Knowing that bash is logging in within .bashrc

eldiener · 10-05-2015, 06:33 PM

Typically the user's .bash_profile script has logic in it to invoke the user's .bashrc if it exists. The .bashrc script is also invoked each time a user opens a terminal. What I want to do is to have the logic of my .bashrc file be different between when it is initially called from .bash_profile and otherwise. Is there something I can test in my .bashrc in order to know that bash is being invoked at log-in time or not ?

goumba · 10-06-2015, 05:26 AM

Something like this may work:

Code:

if [[ $(shopt -q login_shell) ]]
then
    ...
fi

Though I do suppose there's a reason that the distro's default profile doesn't check it to begin with.

SteveG_0001 · 10-06-2015, 05:27 AM

Try somethung like
who|grep `id -un`|wc -l
and test result for greater than 1. It just counts how many times your logged in. Might not work if gui does not maintain utmp.

eldiener · 10-06-2015, 11:40 PM

Quote:

Originally Posted by goumba

Something like this may work:

Code:

if [[ $(shopt -q login_shell) ]]
then
    ...
fi

Though I do suppose there's a reason that the distro's default profile doesn't check it to begin with.

Looks like the right way to do it. Thanks !

eldiener · 10-07-2015, 01:27 AM

Quote:

Originally Posted by goumba

Something like this may work:

Code:

if [[ $(shopt -q login_shell) ]]
then
    ...
fi

Though I do suppose there's a reason that the distro's default profile doesn't check it to begin with.

Unfortunately it does not work. The '$(shopt -q login_shell)' path never gets invoked, even when called from .bash_profile.

pan64 · 10-07-2015, 02:24 AM

How do you know that? (try without -q)

rknichols · 10-07-2015, 07:21 AM

Quote:

Originally Posted by goumba

Something like this may work:

Code:

if [[ $(shopt -q login_shell) ]]
then
    ...
fi

What that is doing is running shopt with the "-q" option, which suppresses the output, and then testing whether that output is non-null. That will never be true.

This should work:

Code:

if shopt -q login_shell
then
    ...
fi

goumba · 10-07-2015, 08:26 AM

Quote:

Originally Posted by rknichols

What that is doing is running shopt with the "-q" option, which suppresses the output, and then testing whether that output is non-null. That will never be true.
[/code]

That's true. We wanted to test the return code. My mistake, doing things rushed again. My apologies to the OP.

In penance, I also offer than the OP may test the first character of argument 0 ($0) for a hyphen ("-").

*slinks under a rock*

eldiener · 10-07-2015, 03:52 PM

Quote:

Originally Posted by rknichols

What that is doing is running shopt with the "-q" option, which suppresses the output, and then testing whether that output is non-null. That will never be true.

This should work:

Code:

if shopt -q login_shell
then
    ...
fi

That does work. Thank you !

eldiener · 10-07-2015, 03:55 PM

Quote:

Originally Posted by goumba

That's true. We wanted to test the return code. My mistake, doing things rushed again. My apologies to the OP.

In penance, I also offer than the OP may test the first character of argument 0 ($0) for a hyphen ("-").

*slinks under a rock*

Apology accepted <g>. Isn't argument 0 the process name ? What does starting with a hyphen mean ?

goumba · 10-07-2015, 07:44 PM

Quote:

Originally Posted by eldiener

Apology accepted <g>. Isn't argument 0 the process name ? What does starting with a hyphen mean ?

In this case, $0 is the name of the shell, if invoked via other than -c, which will be some form of bash. I have gotten both bash and /bin/bash on the same system, the former in a text console, the latter in a terminal under X. However, the following is always true, from the bash man page:

Code:

INVOCATION
       A login shell is one whose first character of argument zero is a -,  or
       one started with the --login option.

So, in a login shell, echo $0 will return "-bash"; otherwise, simply "bash".

pan64 · 10-08-2015, 01:41 AM

Quote:

Originally Posted by goumba

So, in a login shell, echo $0 will return "-bash"; otherwise, simply "bash".

No, unfortunately it is not so simple, just try to execute/source a shell script ($0 will be the name of the script itself).
But $0 will be equal to -bash in case of a login shell.

goumba · 10-08-2015, 07:45 AM

*sgh* Alas, you are right, as like a dummy, I did my testing with bashrc and with interactive shells. The hyphen being the first character is right, however.

rknichols · 10-08-2015, 09:16 AM

But, the OP was talking about doing this in .bashrc, which is sourced by the running shell. $0 and the login_shell option would have their values for that shell. Now you're talking about doing this in some other shell script, which would be interpreted by a different shell instance, not the login shell.

goumba · 10-08-2015, 11:59 AM

Quote:

Originally Posted by rknichols

But, the OP was talking about doing this in .bashrc, which is sourced by the running shell. $0 and the login_shell option would have their values for that shell. Now you're talking about doing this in some other shell script, which would be interpreted by a different shell instance, not the login shell.

I'm not sure if you're referring to my post, as yours is right under mine, but I made no mention of other scripts. :/