Properly escaping output from pipe in xargs


% touch -- safe-name -name-with-dash-prefix "name with space" 
    'name-with-double-quote"' "name-with-single-quote'" 

xargs can’t seem to handle double quotes:

% ls | xargs ls -l 
xargs: unmatched double quote; by default quotes are special to xargs unless you use the -0 option
ls: invalid option -- 'e'
Try 'ls --help' for more information.

If we use the -0 option, it has trouble with name that has dash prefix:

% ls -- * | xargs -0 -- ls -l --
ls: invalid option -- 'e'
Try 'ls --help' for more information.

This is before using other potentially problematic characters like newline, control character, etc.

Asked By: Gerry Lufwansa


For xargs to understand the -0 null-delimited input option, the sending party must also apply the null delimiter to the data that they are sending over.

Else there’s no synchronization between the two.

One option is the GNU find command which can place such delimiters:

find . -maxdepth 1 ! -name . -print0 | xargs -0 ls -ld
Answered By: user218374

As you said, xargs doesn’t like unmatched double quotes unless you use -0 but -0 only makes sense if you feed it null-terminated data. So, this fails:

$ echo * | xargs
xargs: unmatched double quote; by default quotes are special to xargs unless you use the -0 option
name-with-backslash -name-with-dash-prefix

But this works:

$ printf '%s' -- * | xargs -0
-- name-with-backslash -name-with-dash-prefix name-with-double-quote" name-with-single-quote' name with space safe-name

In any case, your basic approach is not really the best way to do this. Instead of fiddling about with xargs and ls and whatnot, just use shell globs instead:

$ for f in *; do ls -l -- "$f"; done
-rw-r--r-- 1 terdon terdon 4142 Aug 11 16:03 a
-rw-r--r-- 1 terdon terdon 0 Aug 11 15:34 'name-with-backslash'
-rw-r--r-- 1 terdon terdon 0 Aug 11 15:34 -name-with-dash-prefix
-rw-r--r-- 1 terdon terdon 0 Aug 11 15:34 'name-with-double-quote"'
-rw-r--r-- 1 terdon terdon 0 Aug 11 15:34 "name-with-single-quote'"
-rw-r--r-- 1 terdon terdon 0 Aug 11 15:34 'name with space'
-rw-r--r-- 1 terdon terdon 0 Aug 11 15:34 safe-name
Answered By: terdon

The POSIX specification does give you an example for that:

ls | sed -e 's/"/"\""/g' -e 's/.*/"&"/' | xargs -E '' printf '<%s>n'

(with filenames being arbitrary sequences of bytes (other than / and NULL) and sed/xargs expecting text, you’d also need to fix the locale to C (where all non-NUL bytes would make valid characters) to make that reliable (except for xargs implementations that have a very low limit on the maximum length of an argument))

The -E '' is needed for some xargs implementations that without it, would understand a _ argument to signify the end of input (where echo a _ b | xargs outputs a only for instance).

With GNU xargs, you can use:

ls | xargs -rd 'n' printf '<%s>n'

(also adding -r (also a GNU extension) for the command not be run if the input is empty).

GNU xargs also has a -0 that has been copied by a few other implementations, so:

ls | tr 'n' '' | xargs -0 printf '<%s>n'

is slightly more portable.

All of those assume the file names don’t contain newline characters. If there may be filenames with newline characters, the output of ls is simply not post-processable. If you get:


That can be either both a a and b files or one file called a<newline>b, there’s no way to tell.

GNU ls has a --quoting-style=shell-always which makes its output unambiguous and could be post-processable, but the quoting is not compatible with the quoting expected by xargs. xargs recognise "...", x and '...' forms of quoting. But both "..." and '...' are strong quotes and can’t contain newline characters (only can escape newline characters for xargs), so that’s not compatible with sh quoting where only '...' are strong quotes (and can contain newline characters) but <newline> is a line-continuation (is removed) instead of an escaped newline.

You can use the shell to parse that output and then output it in a format expected by xargs:

eval "files=($(ls --quoting-style=shell-always))"
[ "${#files[@]}" -eq 0 ] || printf '%s' "${files[@]}" |
  xargs -0 printf '<%s>n'

Or you can have the shell get the list of files and pass it NUL-delimited to xargs. For instance:

  • with zsh:

    print -rNC1 -- *(N) | xargs -r0 printf '<%s>n'
  • with ksh93:

    (set -- ~(N)*; (($# == 0)) || printf '%s' "$@") |
      xargs -r0 printf '<%s>n'
  • with fish:

    begin set -l files *; string join0 -- $files; end |
      xargs -r0 printf '<%s>n'
  • with bash:

      shopt -s nullglob
      set -- *
      (($# == 0)) || printf '%s' "$@"
    ) | xargs -r0 printf '<%s>n'

2023 Edit. Since version 9.0 of GNU coreutils (September 2021), GNU ls now has a --zero option that can be used in conjunction with xargs -r0:

ls --zero | xargs -r0 printf '<%s>n'
Answered By: Stéphane Chazelas

It is extremelly silly to try to parse the output of a command ls that is not designed to be parsed to feed a command which is not designed to deal with several characters (for example: new lines and {}) when the shell does that by itself:

set -- *; for f; do echo "<$f>"; done

set    -- *
for    f
do     ls "$f"

Or, in one command line:

$ set -- *; for f; do echo "<$f>"; done
<name with space>

Note that the output deals (and has n example as last filename) with new-lines perfectly fine.

Or, if the number of files makes the shell slow, use find:

$ find ./ -type f -exec echo '<{}>' ;
<./name with space>

Just mind that find process all dot-files and all sub-directories diferently than the shell.

Answered By: user232326
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