2

Here is what I have:

script_1.sh

echo "HELLO FROM script_1.sh"

./scripts/script_2.sh

echo $MY_VARIABLE

script_2.sh

echo "HELLO FROM script_2.sh"

export MY_VARIABLE="MY_VALUE"

Here is what it logs when I run script_1.sh:

HELLO FROM script_1.sh
HELLO FROM script_2.sh
"" // EMPTY LINE INSTEAD OF "MY_VALUE"

What am I doing wrong?

cbdeveloper
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2 Answers2

6

Variables are exported from a parent to a child, not vice versa. script_2.sh is called in a different shell whose environment doesn't propagate back to the parent shell.

Source the script (using the .) to call it in the same shell. You then don't even need to export the value.

. ./scripts/script_2.sh
choroba
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1

environment variables are inherited down the call chain. they are not returned up to the caller.

in other words: a called script might inherit the variables of the caller. but the caller will not get the variables of the called script.

in you simple example the easiest solution is to "source the script"

. ./scripts/script_2.sh

(the dot is the command to source a script)

sourcing is not a new step in the call chain. instead the sourcer and sourcee share the same environment. for more explanation on the difference of executing and sourcing see here: https://superuser.com/questions/176783/what-is-the-difference-between-executing-a-bash-script-vs-sourcing-it/176788#176788

there are other options but they are more complicated and error prone. it seems that you are starting to learn shell scripting. so learn the difference of sourcing and executing and the implication on the environment for now. if you need other options later then come back and ask another question.

Lesmana
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