How do I check if certain files exist in Bash?

Question

In a bash script, I have to check for the existence of several files.

I know an awkward way to do it, which is as follows, but that would mean that my main program has to be within that ugly nested structure:

if [ -f $FILE1 ]
then
if [ -f $FILE2 ]
then
  echo OK
  # MAIN PROGRAM HERE
fi
fi

The following version does not work:

([ -f $FILE1 ] && [ -f $FILE2 ]) ||  ( echo "NOT FOUND"; exit 1 )
echo OK

It prints

NOT FOUND
OK

Is there an elegant way to do this right?

UPDATE: See the accepted answer. In addition, in terms of elegance I like Jonathan Leffler's answer:

arg0=$(basename $0 .sh)
error()
{
    echo "$arg0: $@" 1>&2
    exit 1
}

[ -f $FILE2 ] || error "$FILE2 not found"
[ -f $FILE1 ] || error "$FILE1 not found"

Answer 1

How about

if [[ ! ( -f $FILE1 && -f $FILE2 ) ]]; then
    echo NOT FOUND
    exit 1
fi

# do stuff
echo OK

See help [[ and help test for the options usable with the [[ style tests. Also read this faq entry.

Your version does not work because (...) spawns a new sub-shell, in which the exit is executed. It therefor only affects that subshell, but not the executing script.

The following works instead, executing the commands between {...} in the current shell.

I should also note that you have to quote both variables to ensure there is no unwanted expansion or word splitting made (they have to be passed as one argument to [).

[ -f "$FILE1" ] && [ -f "$FILE2" ] || { echo "NOT FOUND"; exit 1; }

Answer 2

I think you're looking for:

if [ -f $FILE1 -a -f $FILE2 ]; then
    echo OK
fi

See man test for more details on what you can put inside the [ ].