Linux source does not work in .sh file?

Question

I have a .sh (start_sim.sh) and a .bash (sim_sources.bash) file. The sim_sources.bash file is called from within the start_sim.sh and should set an environment variable $ROBOT to a certain value. However the ROBOT variable never changes when I call ./start_sim.sh. Is there a fundamental mistake in the way I am trying to do this?

start_sim.sh contains:

#!/bin/bash

echo -n "sourcing sim_sources.bash..."
source /home/.../sim_sources.bash
echo "done."

sim_sources.bash contains:

# set the robot id
export ROBOT=robot

EDIT: Could you also propose a way to work around this issue? I would still need to set variables from with in the .bash file.

EDIT2: Thanks for your replys! Finally I ended up solving it with a screen and stuffing commands to it:

echo -n "starting screen..."
screen -dmS "sim_screen"
sleep 2
screen -S "sim_screen" -p 0 -X stuff "source /home/.../sim_sources.bash$(printf \\r)"
sleep 5
screen -S "sim_screen" -p 0 -X stuff "source /home/.../start_sim.sh$(printf \\r)"

mpapis · Accepted Answer · 2013-08-14 14:10:14Z

6

You're setting the ROBOT variable in the start_sim.sh script, but that's not available to parent processes (your spawning shell/command-prompt).

Exporting a variable e.g. export ROBOT=robot makes the variable available to the current process and child processes. When you invoke ./start_sim.sh you're invoking a new process.

If you simply source start_sim.sh in your shell, that script runs as part of your shell process and then your variable will be available.

edited Aug 14, 2013 at 14:10

mpapis

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answered Aug 14, 2013 at 13:38

Brian Agnew

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1 Comment

Marc Over a year ago

Thanks for your hint. But is there any way to work around this? I would need to set the variable from within the script. (There are actually a whole bunch of variables which should be set from the .bash file - however until now source has been called from within the .bashrc).

vegi · Accepted Answer · 2013-08-14 14:16:34Z

4

As Brian pointed out the variables are not available outside of the script.

Here a adapted script that shows this point:

#!/bin/bash

echo -n "sourcing sim_sources.bash..."
. sim_sources.bash
echo $ROBOT
echo "done."

The workaround you are asking for is to start a new shell from the actual shell with the environmental values already set:

#!/bin/bash

echo -n "sourcing sim_sources.bash..."
. sim_sources.bash
echo "done."
bash

This results in:

bash-4.1$ printenv | grep ROBOT
ROBOT=robot

edited Aug 14, 2013 at 14:16

answered Aug 14, 2013 at 13:53

vegi

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Comments

sudhanshu · Accepted Answer · 2018-04-18 05:56:35Z

0

I am on Ubuntu 16.04

I used /bin/sh instead of /bin/bash and it works !

answered Apr 18, 2018 at 5:56

sudhanshu

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