How to pass an argument with space from bash script to bash script?

Question

I have a script the manipulates data, creates arguments and sends them to a second script. One of the arguments contains a space.

script1.sh:

args=()
args+=("A")
args+=("1 2")
args+=("B")
. script2.sh ${args[@]}

script2.sh:

for f in "$@"
do
  echo f=$f
done

I'd like to get "1 2" as single argument, but receive them separately:

f=A
f=1
f=2
f=B

I tried also converting the input in script2 to list in=($@) and iterating over it using for f in ${in[@]} but got the same result.

So, the problem might be in one of the following: building the list of arguments ; passing the built list ; parsing the input ; or iterating over the input.

How to pass an argument with space from bash script to bash script? Thanks.

* I'm using git-bash on Windows.

A combination of "${args[@]}" with for f in "$@" did the trick. Thanks. — AlikElzin-kilaka
– AlikElzin-kilaka, Commented Jul 30, 2015 at 14:55
Just for the record for f; do is the same as for f in "$@"; do. — Etan Reisner
– Etan Reisner, Commented Jul 30, 2015 at 15:15

arco444 · Accepted Answer · 2015-07-30 14:37:22Z

7

You just need to add quotes when passing the array as an argument:

args=()
args+=("A")
args+=("1 2")
args+=("B")
. script2.sh "${args[@]}"

answered Jul 30, 2015 at 14:37

arco444

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3 Comments

choroba Over a year ago

@AlikElzin-kilaka: On the receiving side, use quotes, too: in=("$@")

AlikElzin-kilaka Over a year ago

A combination of "${args[@]}" with for f in "$@" did the trick. Thanks.

arco444 Over a year ago

@AlikElzin-kilaka I was going to suggest the for f in "$@" but in your example it was already quoted so I was puzzled.

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