Like to store all command-line arguments to a bash script into a single variable

Question

Let's say I have a bash script called foo.sh.

I'd like to call it like this

foo.sh Here is a bunch of stuff on the command-line

and I'd like it to store all of that text into a single variable and print it out.

So my output would be:

Here is a bunch of stuff on the command-line

How would I do this?

Answer 1

echo "$*"

would do what you want, namely printing out the entire command-line arguments, separated by a space (or, technically, whatever the value of $IFS is). If you wanted to store it into a variable, you could do

thevar="$*"

If that doesn't answer your question well enough, I'm not sure what else to say...

Answer 2

Have a look at the $* variable. It combines all command line arguments into one.

echo "$*"

This should do what you want.

More info here.

Answer 3

The command echo will take all arguments and print them out:

echo Here is a bunch of stuff on the command-line

You should quote the arguments if you have special characters.

Answer 4

If you want to avoid having $IFS involved, use $@ (or don't enclose $* in quotes)

$ cat atsplat
IFS="_"
echo "     at: $@"
echo "  splat: $*"
echo "noquote: "$*

$ ./atsplat this is a test
     at: this is a test
  splat: this_is_a_test
noquote: this is a test

The IFS behavior follows variable assignments, too.

$ cat atsplat2
IFS="_"
atvar=$@
splatvar=$*
echo "     at: $atvar"
echo "  splat: $splatvar"
echo "noquote: "$splatvar

$ ./atsplat2 this is a test
     at: this is a test
  splat: this_is_a_test
noquote: this is a test

Note that if the assignment to $IFS were made after the assignment of $splatvar, then all the outputs would be the same ($IFS would have no effect in the "atsplat2" example).