Question

I am trying to do an if/then statement, where if there is non-empty output from a ls | grep something command then I want to execute some statements. I am do not know the syntax I should be using. I have tried several variations of this:

if [[ `ls | grep log ` ]]; then echo "there are files of type log";

Answer 1

Well, that’s close, but you need to finish the if with fi.

Also, if just runs a command and executes the conditional code if the command succeeds (exits with status code 0), which grep does only if it finds at least one match. So you don’t need to check the output:

if ls | grep -q log; then echo "there are files of type log"; fi

If you’re on a system with an older or non-GNU version of grep that doesn’t support the -q ("quiet") option, you can achieve the same result by redirecting its output to /dev/null:

if ls | grep log >/dev/null; then echo "there are files of type log"; fi

In this particular case, you can do without grep entirely, since ls returns nonzero if it doesn’t find a specified filename, as in D.Shawley’s answer:

if ls *log* >&/dev/null; then echo "there are files of type log"; fi

But in Zsh, or Bash with the failglob option set, such a command will error out if the wildcard doesn’t match anything, without ever actually running ls. You can take advantage of that behavior to check without needing ls at all:

# Zsh, or Bash with failglob set
if (echo *log*) >&/dev/null; then
  echo "there are files of type log"
fi

Or you could set nullglob instead:

# with 'nullglob' set, in either Bash or Zsh
for f in *log*; do
  echo "There are files of type log"
  break
done

In Ksh, or Bash without either option set, it’s a little more work:

# Ksh or Bash without failglob or nullglob
for f in *log*; do 
  # even if there are no matching files, the body of this loop will run once
  # with $f set to the literal string "*log*", so make sure there's really
  # a file there:
  if [ -e "$f" ]; then 
    echo "there are files of type log"
    break
  fi
done