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bashunixenvironment-variablesenvironment
Niklas
Niklas

Reputation: 1360

Setting environment variable for one program call in bash using env

I am trying to invoke a shell command with a modified environment via the command env.

According to the manual

env HELLO='Hello World' echo $HELLO

should echo Hello World, but it doesn't. If I do

HELLO='Hello World' bash -c 'echo $HELLO'

it prints Hello World as expected (thanks to this answer for this info).

What am I missing here?

Upvotes: 63

Views: 51345

Answers (5)

William Entriken
William Entriken

Reputation: 39253

Here is an easier way to confirm shell is working as expected.

env A=42 env
env

The first command sets A to 42 and runs env. The second command also runs env. Compare the output of both.

Upvotes: 2

Sean
Sean

Reputation: 473

As nos noted, expansion of your variable passed to echo occurs before that variable gets set.

One alternative not yet mentioned in the earlier answers is to use:

$ a=abc eval 'echo $a'
abc

$ echo $a
<blank>

Note that you have to use the a=abc cmd syntax and not env a=abc cmd syntax; apparently env doesn't play nice with the built-in eval.

Upvotes: 2

nos
nos

Reputation: 229098

It's because in your first case, your current shell expands the $HELLO variable before running the commands. And there's no HELLO variable set in your current shell.

env HELLO='Hello World' echo $HELLO

will do this:

  • expand any variables given, in this case $HELLO
  • run env with the 3 arguments 'HELLO=Hello World', 'echo' and '' (an empty string, since there's no HELLO variable set in the current shell)
  • The env command will run and set the HELLO='Hello World' in its environment
  • env will run echo with the argument '' (an empty string)

As you see, the current shell expanded the $HELLO variable, which isn't set.

HELLO='Hello World' bash -c 'echo $HELLO'

will do this:

  • set the variable HELLO='Hello World for the following command
  • run bash with the 2 arguments '-c' and 'echo $HELLO'
  • since the last argument is enclosed in single quotes, nothing inside it is expanded
  • the new bash in turn will run the command echo $HELLO
  • To run echo $HELLO in the new bash sub-shell, bash first expands anything it can, $HELLO in this case, and the parent shell set that to Hello World for us.
  • The subshell runs echo 'Hello World'

If you tried to do e.g. this:

env HELLO='Hello World' echo '$HELLO'
  • The current shell would expand anything it can, which is nothing since $HELLO is enclosed in single quotes
  • run env with the 3 arguments 'HELLO=Hello World', 'echo' and '$HELLO'
  • The env command will run and set the HELLO='Hello World' in its environment
  • env will run echo with the argument '$HELLO'

In this case, there's no shell that will expand the $HELLO, so echo receives the string $HELLO and prints out that. Variable expansion is done by shells only.

Upvotes: 100

deFreitas
deFreitas

Reputation: 4448

This works and is good for me

$ MY_VAR='Hello' ANOTHER_VAR='World!!!' && echo "$MY_VAR $ANOTHER_VAR"
Hello World!!!

Upvotes: 3

Lev Levitsky
Lev Levitsky

Reputation: 65791

I think what happens is similar to this situation in which I was also puzzled.

In a nutshell, the variable expansion in the first case is done by the current shell which doesn't have $HELLO in its environment. In the second case, though, single quotes prevent the current shell from doing the variable expansion, so everything works as expected.

Note how changing single quotes to double quotes prevents this command from working the way you want:

HELLO='Hello World' bash -c "echo $HELLO"

Now this will be failing for the same reason as the first command in your question.

Upvotes: 5

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