How to quote parameters when calling a shell script from within another script using a variable

Question

I have two shell scripts. One script dynamically creates the call to the second script, based on the parameter it received, and then executes the call.

My problem is that the parameters the first script gets may contain spaces, so I must quote the parameter in the call to script2.

This is an example to the problem:

script1.sh:

#!/bin/sh
param=$1
command="./script2.sh \"$param\""
echo $command
$command

script2.sh:

#!/bin/sh
param=$1
echo "the value of param is $param"

When I run:

./script1.sh "value with spaces"

I get:

./script2.sh "value with spaces"
the value of param is "value

Which is of course not what I need.

What is wrong here??

TIA.

EDIT :

I found the solution thanks to the useful link in tripleee's comment. Here it is in case it helps anybody.

In short, in order to solve this, one should use an array for the arguments.

script1.sh:

#!/bin/sh
param=$1

args=("$param")
script_name="./script2.sh"
echo $script_name "${args[@]}"
$script_name "${args[@]}"

tripleee · Accepted Answer · 2011-12-06 12:21:46Z

1

Use "$@" to refer to all command-line parameters with quoting intact.

answered Dec 6, 2011 at 12:21

tripleee

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2 Comments

Dikla Over a year ago

where should I use it? I tried param="$@" but it didn't solve the problem

tripleee Over a year ago

Use it directly, assigning to a named variable is problematic. sh -x ./script2 "$@" if you want to echo the command as well - this is problematic for related reasons. See also mywiki.wooledge.org/BashFAQ/050

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