Pass environment variable to bash script, called from within a function

I mean to have:

1. A bash script scr.sh that takes positional parameters

   #!/bin/bash
   echo "Params = $@"
   echo REMOTE_SERVER=${REMOTE_SERVER}

2. A bash function f defined in another script scr2.sh

   #!/bin/bash
   f() { REMOTE_SERVER=s001 scr.sh "${@}"
   }

I would first

$ source scr2.sh

and then have f available for calling at the command line, but not leaving a trace of what I did with REMOTE_SERVER. For instance, I want

$ f par1 par2
par1 par2
s001
$ echo REMOTE_SERVER=${REMOTE_SERVER}
REMOTE_SERVER=

(actually, if REMOTE_SERVER was set before using f, I want it to keep that value). I couldn't attain this last objective. I always end up with REMOTE_SERVER set. I tried using multiline commands separated with a semicolon, enclosing commands inside f with parenthesis, but it didn't work.

How can I do that?

2 Answers

If you want to set a variable only for a command, prefix the assignment to the command:

REMOTE_SERVER=s001 scr.sh "${@}"

Alternately, use a subshell for the function (then variable assignments won't affect the parent shell). You can create a subshell by wrapping commands in ( ... ) (parentheses), or using parentheses instead of braces for the function body. For example, with:

foo () ( REMOTE_SERVER=s001
)
bar () { REMOTE_SERVER=s001
}
foo; echo "REMOTE_SERVER=$REMOTE_SERVER"; bar; echo "REMOTE_SERVER=$REMOTE_SERVER"; 

My output would be:

REMOTE_SERVER=
REMOTE_SERVER=s001

Using

#!/bin/bash
f() { ( REMOTE_SERVER=s001 scr.sh "${@}" )
}

works fine. This is likely equivalent to the accepted answer, with the extra flexibility of including parts of the code inside the local shell and parts outside.