Bash: pass variable as a single parameter / shell quote parameter

Question

I'm writing a bash script which has to pass a variable to another program:

./program $variable

The problem is, it is absolutely necessary for $variable to be passed as a single parameter, which isn't the case if it contains whitespace.

variable=Hello World
./program $variable
-> program receives two arguments: 'Hello' 'World'

Quoting it doesn't do anything at all (well done, bash devs):

variable=Hello World
./program "$variable"
-> program receives: 'Hello' 'World'

Double quoting it does crazy stuff (well done, bash devs):

variable=Hello World
./program "'$variable'"
-> program receives: "'Hello" "World'"

Is there an easy way to do this? Heck, is there a way to do this at all?

Update: Okay, since the problem doesn't seem to be bash, here's some additional info. The program I'm passing arguments to is a python script. Without modifying the arguments in any way, I get

print sys.argv
-> ['/usr/bin/program', "'Hello", "World'"]

How can I fix that?

Edit: No, I haven't tried

variable="Hello World"

because I never declare $variable. It's not being declared inside my bash function and I'm not allowed to modify it.

Edit: Okay, I got it to work that way.

local temp="$variable"
./program "$temp"

I'd like to know why it works that way and not any other way, though.

the problem is in the "program" ; ./program "$variable" passes one argument which can contain spaces, what the program is doing? — Nahuel Fouilleul
– Nahuel Fouilleul, Commented Oct 22, 2012 at 8:34
@sarathi: No, since I actually never declare $variable. I just pass it on. — Aran-Fey
– Aran-Fey, Commented Oct 22, 2012 at 8:41
How is variable declared, and where? The bug might be there. — cdarke
– cdarke, Commented Oct 22, 2012 at 9:02

Vijay · Accepted Answer · 2012-10-22 08:40:24Z

1

did you try with var="hello world"?

i tried this in my solaris box.

> setenv var "hello world"
> cat temp.sh
#!/bin/sh

echo $1
echo $2
> ./temp.sh "$var"
hello world

>

as you can see the $2 is not printed.$var is considered as only one argument.

answered Oct 22, 2012 at 8:40

Vijay

67.7k94 gold badges238 silver badges327 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

morgents · Accepted Answer · 2012-10-22 08:52:20Z

1

When you call your script pass the arguments within quotes.

Example script:

#!/bin/bash
for arg in "$@"; do
 echo "arg: $1";
 shift;
done

When you call it with:

./program "parameter with multiple words" parameter2 parameter3 "another parameter"

The output should be:

arg: parameter with multiple words
arg: parameter2
arg: parameter3
arg: another parameter

edited Oct 22, 2012 at 8:52

answered Oct 22, 2012 at 8:40

morgents

5055 silver badges11 bronze badges

Comments

Daniel Voina · Accepted Answer · 2012-10-22 08:58:51Z

1

Have a look on http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html .

The problem is that the expansion of variables is done before of the command line parameters hence your behavior.

You might work it arround with setting IFS to something weird as

IFS='###' V='foo bar baz'; ./program $V

answered Oct 22, 2012 at 8:58

Daniel Voina

3,2501 gold badge30 silver badges33 bronze badges

1 Comment

Aran-Fey Over a year ago

Thanks for the link and explanation. Looking at that guide, it looks like I'll never get to like bash.

Nahuel Fouilleul · Accepted Answer · 2012-10-22 08:40:21Z

0

The problem seems to be inside the "program"

variable="Hello World"    # quotes are needed because of the space
./program "$variable"     # here quotes again

and inside the program

echo "program received $# arguments:" 
i=1
for arg in "$@"      # again quotes needed
do echo "arg $((i=i+1)): '$arg'"    # and again
done

answered Oct 22, 2012 at 8:40

Nahuel Fouilleul

19.5k2 gold badges36 silver badges37 bronze badges

1 Comment

yesennes Over a year ago

Great! What can you do if the variable contains spaces and double quotes?

ShankarG · Accepted Answer · 2012-10-22 08:32:39Z

0

This is almost certainly a problem in the way you are reading the variable in your program.
For instance suppose this is your script (just one line for testing):

echo "$1"

Let's call it echo.sh. If you run echo.sh "test best", you will get test best.

But if your program says

echo $1

you might get behaviour like what you're seeing.

answered Oct 22, 2012 at 8:32

ShankarG

1,16512 silver badges29 bronze badges

Collectives™ on Stack Overflow

Bash: pass variable as a single parameter / shell quote parameter

5 Answers 5

Comments

Comments

1 Comment

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

Comments

1 Comment

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related