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 double quote variable expansion

I'm trying to pass a string containing spaces to an external program
(dialog) from a shell script.  Hopefully I can explain this well enough
so you don't need to know the 'dialog' program.  Basically, I present a
menu with 2 or 3 options.  The third option may or may not be available
depending on the value of $OPT3.  If it is not available, I don't want
to put it in the menu.  The problem I am running into is how to pass the
10th and 11th options to the external program.  See code below:


MSG3="Third option"

dialog --menu "Menu" 10 60 5 \
         "One" "First option" \
         "Two" "Second option" \
         ${OPT3+`echo "Three" \""$MSG3"\"`}

echo done

When I run with -x, I get the following:
$ sh -x /tmp/
+ MSG3=Third option
++ echo Three '"Third option"'
+ dialog --menu Menu 10 60 5 One 'First option' Two 'Second option'
Three '"Third' 'option"'

Error: Expected 2 arguments, found only 1.
Use --help to list options.

+ echo done

I understand why the error occurs (the output from 'dialog').  I would
like to understand how I can cause the script to execute dialog like so:

+ dialog --menu Menu 10 60 5 One 'First option' Two 'Second option'
Three 'Third option'

Can someone please enlighten me on this?


 Sat, 04 Oct 2008 05:06:22 GMT   
 double quote variable expansion
On 2006-04-17, Jason Roscoe wrote:

         ${OPT3+"Three" "$MSG3"}

- Show quoted text -

   Chris F.A. Johnson, author   |    <>
   Shell Scripting Recipes:     |  My code in this post, if any,
   A Problem-Solution Approach  |          is released under the
   2005, Apress                 |     GNU General Public Licence

 Sat, 04 Oct 2008 05:21:44 GMT   
 double quote variable expansion
That works.  Thanks Chris.  That must be the only thing I didn't already

 Sat, 04 Oct 2008 05:31:31 GMT   
   [ 3 post ] 

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