISTORY EXPANSION below), the history expansion character, usually !, must be quoted to prevent history expansion.

and

Non-interactive shells do not perform history expansion by default.

see man bash, search for QUOTING and HISTORY EXPANSION for full details.

Answer 2

In general you don't have to escape and quote anything. The secure way to write this would be:

FILE="/usr/local/var/www/htdocs/${FILENAME}!3.zip"
UNZIPPEDFOLDER="$("$UNZIP" "$FILE" | "$GREP" -m1 'creating:' | "$CUT" -d' ' -f5-)"
echo "$UNZIPPEDFOLDER"