Test if a command outputs an empty string

TL;DR

if [[ $(ls -A | head -c1 | wc -c) -ne 0 ]]; then ...; fi

Thanks to netj for a suggestion to improve my original:
if [[ $(ls -A | wc -c) -ne 0 ]]; then ...; fi

This is an old question but I see at least two things that need some improvement or at least some clarification.

First problem

First problem I see is that most of the examples provided here simply don't work. They use the ls -al and ls -Al commands - both of which output non-empty strings in empty directories. Those examples always report that there are files even when there are none.

For that reason you should use just ls -A - Why would anyone want to use the -l switch which means "use a long listing format" when all you want is test if there is any output or not, anyway?

So most of the answers here are simply incorrect.

Second problem

The second problem is that while some answers work fine (those that don't use ls -al or ls -Al but ls -A instead) they all do something like this:

  1. run a command
  2. buffer its entire output in RAM
  3. convert the output into a huge single-line string
  4. compare that string to an empty string

What I would suggest doing instead would be:

  1. run a command
  2. count the characters in its output without storing them
    • or even better - count the number of maximally 1 character using head -c1
      (thanks to netj for posting this idea in the comments below)
  3. compare that number with zero

So for example, instead of:

if [[ $(ls -A) ]]

I would use:

if [[ $(ls -A | wc -c) -ne 0 ]]
# or:
if [[ $(ls -A | head -c1 | wc -c) -ne 0 ]]

Instead of:

if [ -z "$(ls -lA)" ]

I would use:

if [ $(ls -lA | wc -c) -eq 0 ]
# or:
if [ $(ls -lA | head -c1 | wc -c) -eq 0 ]

and so on.

For small outputs it may not be a problem but for larger outputs the difference may be significant:

$ time [ -z "$(seq 1 10000000)" ]

real    0m2.703s
user    0m2.485s
sys 0m0.347s

Compare it with:

$ time [ $(seq 1 10000000 | wc -c) -eq 0 ]

real    0m0.128s
user    0m0.081s
sys 0m0.105s

And even better:

$ time [ $(seq 1 10000000 | head -c1 | wc -c) -eq 0 ]

real    0m0.004s
user    0m0.000s
sys 0m0.007s

Full example

Updated example from the answer by Will Vousden:

if [[ $(ls -A | wc -c) -ne 0 ]]; then
    echo "there are files"
else
    echo "no files found"
fi

Updated again after suggestions by netj:

if [[ $(ls -A | head -c1 | wc -c) -ne 0 ]]; then
    echo "there are files"
else
    echo "no files found"
fi

Additional update by jakeonfire:

grep will exit with a failure if there is no match. We can take advantage of this to simplify the syntax slightly:

if ls -A | head -c1 | grep -E '.'; then
    echo "there are files"
fi

if ! ls -A | head -c1 | grep -E '.'; then
    echo "no files found"
fi

Discarding whitespace

If the command that you're testing could output some whitespace that you want to treat as an empty string, then instead of:

| wc -c

you could use:

| tr -d ' \n\r\t ' | wc -c

or with head -c1:

| tr -d ' \n\r\t ' | head -c1 | wc -c

or something like that.

Summary

  1. First, use a command that works.

  2. Second, avoid unnecessary storing in RAM and processing of potentially huge data.

The answer didn't specify that the output is always small so a possibility of large output needs to be considered as well.


For those who want an elegant, bash version-independent solution (in fact should work in other modern shells) and those who love to use one-liners for quick tasks. Here we go! 

ls | grep . && echo 'files found' || echo 'files not found'

(note as one of the comments mentioned, ls -al and in fact, just -l and -a will all return something, so in my answer I use simple ls


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.


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!

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