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?
bash v3.2.48(1)-release osx 10.8.3
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.
4.2 is the earliest version to have -g for declare. It's a really new feature.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
-x then? man says: Mark names for export to subsequent commands via the environment??
subsequent commands via the environment in the present context i.e. inside that function.declare -x is absolutely identical to no use of declare -x at all, in all cases?