列出文件夹，但如果文件夹与 Bash 中的列表匹配，则排除

Question

I am playing around with Bash for the first time.

This script will list all folders, however I want to modify it to not list a certain folder if possible.

Working:

_list_banners () {
  for f in * ; do [ -d "$f" ] && echo ${_handle_banner_type}${_handle_message} $f ; done
}

This works and lists the folders, but I want to exclude for example folder MASTER from the list.

Attempted:

_list_banners () {
  for f in * ;
    if [ $f != "MASTER" ]; then
      do [ -d "$f" ] && echo ${_handle_banner_type}${_handle_message} $f
    fi
  ; done
}

But when I run it, I get the following error instead of my file list:

$ bash myscript
myscript: line 3: syntax error near unexpected token `if'
myscript: line 3: `    if [ $f != "MASTER" ]; then'

Answer 1

do is part of the for loop and shouldn't be moved inside the if block:

_list_banners () {
  for f in * ;
  do 
    if [ "$f" != "MASTER" ]; then
      [ -d "$f" ] && echo "${_handle_banner_type}${_handle_message} $f"
    fi
  done
}

Answer 2

You could move both the directory check and the exclusion into your glob (requires shopt -s extglob , and shopt -s nullglob would be a good idea to avoid unexpected behaviour when nothing matches):

shopt -s extglob
shopt -s nullglob

_list_banners () {
    for f in !(MASTER)/; do 
        echo "${_handle_banner_type}${_handle_message} $f"
    done
}

!(MASTER) is a glob that excludes MASTER , and appending / only returns directories.

If you want just the directory name without the trailing slash, you can use ${f%/} instead of just $f on the echo line.