In a Bash if condition, how to check whether any files matching a simple wildcard expression exist?

Question

Stupidly, I had been using a condition like this as part of a script:

if [ $(ls FOO* 2> /dev/null) ] # if files named "FOO*" were downloaded 
then
    echo "Files found" 
    # ... process and email results
else
    echo "Not found"
    # ... email warning that no files were found (against expectations)
fi

That works for zero and one files named FOO*, but fails if there are more than one. From logs I found several different error messages stemming from this:

[: FOO_20131107_082920: unary operator expected
[: FOO_20131108_070203: binary operator expected
[: too many arguments

My question is: what is the correct way to check, in a Bash if condition, whether one or more files whose name begins with FOO exist?

GNU bash, version 4.2.25(1)-release (x86_64-pc-linux-gnu)

Answer 1

This happens because your command substitution for ls outputs whitespace, and it ultimately undergoes word splitting before being passed to [. A less breakable way would be to put the files in an array, and then check that the array has at least one member.

shopt -s nullglob

files=( FOO* )
if (( ${#files[@]} )); then
    # there were files
fi

This works because (( by default returns true if the value does not equal 0, and ${#files[@]} gets the number of items in the array (which will be >0 if there are files matching the glob).

You could also do something like this, as long as nullglob is not set:

if ls FOO* >/dev/null 2>&1; then
    # there were files
fi

This just checks the exit code of ls, which will be 1 if you passed a filename that doesn't exist (the literal FOO*, if nothing is matched (unless, of course, you are evil and there is a file named FOO*, in which case it will return 0 :-) )).

Note that both of these also match directories. If you really only want to match regular files, you need to test that:

for file in FOO*; do
    if [[ -f $file ]]; then
        # file found, do some stuff and break
        break
    fi
done

Answer 2

The multi-line answers above didn't suit my desire for a one-line solution and not modifying the environment. Here is a generic one-liner that may work for you:

echo $(ls FOO* 2>/dev/null | wc -w)

the /dev/null is because ls throws an error if there's no file. This just ignores ls and counts the number of files found based on the number of "words" (really filenames). You can the use it in a simple if statement:

if [ 0 -lt $(ls FOO* 2>/dev/null | wc -w) ]; then
 echo "Some FOO is there."
fi

This approach has the flexibility of not just checking for whether any matching files exist but also optionally checking for how many matching files exist (just change 0 and/or -lt) and then you proceed further on that basis.

I haven't checked against files/folders that have wildcards or spaces in them; I suspect the above method would give wrong counts on spaces in particular, which is only an issue if an accurate count is required beyond 0. One fix I tried for this is to use $(ls -l FOO* 2>/dev/null | wc -l), which will break the list down by lines and therefore should be more accepting of white space.

Answer 3

There is another option, using case,

FILES=FOO*
FILES=$(echo $FILES)
case $FILES in FOO\*) echo "Not found" ;; esac

This works as case expects only one argument to test.