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Old 09-06-2007, 04:07 PM   #1
antis
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shell script, Passing string as argument?


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.
 
Old 09-06-2007, 04:11 PM   #2
gnashley
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Doesn't this work? VAR="this is a test string"
 
Old 09-06-2007, 04:23 PM   #3
antis
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Quote:
Originally Posted by gnashley View Post
Doesn't this work? VAR="this is a test string"
No, still just results in "this"...
 
Old 09-06-2007, 04:58 PM   #4
marozsas
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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"
 
Old 09-07-2007, 02:03 AM   #5
antis
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Registered: Nov 2004
Location: sweden
Distribution: Arch
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Quote:
Originally Posted by marozsas View Post
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!
 
  


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