How can I check if a directory exists in a Bash shell script?

Solution 1:

To check if a directory exists in a shell script, you can use the following:

if [ -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY exists.
fi

Or to check if a directory doesn't exist:

if [ ! -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY doesn't exist.
fi

However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check. E.g. running this:

ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then 
  rmdir "$SYMLINK" 
fi

Will produce the error message:

rmdir: failed to remove `symlink': Not a directory

So symbolic links may have to be treated differently, if subsequent commands expect directories:

if [ -d "$LINK_OR_DIR" ]; then 
  if [ -L "$LINK_OR_DIR" ]; then
    # It is a symlink!
    # Symbolic link specific commands go here.
    rm "$LINK_OR_DIR"
  else
    # It's a directory!
    # Directory command goes here.
    rmdir "$LINK_OR_DIR"
  fi
fi

Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.

If the variables contain spaces or other unusual characters it will probably cause the script to fail.

Solution 2:

Remember to always wrap variables in double quotes when referencing them in a Bash script. Kids these days grow up with the idea that they can have spaces and lots of other funny characters in their directory names. (Spaces! Back in my days, we didn't have no fancy spaces! ;))

One day, one of those kids will run your script with $DIRECTORY set to "My M0viez" and your script will blow up. You don't want that. So use this.

if [ -d "$DIRECTORY" ]; then
    # Will enter here if $DIRECTORY exists, even if it contains spaces
fi