source bash script in bash -c ""

Question

I am trying to source a bash script inside a bash -c "..." command but it doesn't work.

If I run the command outside of the bash -c "..." it works.

I need to use bash -c "..." because I want to ensure people are using Bash (no sh, tcsh, etc.)

working:

bash> source /neuro/arch/scripts/neuro-fs stable;echo $XAPPLRESDIR;
/neuro/arch/x86_64-Linux/packages/matlab/MATLAB_Compiler_Runtime/current/X11/app-default

not working:

bash> bash -c "source /neuro/arch/scripts/neuro-fs stable;echo $XAPPLRESDIR;"

In the Bash script, we define XAPPLRESDIR as follows:

export XAPPLRESDIR=/neuro/arch/x86_64-Linux/packages/matlab/MATLAB_Compiler_Runtime/current/X11/app-defaults

#!/bin/bash at the beginning of the script or not doesn't seem to make any difference.

Is export a "bad" practice in this case?

linuts · Accepted Answer · 2014-05-08 13:44:54Z

3

The variable expansion in the "bash -c" is happening before "bash -c" is run; you need to do something like:

bash -c "source /neuro/arch/scripts/neuro-fs stable;echo \$XAPPLRESDIR;"

answered May 8, 2014 at 13:44

linuts

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