如何将当前路径的目录名称(带空格)作为参数传递给 bash 脚本？

Question

我可以在ubuntu-bash和windows/cygwin-bash上重现这个问题。这个问题发生在文件名和目录名中有空格的情况下。这是我的脚本，名为pass-dir-names-with-spaces.sh:

for dir in "$@"
do
echo dir = $dir
done

这很好：

./pass-dir-names-with-spaces.sh *
dir = pass-file-names-with-spaces.sh
dir = this is a directory with spaces in it
dir = this is a file name with spaces.txt
dir = this is another file name with spaces.txt

我在每一行都得到完整的文件名，即使其中有空格。有时这是完全可以接受的，因为我想要文件名和目录名。然而，有时我想用perl实现一个过滤器来删除文件名。

backtick/perl方法不起作用！这是我第一次尝试逃离太空。

./pass-dir-names-with-spaces.sh `perl -e ' my @m  = (); my $c = shift @ARGV; opendir $h, $c or die "cannot opendir $c"; @a= map { push @m, $_ if  -d $_ } grep(!/^..$/,readdir $h); closedir $h; foreach (@m){s/ /\\ /g; print " $_ " } '  .`
dir = .
dir = this
dir = is
dir = a
dir = directory
dir = with
dir = spaces
dir = in
dir = it

我试着用引号代替：

./pass-dir-names-with-spaces.sh `perl -e ' my @m  = (); my $c = shift @ARGV; opendir $h, $c or die "cannot opendir $c"; @a= map { push @m, $_ if  -d $_ } grep(!/^..$/,readdir $h); closedir $h; foreach (@m){ print " "$_" " } '  .`
dir = "."
dir = "this
dir = is
dir = a
dir = directory
dir = with
dir = spaces
dir = in
dir = it"

看来bash忽略了我的语录！看来"$@"行为不端！如何过滤掉文件名？

感谢

齐格弗里德

Answer 1

传递目录名的方式不对。

您需要像这样重新构建调用(perl脚本的简化版本(：

readarray -t -d '' dirs < <(perl -e 'print "$_x00"')
./pass-dir-names-with-spaces.sh "${dirs[@]}"

Answer 2

• set --：清空参数数组
• for d in *; do：将每个目录条目作为变量d进行迭代
• [ -d "$d" ] && set -- "$@" "$d"：如果条目d是目录，则将其添加到参数数组中
• ./pass-dir-names-with-spaces.sh "$@"：最后将参数数组传递给脚本或命令

set --
for d in *; do
[ -d "$d" ] && set -- "$@" "$d"
done
./pass-dir-names-with-spaces.sh "$@"

现在，为了反映@Shellter的建议，让我们使用find:

find . -mindepth 1 -maxdepth 1 -type d -exec ./pass-dir-names-with-spaces.sh {} +

或者，如果find对于-exec command {} +语法来说太旧：

find . -mindepth 1 -maxdepth 1 -type d -printf '%f' |
xargs -0 ./pass-dir-names-with-spaces.sh