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aris melachroinos
aris melachroinos

Reputation: 180

variable within functions-Linux Shell Script

i am facing an issue with this simple code , where i call a function , but the variable $X of that function doesnt seem to pass outside, since the printout message is always "Your interface is ." ... Am i missing something?

#! /bin/bash
function choose
{
echo -e " Choose your interface:"
echo -e " 1) WLan0"
echo -e " 2) WLan0mon"
echo -e " Choose: "
read -e X
if [ "$X" = "1" ]
then
    X="wlan0"
elif [ "$X" = "2" ]
then
    X="wlan0mon"
fi
}


(choose)
echo -e "Your interface is $X."

Upvotes: 0

Views: 89

Answers (1)

Phil Grigsby
Phil Grigsby

Reputation: 271

Running (choose) using the parentheses executes it in a subshell.

From Advanced Bash-Scripting Guide:

Variables in a subshell are not visible outside the block of code in the subshell. They are not accessible to the parent process, to the shell that launched the subshell. These are, in effect, local variables.

Removing the parentheses around choose will make variable X visible to your echo.

Upvotes: 2

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