~/.bashrc displays entries depending on how it is called

Herbert Qerbert

6/2/24, 2:53 PM

I am trying to export a new directory "foo" to $PATH using a shell script. For now I used a shell script install.sh that exports the path like this:

<install.sh> 
#! /bin/sh

echo export PATH=$PATH:<path/foo> >> ~/.bashrc

For some reason, if I now try to display the ~/.bashrc file, I do get different outputs depending on how I print it. For example, accessing it in a Terminal via vi ~/.bashrc will display the file without my exported line. However, if I load it via a shell script (the same command but in a .sh file), my exported line is shown. I have no other way to see this line, except through this method. I already checked different versions of the command with different shebangs (bin/sh vs bin/bash) and with/without sudo. Where is my added line stored? I already checked /etc/skel/.bashrc and ~/.bashrc.

(I know that it adds a new export line every time the script gets called. Using Ubuntu 22.04.2 LTS).

EDIT:

calling the install.sh script with sudo script sets the $PATH variable as root and calling it as user returns the user $PATH variable as user which caused this behaviour.

0 + 3

bash

environment-variables

bashrc

mook765

6/2/24, 3:25 PM

Do you run the terminal command and the script as the same user?

raj

6/2/24, 4:31 PM

Type the command `echo ~` directly in your terminal, and then run the same command in a shell script. See if there is a difference in the directory name that is printed.

Herbert Qerbert

6/2/24, 4:48 PM

Thank you for the quick help. This seems to resolve the issue. The script set the $PATH variable as root and calling it as user returns the user $PATH variable.

Elon Musk

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EN: ~/.bashrc displays entries depending on how it is called

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