Trying to run a function defined in my .bashrc using "bash -c ". I end up with the error "command not found". How do I get "bash -c" to load my init file?
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2 Answers
You can make it into an interactive shell with -i, then your ~/.bashrc
will be read:
bash -i -c "echo \$EDITOR"
The other thing you can do is source the file explicit. If you have /var/tmp/test with content:
export XXX=123
and you do
bash -c "source /var/tmp/test; echo \$XXX"
you will get 123 echoed.
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This doesn't work either: the
-ifails with this same issue ( stackoverflow.com/a/27581210/293735 ), and I cannot use the-lsince that'd reload the.profile(which I want to avoid) and the explicit source doesn't work (bash 4.3.30 on Ubuntu, maybe it's platform specific?)– berdarioCommented Feb 24, 2015 at 23:14 -
@berdario the
-iflag does work, it will run an interactive shell which reads~/.bahshrcand, since it is interactive, also enables aliases. How exactly did it fail for you? You're linking to an answer on SO that does not describe failure and the question is very specific and is about calling bash from within a perl script. Is that what you're doing? In any case, I couldn't repeat the OP's issue, it worked as expected even within Perl.– terdon ♦Commented Feb 26, 2015 at 13:08 -
@terdon invoking
bash -istops the outer process, this happens with anything as the caller process (I personally tried with Python and Ruby)... but indeed something weird is happening: I could reproduce it on Ubuntu (I'll try again this evening at home), but I cannot reproduce it on Nixos and Kali– berdarioCommented Feb 27, 2015 at 13:58 -
@berdario it sounds like your issue is quite specific. I suggest you ask a new question and link back to this one explaining how the answers here don't work for you.– terdon ♦Commented Feb 27, 2015 at 13:59
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Well, I think that this answer is wrong (and yours might be correct, but I want to double check on my machine before jumping to any conclusion).– berdarioCommented Feb 27, 2015 at 14:01
Another option would be to set the $BASH_ENV variable:
When bash is started non-interactively, to run a shell script, for
example, it looks for the variable BASH_ENV in the environment, expands
its value if it appears there, and uses the expanded value as the name
of a file to read and execute. Bash behaves as if the following com‐
mand were executed:
if [ -n "$BASH_ENV" ]; then . "$BASH_ENV"; fi
but the value of the PATH variable is not used to search for the file
name.
So, you could do:
BASH_ENV=~/.bashrc && bash -c 'your_function'
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bash --rcfile ~/.bashrc -c 'echo $FOO'doesn't printbaras expected– berdarioCommented Feb 24, 2015 at 23:05 -
@berdario no, it doesn't. Thanks for pointing it out, I could have sworn I had tested it. The problem is that the
--rcfileis only used when launching an interactive shell andbash -claunches a non-interactive one. SettingBASH_ENVdoes work though, did you try that?– terdon ♦Commented Feb 26, 2015 at 12:59 -
This doesn't work either: with
export FOO=barin the.bashrc,> echo $BASH_ENV\n /home/dario/.bashrc\n > bash -c 'echo $FOO'\n: no output except the newline– berdarioCommented Feb 27, 2015 at 22:46 -
1@berdario I'm not sure exactly what you did but try this:
BASH_ENV=~/.bashrc bash -c 'echo $FOO'. I just tested that on my system withFOO=barin~/.bashrcand it worked as expected. If it doesn't, please ask a new question and link back to this one because your issue will be different.– terdon ♦Commented Feb 27, 2015 at 23:06 -
Upon running
set -xe && source ~/.bashrcI realized that apparently bash can silently swallow some errors when sourcing... these errors will make the sourcing silently fail. Since this got out of hand I opened a new question: unix.stackexchange.com/questions/187738/…– berdarioCommented Mar 2, 2015 at 15:56