Passing arguments with SPACES to bash script

theEditor876 · 06-28-2014, 06:04 AM

Hey guys!

Lets say i have a shell script looking like this:

Code:

#!/bin/bash
var1="wehavea spacehere"
var2="11.02.2013"

./dosomething.sh $var1 $var2

dosomething.sh:

Code:

#!/bin/bash

whatever=123

./someprogram $1 $2 $whatever

when i pass $var1 and $var2 to dosomething.sh, $var1 will get split up in 2 arguments!
calling dosomething.sh like this:

Code:

./dosomething.sh "$var1" "$var2"

does not work, either.

Can somebody tell me how to properly pass arguments with a space in between them?
Thank you!

mark_alfred · 06-28-2014, 07:30 AM

I'm not sure, but perhaps a backward slash would work.

Code:

#!/bin/bash
var1="wehavea\ spacehere"
var2="11.02.2013"

./dosomething.sh $var1 $var2

rknichols · 06-28-2014, 07:41 AM

Quote:

Originally Posted by theEditor876

dosomething.sh:

Code:

#!/bin/bash

whatever=123

./someprogram $1 $2 $whatever

when i pass $var1 and $var2 to dosomething.sh, $var1 will get split up in 2 arguments!
calling dosomething.sh like this:

Code:

./dosomething.sh "$var1" "$var2"

does not work, either.

Can somebody tell me how to properly pass arguments with a space in between them?

You are passing the argument correctly there, but dosomething.sh isn't using it correctly. You need to quote the arguments there, too:

Code:

./someprogram "$1" "$2" $whatever