I am looking to create a bash function to filter all the dotfiles (no directories) in a selected directory. I only need the file name, not the full path.
For the moment, i only have this command:
find . -maxdepth 1 -type f -print0
which prints all the files excluding the dirs. Now i still have to exclude the non-dotfiles. So i've try to pipe the output to grep, like so:
find . -maxdepth 1 -type f -print0 | grep "^\."
and it didn't seem to work. ->
Binary file (standard input) matches
Do you guys have an elegant solution to this?
Thanks!
If you want only dot-files:
find . -maxdepth 1 -type f -name '.*' -printf '%f\0'
The test -name '.*' selects dot files. Since -name accepts globs, . means a literal period and * means any number of any character.
The action -printf '%f\0' will print NUL-separated filenames without the path.
If your name selection criteria becomes more complex, find also offers -regex which selects files based on regular expressions. GNU find understands several different dialects of regular expression. These can be selected with -regextype. Supported dialects include emacs (default), posix-awk, posix-basic, posix-egrep, and posix-extended.
BSD find does not offer -printf. In its place, try this:
find . -maxdepth 1 -type f -name '.*' -exec basename {} \;
Note that this will be safe all file names, even those containing difficult characters such as blanks, tabs or newlines.
If you want to get all dot files and directories into an array, it is simple:
all=(.*).
That is safe for all file names.
If you want to get only regular files, not directories, then use bash:
a=(); while IFS= read -r -d ''; do a+=("$(basename "$REPLY")"); done < <( find $HOME -maxdepth 1 -type f -name '.*' -print0 )
This is also safe for all file names.