SHELL

shell script to check the directory exists

Directory="/opt"
if [ -d "$Directory" ];
then
	echo -e "it's exits
"
fi
### To check if it's not exists
if [ ! -d "$Directory" ];
then 
	echo -e "It's not there
"
fi

linux command if directory exists

DIR="/etc/httpd/"
if [ -d "$DIR" ]; then
  ### Take action if $DIR exists ###
  echo "Installing config files in ${DIR}..."
else
  ###  Control will jump here if $DIR does NOT exists ###
  echo "Error: ${DIR} not found. Can not continue."
  exit 1
fi

Check is directory exist shell

DIR="/etc/httpd/"
if [ -d "$DIR" ]; then
  # Take action if $DIR exists. #
  echo "Installing config files in ${DIR}..."
fi

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

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.