Reading an argument list with a glob from a file

Question

I am trying to read arguments of ll command from file. I have file args.txt with content some?. I have files with names some1, some2 and some3 and some other files. when I execute cat args.txt | ll I get the same result as when I execute ll.

Can someone explain me why is that and how can I achieve the desired result. Thanks in advance.

Answer 1

^{Note: I'm assuming that ll is an alias for ls -l or some variation.
As Charles Duffy points out, (by default) only interactive shells know about aliases, and scripts as well as tools such as xargs are unaware of them.}

What you send through a pipeline to a command is received via its stdin stream, which is distinct from a command's arguments (options such as -l and operands such as some?).

Therefore, an explicit transformation is needed to convert the contents of file args.txt to an argument you can pass to ls -l:

ls -l $(cat args.txt)   # In Bash you can simplify to: ls -l $(< args.txt)

Note that the command substitution ($(...)) is deliberately unquoted so as to ensure that globbing (filename expansion) is applied to the contents of args.txt, as desired.

Yours is a rare case where this technique - which is brittle - is actually needed.

To illustrate the brittleness: If your globbing pattern were "some file"?, for instance (you'd need the double quotes for the pattern to still be recognized as a single argument), the command wouldn't work anymore, because the " characters lose their syntactic function when they're the result of a command substitution (or variable expansion).

---

The standard utility for transforming stdin or file content into arguments is xargs. However, in your particular case it is not an option, because your file contains a glob (filename pattern) that must be expanded by the shell, but xargs only invokes external utilities, without involving a shell:

$ xargs ls -l < args.txt # !! Does NOT work as intended here.
ls: file?: No such file or directory

Filename file? was passed literally to ls (and no file actually named file? exists) - no globbing happened, because no shell was involved.

Answer 2

The following is somewhat overkill, but goes to pains to be correct and consistent regardless of shell configuration, and to work with all possible filenames (even those with whitespace):

# note f() ( ) instead of f() { }; this runs in a subshell, so its changes to IFS or
# shopt settings don't modify behavior of the larger shell.
globfiles() (
  set +f                      ## ensure that globbing is enabled
  shopt -u nullglob failglob  ## ensure that non-default options on how to handle failed
                              ## ...glob attempts are both disabled
  IFS=                        ## disable string-splitting
  while IFS= read -r -d '' filename; do
    printf '%s\0' $filename   ## unquoted use -> expand as glob
                              ## ...expands format string once per argument, hence once per
                              ## ...file found. (No string-splitting due to prior IFS=)
  done
)

...thereafter used as (if your input file is newline-delimited):

tr '\n' '\0' <args.txt | globfiles | xargs -0 ls -l --

...or (if your input file is NUL-delimited -- which is ideal, as this permits filenames containing literal newlines to be referenced):

globfiles <args.0sv | xargs -0 ls -l --