如何在 Bash 中使用 ls 命令仅显示目录中的隐藏文件

Question

I am trying to display the hidden files from a directory using the ls command alone, except for . and .. .

ls -lAd .*

But the command returns . and .. directory names also in the output . What's wrong in this command even though I mentioned "A" option with ls .

Answer 1

您可以使用GLOBIGNORE内部变量。

$ GLOBIGNORE=.:.. ls -ld .*

Answer 2

This is here for reference but the answer is given by @anishsane below .

Your glob is wrong:

.*

will be expanded to . , and .. . You can either change the glob to something like:

.??*

but this will not match hidden files like: .a , .b . Or .[^.]* but that will not match files like ..a , ..b , combining both might be an option:

ls -lAd .[^.]*  ..?*

But that will most likely yeild: ls: error: ..?* no such file or directory . Enabling nullglob will prevent this:

shopt -s nullglob
ls -lAd .[^.]*  ..?*

This will however expand to ls -lAd if no hidden files are in the directory, which will show current working directory ( . )

Alternative you can pipe the output from ls to awk , but that will most likely get rid of the colors:

ls -lA | awk '$9 ~ /^\./'

Also consider reading: Why you shouldn't parse the output of ls(1)

Answer 3

ls -Al --ignore="[^.]*"

where you ignore all files and directories which NOT start with a .

Extra : You can call this for any directory, which comes in handy if you use it as an alias in your .bashrc

alias lsh='ls -Al --ignore="[^.]*"'

Just do

lsh /path/directory