Get name of a variable as input and change the variable with that name

Question

I get the name of a variable from the script user as the first argument and I echo the value of said variable back to the console:

#!/bin/bash
variablename=$1
echo "The value of $variablename is: " ${!variablename}

This works great!

What I can't get to work is if I want to change that variable into the value of the second argument from the user. An example with a syntax error is:

#!/bin/bash
variablename=$1
echo "The value of $variablename is: " ${!variablename}
echo "I will now try to change the value into $2."
(!variablename}=$2
# The above line generates: {!variablename}=2: command not found

In other words: If it is possible, how do I get a variable name from the user and both read (already solved) and write to said variable?

Answer 1

I had a flash all of a sudden, minutes after asking for help, and I think I have a solution:

#!/bin/bash
variablename=$1
echo "The value of $variablename is: " ${!variablename}
echo "I will now try to change the value into $2."
eval "$variablename=$2"
echo "Success! $variablename now has the value ${!variablename}!"

It works. Is there a better way?

Answer 2

Your solution will work but the variable will only have that value within your script. It won't affect the "calling" application's variable.

When you run a script or program it spawns a new shell with a copy of the environment. This is by design so your script doesn't modify things like PATH.

If you want your script to be able to modify environment variables for the calling shell you need to either "source scriptname" or ". scriptname". This is how .bashrc works.