Propagate all arguments in a Bash shell script

Question

I am writing a very simple script that calls another script, and I need to propagate the parameters from my current script to the script I am executing.

For instance, my script name is foo.sh and calls bar.sh.

foo.sh:

bar $1 $2 $3 $4

How can I do this without explicitly specifying each parameter?

Answer 1

Use "$@" instead of plain $@ if you actually wish your parameters to be passed the same.

Observe:

$ cat no_quotes.sh
#!/bin/bash
echo_args.sh $@

$ cat quotes.sh
#!/bin/bash
echo_args.sh "$@"

$ cat echo_args.sh
#!/bin/bash
echo Received: $1
echo Received: $2
echo Received: $3
echo Received: $4

$ ./no_quotes.sh first second
Received: first
Received: second
Received:
Received:

$ ./no_quotes.sh "one quoted arg"
Received: one
Received: quoted
Received: arg
Received:

$ ./quotes.sh first second
Received: first
Received: second
Received:
Received:

$ ./quotes.sh "one quoted arg"
Received: one quoted arg
Received:
Received:
Received:

Answer 2

For bash and other Bourne-like shells:

bar "$@"

Answer 3

Use "$@" (works for all POSIX compatibles).

[...] , bash features the "$@" variable, which expands to all command-line parameters separated by spaces.

From Bash by example.

Answer 4

I realize this has been well answered but here's a comparison between "$@" $@ "$*" and $*

Contents of test script:

# cat ./test.sh
#!/usr/bin/env bash
echo "================================="

echo "Quoted DOLLAR-AT"
for ARG in "$@"; do
    echo $ARG
done

echo "================================="

echo "NOT Quoted DOLLAR-AT"
for ARG in $@; do
    echo $ARG
done

echo "================================="

echo "Quoted DOLLAR-STAR"
for ARG in "$*"; do
    echo $ARG
done

echo "================================="

echo "NOT Quoted DOLLAR-STAR"
for ARG in $*; do
    echo $ARG
done

echo "================================="

Now, run the test script with various arguments:

# ./test.sh  "arg with space one" "arg2" arg3
=================================
Quoted DOLLAR-AT
arg with space one
arg2
arg3
=================================
NOT Quoted DOLLAR-AT
arg
with
space
one
arg2
arg3
=================================
Quoted DOLLAR-STAR
arg with space one arg2 arg3
=================================
NOT Quoted DOLLAR-STAR
arg
with
space
one
arg2
arg3
=================================

Answer 5

A lot answers here recommends $@ or $* with and without quotes, however none seems to explain what these really do and why you should that way. So let me steal this excellent summary from this answer:

+--------+---------------------------+
| Syntax |      Effective result     |
+--------+---------------------------+
|   $*   |     $1 $2 $3 ... ${N}     |
+--------+---------------------------+
|   $@   |     $1 $2 $3 ... ${N}     |
+--------+---------------------------+
|  "$*"  |    "$1c$2c$3c...c${N}"    |
+--------+---------------------------+
|  "$@"  | "$1" "$2" "$3" ... "${N}" |
+--------+---------------------------+

Notice that quotes makes all the difference and without them both have identical behavior.

For my purpose, I needed to pass parameters from one script to another as-is and for that the best option is:

# file: parent.sh
# we have some params passed to parent.sh 
# which we will like to pass on to child.sh as-is

./child.sh $*

Notice no quotes and $@ should work as well in above situation.

Answer 6

#!/usr/bin/env bash
while [ "$1" != "" ]; do
  echo "Received: ${1}" && shift;
done;

Just thought this may be a bit more useful when trying to test how args come into your script

Answer 7

If you include $@ in a quoted string with other characters the behavior is very odd when there are multiple arguments, only the first argument is included inside the quotes.

Example:

#!/bin/bash
set -x
bash -c "true foo $@"

Yields:

$ bash test.sh bar baz
+ bash -c 'true foo bar' baz

But assigning to a different variable first:

#!/bin/bash
set -x
args="$@"
bash -c "true foo $args"

Yields:

$ bash test.sh bar baz
+ args='bar baz'
+ bash -c 'true foo bar baz'

Answer 8

My SUN Unix has a lot of limitations, even "$@" was not interpreted as desired. My workaround is ${@}. For example,

#!/bin/ksh
find ./ -type f | xargs grep "${@}"

By the way, I had to have this particular script because my Unix also does not support grep -r

Answer 9

Sometimes you want to pass all your arguments, but preceded by a flag (e.g. --flag)

$ bar --flag "$1" --flag "$2" --flag "$3"

You can do this in the following way:

$ bar $(printf -- ' --flag "%s"' "$@")

note: to avoid extra field splitting, you must quote %s and $@, and to avoid having a single string, you cannot quote the subshell of printf.

Answer 10

bar "$@" will be equivalent to bar "$1" "$2" "$3" "$4"

Notice that the quotation marks are important!

"$@", $@, "$*" or $* will each behave slightly different regarding escaping and concatenation as described in this stackoverflow answer.

One closely related use case is passing all given arguments inside an argument like this:

bash -c "bar \"$1\" \"$2\" \"$3\" \"$4\"".

I use a variation of @kvantour's answer to achieve this:

bash -c "bar $(printf -- '"%s" ' "$@")"

Answer 11

Works fine, except if you have spaces or escaped characters. I don't find the way to capture arguments in this case and send to a ssh inside of script.

This could be useful but is so ugly

_command_opts=$( echo "$@" | awk -F\- 'BEGIN { OFS=" -" } { for (i=2;i<=NF;i++) { gsub(/^[a-z] /,"&@",$i) ; gsub(/ $/,"",$i );gsub (/$/,"@",$i) }; print $0 }' | tr '@' \' )

Answer 12

"${array[@]}" is the right way for passing any array in bash. I want to provide a full cheat sheet: how to prepare arguments, bypass and process them.

pre.sh -> foo.sh -> bar.sh.

#!/bin/bash

args=("--a=b c" "--e=f g")
args+=("--q=w e" "--a=s \"'d'\"")

./foo.sh "${args[@]}"

#!/bin/bash

./bar.sh "$@"

#!/bin/bash

echo $1
echo $2
echo $3
echo $4

result:

--a=b c
--e=f g
--q=w e
--a=s "'d'"