Using file contents as command line arguements in BASH

Question

I'd like to know how to use the contents of a file as command line arguments, but am struggling with syntax.

Say I've got the following:

# cat > arglist
src/file1 dst/file1
src/file2 dst/file2
src/file3 dst/file3

How can I use the contents of each line in the arglist file as arguments to say, a cp command?

Answer 1

You can use pipe (|) :

cat file | echo

or input redirection (<)

cat < file

or xargs

xargs sh -c 'emacs "$@" < /dev/tty' emacs

Then you may use awk to get arguments:

cat file | awk '{ print $1; }'

Hope this helps..

Answer 2

Using read (this does assume that any spaces in the filenames in arglist are escaped):

while read src dst; do cp "$src" "$dst"; done < argslist

If the arguments in the file are in the right order and filenames with spaces are quoted, then this will also work:

while read args; do cp $args; done < argslist

Answer 3

the '-n' option for xargs specifies how many arguments to use per command :

$ xargs -n2 < arglist echo cp

cp src/file1 dst/file1
cp src/file2 dst/file2
cp src/file3 dst/file3

Answer 4

if your purpose is just to cp those files in thelist

$ awk '{cmd="cp "$0;system(cmd)}' file

Answer 5

Use for loop with IFS(Internal Field Separator) set to new line

OldIFS=$IFS # Save IFS
$IFS=$'\n' # Set IFS to new line
for EachLine in `cat arglist"`; do
Command="cp $Each"
`$Command`;
done
IFS=$OldIFS