Passing arguments with spaces via list in bash

Question

I am interesting in calling some function using dynamically generated arguments in a bash script.

However, if some of the arguments had spaces, it does not seem to work.

Let's have this test script:

#!/bin/bash
# file uu.sh
[[ -z $1 ]] || echo '$1' $1
[[ -z $2 ]] || echo '$2' $2
[[ -z $3 ]] || echo '$3' $3
[[ -z $4 ]] || echo '$4' $4
[[ -z $5 ]] || echo '$5' $5
[[ -z $6 ]] || echo '$6' $6
[[ -z $7 ]] || echo '$7' $7
[[ -z $8 ]] || echo '$8' $8
[[ -z $9 ]] || echo '$9' $9
[[ -z $0 ]] || echo '$0' $0

Any my script:

#!/bin/bash
# file vv.sh
ARR=(-x)
ARR+=($1)
ARR+=("$1")
ARR+=("'$1'")
ARR+=("\"$1\"")
bash uu.sh ${ARR[*]}
echo
bash uu.sh "${ARR[*]}"

When calling bash vv.sh "a b" I get the following result:

$2 -x
$2 a
$3 b
$4 a
$5 b
$6 'a
$7 b'
$8 "a
$9 b"
$0 uu.sh

$1 -x a b a b 'a b' "a b"
$0 uu.sh

I am expecting a way to pass the variables to uu.sh such as the result would be:

$1 -x
$2 a b
$0 uu.sh

(Which I can get directly by calling bash uu.sh -x a\ b, or bash uu.sh -x "a b", or bash uu.sh -x 'a b')

You want "$1" and not $1. Also you could use "$@" for all args.. — clt60
– clt60, Commented Apr 6, 2017 at 17:34
The problem is not in uu.sh (and, any how, I tested with the quotation marks, and I still see the same problem) — Carlos Eugenio Thompson Pinzón
– Carlos Eugenio Thompson Pinzón, Commented Apr 6, 2017 at 17:38

Charles Duffy · Accepted Answer · 2017-04-06 17:48:54Z

5

You should have vv.sh like this to be able to pass all the arguments after -x in the 2nd argument to uu.sh:

#!/bin/bash
# file vv.sh
ARR=(-x)    # initialize array with "-x" as first element
ARR+=("$1") # append 1st argument in second element of array
bash uu.sh "${ARR[@]}" # call uu.sh with quoted and [@]

Then use it as (see quoted arguments):

bash vv.sh "a b"

...which will emit as output:

$1 -x
$2 a b
$0 uu.sh

edited Apr 6, 2017 at 17:48

Charles Duffy

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answered Apr 6, 2017 at 17:37

anubhava

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2 Comments

Carlos Eugenio Thompson Pinzón Over a year ago

I'm not interested in passing all arguments from vv.sh, just the first argument (or just the third, or some other variable)

Carlos Eugenio Thompson Pinzón Over a year ago

However the @ (and quotes) did the trick! bash uu.sh "${ARR[@]}"

clt60 · Accepted Answer · 2017-04-06 17:44:49Z

Here is a difference between $* and $@ and also, you should always quote the variables (unless you know why do not want) :)

#uu.sh
[[ -z "$1" ]] || echo '$1' "$1"
[[ -z "$2" ]] || echo '$2' "$2"
[[ -z "$3" ]] || echo '$3' "$3"
[[ -z "$4" ]] || echo '$4' "$4"
[[ -z "$5" ]] || echo '$5' "$5"
[[ -z "$6" ]] || echo '$6' "$6"
[[ -z "$7" ]] || echo '$7' "$7"
[[ -z "$8" ]] || echo '$8' "$8"
[[ -z "$9" ]] || echo '$9' "$9"
[[ -z "$0" ]] || echo '$0' "$0"

and the vv.sh

ARR=(-x)
ARR+=($1)
ARR+=("$1")
ARR+=("'$1'")
ARR+=("\"$1\"")

echo 'using quoted ARR[@]'
printf "=%s=\n" "${ARR[@]}"

echo 'using unquoted ARR[@]'
printf "=%s=\n" ${ARR[@]}

echo 'using quoted ARR[*]'
printf "=%s=\n" "${ARR[*]}"

echo 'using unquoted ARR[*]'
printf "=%s=\n" ${ARR[*]}

echo "running UU"
bash uu.sh "${ARR[@]}"

output:

using quoted ARR[@]
=-x=
=A=
=B=
=A B=
='A B'=
="A B"=
using unquoted ARR[@]
=-x=
=A=
=B=
=A=
=B=
='A=
=B'=
="A=
=B"=
using quoted ARR[*]
=-x A B A B 'A B' "A B"=
using unquoted ARR[*]
=-x=
=A=
=B=
=A=
=B=
='A=
=B'=
="A=
=B"=
running UU
$1 -x
$2 A
$3 B
$4 A B
$5 'A B'
$6 "A B"
$0 uu.sh

ilkkachu · Accepted Answer · 2017-04-06 17:47:10Z

3

"${array[*]}" and "$*" expand to a single word with the array members or positional parameters joined with spaces. That's almost never useful. Instead, use "${array[@]}" or "$@", which expand all the array members to separate words. (See Arrays and Special parameters in the manual.)

So when you do bash uu.sh "${ARR[*]}" the command is passed one string as parameter, and with bash uu.sh ${ARR[*]} there are no quotes so all contents of the array get split on whitespace.

answered Apr 6, 2017 at 17:47

ilkkachu

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