Shell Programming and Scripting

I have a script which I call and pass a text string to it. This string is then is assigned to a variable in the script. I then call another script and pass that variable to the second script, but when I do, the quotes are lost and the second script gets a total of three variables 'my', 'lovely' and 'string' instead of one whole "my lovely string".
script1.sh

Code:

#!/bin/ksh
var1=$1
echo var1 is $var1
./script2.sh \"$var1\"

script2.sh

Code:

#!/bin/ksh

if [ ! "$1" = "" ]
then
        echo var1 is $1
fi

The whole sequence when run in debug mode looks like this:
Note the extra single quotes in when script2.sh is called. So my question is is there a way to preserve my original string as a whole while passing it on to script2.sh? An alternative would be to replace spaces with underscores, but that would be lame and I'd like to be able to support spaces in the string. Any pointers are appreciated.

Why do you escape double quotes in ./script2.sh \"$var1\" ? Without backslashes everything should work. But the idiomatic way to pass shell script arguments to another command is:

Code:

./script2 "$@"

And extra single quotes are the result of working 'set -x'.

Those aren't extra single quotes. That's just the shell tracing making explicit what's going on.

If you want to pass the value of $var1 with literal quotes to the second script, you need to do the following:

Code:

./script2.sh \""$var1"\"

Aside from the backslash-escaped set of quotes that you intend to pass literally, you must use an unescaped pair to protect $var1 from subsequent field splitting and file globbing.

Regards,
Alister

Thanks, this combination of quotes/escapes worked!