Bash function, export function argument as environment variable

Question

I wrote a bash function to export an environment variable. First function argument is a variable name, second is a variable value. I want to echo it to show what value was exported:

#!/bin/bash

env_var_export()
{
    export $1=$2

echo ""
echo "  export $1=$$1"
echo ""
}

env_var_export var defaultVal456

I mean, echo should print: export var=defaultVal456. Any help? I know I can do this:

echo ""
echo "  export $1=$2"
echo ""

but its not the solution to my problem.

FatalError · Accepted Answer · 2013-04-05 15:02:06Z

3

$$ is a special variable that expands to the shell's pid, and that's what is going to be evaluated in your echo. You should instead use an indirect reference like this:

echo ""
echo "  export $1=${!1}"
echo ""

This syntax will take the variable named in $1 and then lookup the value based on that name (i.e. indirection).

answered Apr 5, 2013 at 15:02

FatalError

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2 Comments

Daniel Cazares Over a year ago

Took me forever to find this reply! Thank you for making this answer 3 years ago.

FatalError Over a year ago

Glad it helped! :)

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