Bashscripts.org

I have a file named default.conf like this
A=1
B=2
C=3

And I have a script that source defaults.conf via find
find . -name default.conf |\
while read file; do
for var in $(cat $file | cut -d = -f 1); do
export $var=undef;
done
source $file;
done;
echo $A

The command echo $A return nothing.

Anybody have a valid explanation to this?

PS: My problem is was simplifiquied here

Hmmm...should do what you want.

I haven't fully investigated but it seems this is a weird bashism down to how sub shells are managed.



The above piece of code opens a subshell due to the pipe. This subshell can have weird effects on variables. I know i'm not explaining this very well but essentially there is a way around the issue by using the following format;



The above format removes the need for the extra subshell so should work as expected.
It took me AGES to work this out when I first met it.
If you re-factor your code to work in that way you should be able to get around the issue.

Of course there's no reason why Watael's suggestion should not work