Cannot successfully source .bashrc from a shell script

Question

Normally we can source ~/.bashrc file using this command

source ~/.bashrc

but if I write this in a shell script and execute it, nothing happens. Why?
Is there any way to do this?

My script:

#!/bin/bash
chmod a+x ~/.bashrc
source ~/.bashrc

Also tried . (dot) instead of source. Same result.

Answer 1

I want to complement ravi's answer:

This behavior is specific to Ubuntu (and probably most derived distros), since your default ~/.bashrc file starts with a short-circuit, Ubuntu 18.04, for example:

# If not running interactively, don't do anything
case $- in
    *i*) ;;
      *) return;;
esac

That will stop the evaluation of the file if it is running in a non-interactive shell, which is the case of your script since all scripts are run in a non-interactive shell, and subsequently every file you source will inherit this property.

eval hack

I found out an ugly hack to workaround Ubuntu specifically, using eval instead of source:

eval "$(cat ~/.bashrc | tail -n +10)"

It simply skips the few first lines and evaluates the rest of the ~/.bashrc so the rest is evaluated and modifies the current execution.

Be aware it is a magic number and might not work across Ubuntu versions; but may be a good solution if your are crafting scripts for more-or-less known systems.

A fancier solution might involve using regex to target the specific bits that stop the evaluation.

Shebang alternative

Another alternative that might work better in some scenarios is forcing the script to run in an interactive shell by adding a flag in the shebang:

#!/bin/bash -i

Be aware of a few things:

• It is a better practice to use the #!/usr/bin/env bash form but this way you cannot start the shell with arguments.
• Using the -i has it's own set of consequences, among them, programs will prompt for user interaction and this is usually not intended for scripts, for example, installing deb packages might stop the script at dpkg configure prompts.
• I initially tried to use set -i and set +i to turn the feature on and off where I needed it, but this doesn't work.

Answer 2

A shell script is run in its own shell instance. All the variable settings, function definitions and such only affect this instance (and maybe its children) but not the calling shell so they are gone after the script finished.

By contrast the source command doesn't start a new shell instance but uses the current shell so the changes remain.

If you want a shortcut to read your .bashrc use a shell function or an alias instead of a shell script, like

alias brc='source ~/.bashrc'

Answer 3

Your .bashrc usually starts:

# If not running interactively, don't do anything
[ -z "$PS1" ] && return

Since your script does not have PS1 set (because it is not interactive), it doesn't reset path because it exits early . To demonstrate, modify your script:

    #!/bin/bash
    chmod a+x ~/.bashrc
    PS1='$ '
    source ~/.bashrc

this will now allow your scripts to work with the new .bashrc. Note: once your script exits , the env will be set to what it was before starting the script . The changes will be reflected the next time a terminal is started.

Answer 4

Try:

exec bash

This should reload ~/.bashrc, ~/.bash_aliases, etc.

Answer 5

None of the other methods worked for me [source /path/to/file vs . ./path/to/file, alias, etc...], until, thanks to this tutorial I found that using the:

#!/usr/bin/env bash shebang

instead of the simpler #!/usr/bin/env one lets arguments pass on to the interpreter, which I think is the key here – see this document for more info.

In any event, if source commands in any form aren't working for you, try checking your shebang, that might be the problem :)

Answer 6

I just copied the contents of the original ~/.bashrc file, deleted it, created a new ~/.bashrc and pasted the contents. And, it worked.