Test if a command outputs an empty string

Question

How can I test if a command outputs an empty string?

Answer 1

if [ -z "$(ls -lA)" ]; then
  echo "no files found"
else
  echo "There are files"
fi

This will run the command and check whether the returned output (string) has a zero length. You might want to check the 'test' manual pages for other flags.

Use the "" around the argument that is being checked, otherwise empty results will result in a syntax error as there is no second argument (to check) given!

Note: that ls -la always returns . and .. so using that will not work, see ls manual pages. Furthermore, while this might seem convenient and easy, I suppose it will break easily. Writing a small script/application that returns 0 or 1 depending on the result is much more reliable!

Answer 2

Bash Reference Manual

6.4 Bash Conditional Expressions

-z string
     True if the length of string is zero.

-n string
string
     True if the length of string is non-zero.

You can use shorthand version:

if [[ $(ls -A) ]]; then
  echo "there are files"
else
  echo "no files found"
fi

Answer 3

Sometimes you want to save the output, if it's non-empty, to pass it to another command. If so, you could use something like

list=`grep -l "MY_DESIRED_STRING" *.log `
if [ $? -eq 0 ]
then
    /bin/rm $list
fi

This way, the rm command won't hang if the list is empty.

Answer 4

Here's an alternative approach that writes the std-out and std-err of some command a temporary file, and then checks to see if that file is empty. A benefit of this approach is that it captures both outputs, and does not use sub-shells or pipes. These latter aspects are important because they can interfere with trapping bash exit handling (e.g. here)

tmpfile=$(mktemp)
some-command  &> "$tmpfile"
if [[ $? != 0 ]]; then
    echo "Command failed"
elif [[ -s "$tmpfile" ]]; then
    echo "Command generated output"
else
    echo "Command has no output"
fi
rm -f "$tmpfile"

Answer 5

_{Previously, the question asked how to check whether there are files in a directory. The following code achieves that, but see rsp's answer for a better solution.}

---

Empty output

Commands don’t return values – they output them. You can capture this output by using command substitution; e.g. $(ls -A). You can test for a non-empty string in Bash like this:

if [[ $(ls -A) ]]; then
    echo "there are files"
else
    echo "no files found"
fi

Note that I've used -A rather than -a, since it omits the symbolic current (.) and parent (..) directory entries.

Note: As pointed out in the comments, command substitution doesn't capture trailing newlines. Therefore, if the command outputs only newlines, the substitution will capture nothing and the test will return false. While very unlikely, this is possible in the above example, since a single newline is a valid filename! More information in this answer.

---

Exit code

If you want to check that the command completed successfully, you can inspect $?, which contains the exit code of the last command (zero for success, non-zero for failure). For example:

files=$(ls -A)
if [[ $? != 0 ]]; then
    echo "Command failed."
elif [[ $files ]]; then
    echo "Files found."
else
    echo "No files found."
fi

More info here.

Answer 6

As Jon Lin commented, ls -al will always output (for . and ..). You want ls -Al to avoid these two directories.

You could for example put the output of the command into a shell variable:

v=$(ls -Al)

An older, non-nestable, notation is

v=`ls -Al`

but I prefer the nestable notation $( ... )

The you can test if that variable is non empty

if [ -n "$v" ]; then
    echo there are files
else
    echo no files
fi

And you could combine both as if [ -n "$(ls -Al)" ]; then