shell script, Passing string as argument?

antis · 09-06-2007, 04:07 PM

Hi,

I have a function in a ksh script to which I'm trying to pass a string as an argument. Like this:

Code:

function myFunct {
echo $1
}

VAR='this is a test string'

myFunct $VAR

The problem is that when I pass the variable $VAR to the function the argument is split into $1, $2, $3, $4, $5 which causes the "echo $1" to only print "this". This is not the case if I pass the string directly to the function, like this:

Code:

myFunct 'this is a test string'

This way "echo $1" prints the whole string.

How would I go about using the variable and get the function to treat it as one whole string? I've tried every qutoing way that I know of but can't get it to work.

gnashley · 09-06-2007, 04:11 PM

Doesn't this work? VAR="this is a test string"

antis · 09-06-2007, 04:23 PM

Quote:

Originally Posted by gnashley

Doesn't this work? VAR="this is a test string"

No, still just results in "this"...

marozsas · 09-06-2007, 04:58 PM

Protect the argument when passing it to your function:

Code:

myFunct "$VAR"

This is true for every *nix command, not just for your own functions/programs, e.g.

Code:

rm "a filename with spaces on it"

antis · 09-07-2007, 02:03 AM

Quote:

Originally Posted by marozsas

Protect the argument when passing it to your function:

Code:

myFunct "$VAR"

This is true for every *nix command, not just for your own functions/programs, e.g.

Code:

rm "a filename with spaces on it"

Oh my... that was simple.
Time to read up on the differences between hard and soft quotes I guess.
Thanks!