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I want to get a variable from another script, as demonstrated in
http://stackoverflow.com/questions/5228345/bash-script-how-to-reference-a-file-for-variables

However, the answer uses source command only available in bash. I want to do this in a portable way.

I have also tried

a.sh

export VAR="foo"
echo "executing a"

b.sh

#!/bin/sh
./a.sh
echo $VAR

But of course that does not work either. How to do this?

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2 Answers 2

up vote 2 down vote accepted

First of all, be aware that var and VAR are different variables.

To answer your question the . command is not bash-specific:

# a.sh
num=42
# b.sh
. ./a.sh
echo $num

The variables in "a" do not need to be exported.

http://www.gnu.org/software/bash/manual/bashref.html#Bourne-Shell-Builtins

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Thank you and excuse my typo. I was pretty surprised with the difference between ./a.sh and . ./a.sh. Care to explain the difference? –  Eero Aaltonen Jun 12 '13 at 7:29
    
The dot command (. or source) evaluates the script in your current shell. Executing the script first spawns a subshell, and any environment changes in the subshell are lost when the subshell exits -- a child process cannot alter the environment of its parent. –  glenn jackman Jun 12 '13 at 9:01
    
bash has a handy help builtin to access the manual for a specific command -- see help . –  glenn jackman Jun 12 '13 at 9:02

Environment variables are only inherited from parent to child and not the other way round. In your example, b.sh calls a.sh, so a runs as a child of b. When a.sh exports var, it won't be seen by b.sh. Amend the logic so that the parent process exports the variable, e.g.

a.sh:

echo In a.sh...
VAR="test"
export VAR
./b.sh

b.sh:

echo In b.sh...
echo $VAR
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Thank you, but I actually have sharedDefs.sh and two separate targets depending on it, so I need to have the dependencies this way. –  Eero Aaltonen Jun 12 '13 at 7:31

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