Export variable from Bash

Question

In bash I have a function which should export some variables, but it doesn't.

Here an example:

function f() {

  # ...
  declare -rx TIMESTAMP_FORMAT='+%Y-%m-%d %H:%M:%S'
  # ...

}

What is the correct way of doing this from within a function?

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My environment

bash v3.2.48(1)-release osx 10.8.3

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Answer 1

to declare the variable as global use the option -g. Like this:

function f() {

  # ...
  declare -grx TIMESTAMP_FORMAT='+%Y-%m-%d %H:%M:%S'
  # ...

}

quote from help declare :

-d create global variables when used in a shell function; otherwise ignored

---

Update After reading your last comment:

Available outside the function and available in subshells

... it looks like export is the command you are looking for. Check this example:

function foo {
    export BAR="test"
}

# call function
foo

# will output 'test'
echo "$BAR"

# as the var is in environemt:
export -p | grep "BAR"

# ... it will be available from sub shells as well

Note: The export solution is missing the -r (read-only) feature from declare. If you can live with this, export should be ok.

Answer 2

Don't use declare and your variable will be available outside function as well.

function f() {
  # ...
  TIMESTAMP_FORMAT='+%Y-%m-%d %H:%M:%S'
  # ...
}

f # call your function
echo "OUTSIDE: " $TIMESTAMP_FORMAT