How should I export http_proxy variable?

Question

I'm trying to write a simple script that will set proxy settings. Actually I just need to export http_proxy ftp_proxy https_proxy ... variables with export command. But it's not working when I run it manually from the shell because export affect only current shell and subshells, but no others. Also I don't want to call it from .bashrc because it's not my default proxy settings.

So how should I export http_proxy variable to make effect globally?

Answer 1

Back in the day I was also sick of setting and then unsetting the proxy settings after my work was done. I always wished if there was a command simple command to do the set and unset function for me.

Then I figured that if I create a new function in my .bashrc I can call it from the command line by using the bash-tab-completion. Saves even more time.

This is what I did:

$ vi ~/.bashrc
function setproxy() {
    export {http,https,ftp}_proxy='http://proxy-serv:8080'
}

function unsetproxy() {
    unset {http,https,ftp}_proxy
}

$ . ~/.bashrc

Now I just do:

$ setproxy

or

$ setp<TAB> and <ENTER>

and it sets the proxy for me. Hope this helps.

Answer 2

Instead of doing this in a script, make this a function. You can declare this function in your .bashrc:

function set_custom_proxy() {
  export http_proxy='http://myproxy:3128'
}

Then run this in the current shell:

echo $http_proxy
set_custom_proxy
echo $http_proxy

It works as a variable modification in a function is not local to the function.

EDIT:

FYI: to use a local variable in a function, you need to use the local keyword:

atest="Hello World"
btest="Hello World2"
function my_func() {
  local atest;
  atest="Hello World3"
  btest="Hello World4"
  echo $atest
  echo $btest
}
echo $atest
echo $btest
my_func
echo $atest
echo $btest

Answer 3

Since you can't access .bashrc, you can use source command which will run in the current shell's context and all the variables you set will be available.

source ./script