How do I set an environment variable on the command line and have it appear in commands?

If I run

export TEST=foo
echo $TEST

It outputs foo.

If I run

TEST=foo echo $TEST

It does not. How can I get this functionality without using export or a script?

Asked By: ashleysmithgpu

||

You can get this working by using:

TEST=foo && echo $TEST
Answered By: pradeepchhetri

This is because the shell expands the variable in the command line before it actually runs the command and at that time the variable doesn’t exist. If you use

TEST=foo; echo $TEST

it will work.

export will make the variable appear in the environment of subsequently executed commands (for on how this works in bash see help export). If you only need the variable to appear in the environment of one command, use what you have tried, i.e.:

TEST=foo your-application

The shell syntax describes this as being functionally equivalent to:

export TEST=foo
your-application
unset TEST

See the specification for details.

Interesting part is, that the export command switches the export flag for the variable name. Thus if you do:

unset TEST
export TEST
TEST="foo"

TEST will be exported even though it was not defined at the time when it was exported. However further unset should remove the export attribute from it.

Answered By: peterph

I suspect you want to have shell variables to have a limited scope, rather than environment variables. Environment variables are a list of strings passed to commands when they are executed.

In

var=value echo whatever

You’re passing the var=value string to the environment that echo receives. However, echo doesn’t do anything with its environment list┬╣ and anyway in most shells, echo is built in and therefore not executed.

If you had written

var=value sh -c 'echo "$var"'

That would have been another matter. Here, we’re passing var=value to the sh command, and sh does happen to use its environment. Shells convert each of the variables they receive from their environment to a shell variable, so the var environment variable sh receives will be converted to a $var variable, and when it expands it in that echo command line, that will become echo value. Because the environment is by default inherited, echo will also receive var=value in its environment (or would if it were executed), but again, echo doesn’t care about the environment.

Now, if as I suspect, what you want is to limit the scope of shell variables, there are several possible approaches.

Portably (Bourne and POSIX):

(var=value; echo "1: $var"); echo "2: $var"

The (…) above starts a sub-shell (a new shell process in most shells), so any variable declared there will only affect that sub-shell, so I’d expect the code above to output "1: value" and "2: " or "2: whatever-var-was-set-to-before".

With most Bourne-like shells (see List of shells that support `local` keyword for defining local variables), you can use functions and the "local" builtin:

f() {
  local var
  var=value
  echo "1: $var"
}
f
echo "2: $var"

With zsh, you can use inline functions:

(){ local var=value; echo "1: $var"; }; echo "2: $var"

or:

function { local var=value; echo "1: $var"; }; echo "2: $var"

With bash and zsh (but not ash, pdksh or AT&T ksh), this trick also works:

var=value eval 'echo "1: $var"'; echo "2: $var"

A variant that works in a few more shells (dash, mksh, yash) but not zsh (unless in sh/ksh emulation):

var=value command eval 'echo "1: $var"'; echo "2: $var"

(using command in front of a special builtin (here eval) in POSIX shells removes their specialness (here that variables assignments in front of them remain in effect after they have returned))


┬╣ Stricktly speaking, that’s not completely true. Several implementations will care about the localisation environment variables (LANG, LOCPATH, LC_*…), the GNU implementation cares about the POSIXLY_CORRECT environment variable (compare env echo --version with POSIXLY_CORRECT=1 env echo --version on a GNU system).

Answered By: Stéphane Chazelas

You’re doing it correctly, but the bash syntax is easy to misinterpret: you could think that echo $TEST causes echo to fetch TEST env var then print it, it does not. So given

export TEST=123

then

TEST=456 echo $TEST

involves the following sequence:

  1. The shell parses the whole command line and executes all variable substitutions, so the command line becomes

    TEST=456 echo 123
    
  2. It creates the temp vars set before the command, so it saves the current value of TEST and overwrites it with 456; the command line is now

    echo 123
    
  3. It executes the remaining command, which in this case prints 123 to stdout (so shell command that remains didn’t even use the temp value of TEST)

  4. It restores the value of TEST

Use printenv instead, as it does not involve variable substitution:

>> export TEST=123
>> printenv TEST
123
>> TEST=456 printenv TEST
456
>> printenv TEST && TEST=456 printenv TEST && TEST=789 printenv TEST && printenv TEST
123
456
789
123
>>
Answered By: Oliver

As you’ve already discovered, in Bash, TEST=456 echo $TEST doesn’t work like you expected, because you are accessing the variable on the same line that’s currently being interpreted by the shell. But rest assured, the variable has been set for the duration of the current command. To verify that, you could do something like this:

show () { echo "${!1}" ;}
TEST=456 show TEST
Answered By: Pourko
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