How do I use the lines of a file as arguments of a command?

Question

If your shell is bash (amongst others), a shortcut for $(cat afile) is $(< afile), so you'd write:

mycommand "$(< file.txt)"

Documented in the bash man page in the 'Command Substitution' section.

Alterately, have your command read from stdin, so: mycommand < file.txt

As already mentioned, you can use the backticks or $(cat filename).

What was not mentioned, and I think is important to note, is that you must remember that the shell will break apart the contents of that file according to whitespace, giving each "word" it finds to your command as an argument. And while you may be able to enclose a command-line argument in quotes so that it can contain whitespace, escape sequences, etc., reading from the file will not do the same thing. For example, if your file contains:

a "b c" d

the arguments you will get are:

a
"b
c"
d

If you want to pull each line as an argument, use the while/read/do construct:

while read i ; do command_name $i ; done < filename

command `< file`

will pass file contents to the command on stdin, but will strip newlines, meaning you couldn't iterate over each line individually. For that you could write a script with a 'for' loop:

for line in `cat input_file`; do some_command "$line"; done

Or (the multi-line variant):

for line in `cat input_file`
do
    some_command "$line"
done

Or (multi-line variant with $() instead of ``):

for line in $(cat input_file)
do
    some_command "$line"
done

References:

1. For loop syntax: https://www.cyberciti.biz/faq/bash-for-loop/

You do that using backticks:

echo World > file.txt
echo Hello `cat file.txt`

If you want to do this in a robust way that works for every possible command line argument (values with spaces, values with newlines, values with literal quote characters, non-printable values, values with glob characters, etc), it gets a bit more interesting.

---

To write to a file, given an array of arguments:

printf '%s\0' "${arguments[@]}" >file

...replace with "argument one", "argument two", etc. as appropriate.

---

To read from that file and use its contents (in bash, ksh93, or another recent shell with arrays):

declare -a args=()
while IFS='' read -r -d '' item; do
  args+=( "$item" )
done <file
run_your_command "${args[@]}"

---

To read from that file and use its contents (in a shell without arrays; note that this will overwrite your local command-line argument list, and is thus best done inside of a function, such that you're overwriting the function's arguments and not the global list):

set --
while IFS='' read -r -d '' item; do
  set -- "$@" "$item"
done <file
run_your_command "$@"

Note that -d (allowing a different end-of-line delimiter to be used) is a non-POSIX extension, and a shell without arrays may also not support it. Should that be the case, you may need to use a non-shell language to transform the NUL-delimited content into an eval-safe form:

quoted_list() {
  ## Works with either Python 2.x or 3.x
  python -c '
import sys, pipes, shlex
quote = pipes.quote if hasattr(pipes, "quote") else shlex.quote
print(" ".join([quote(s) for s in sys.stdin.read().split("\0")][:-1]))
  '
}

eval "set -- $(quoted_list <file)"
run_your_command "$@"

If all you need to do is to turn file arguments.txt with contents

arg1
arg2
argN

into my_command arg1 arg2 argN then you can simply use xargs:

xargs -a arguments.txt my_command

You can put additional static arguments in the xargs call, like xargs -a arguments.txt my_command staticArg which will call my_command staticArg arg1 arg2 argN

Here's how I pass contents of a file as an argument to a command:

./foo --bar "$(cat ./bar.txt)"