I'm trying to see what the output of a command would be if I were in a login shell, without having to go into a login shell. I've tried several variations of
zsh --login -c "alias"
But none of my aliases get shown; are --login and -c incompatible?
To test the difference between zsh --login -c "alias" and a normal login shell, you can/should add the -x option to see what the shell is up to.
When I run zsh -x --login -c "alias", then it processes /etc/zprofile.
When I run zsh -x --login, then it processes /etc/zprofile and /etc/zshrc.
I don't normally use zsh, so I don't have any personalized profile or start up file for it, but it seems plausible that it might look for (but, in my case, not find) ~/.zprofile and ~/.zshrc too.
I created trivial versions of those files:
$ echo "echo in .zprofile" > ~/.zprofile
$ echo "echo in .zshrc" > ~/.zshrc
and sure enough, they're processed. Further, the -c command with --login processed the .zprofile but did not process the .zshrc file.
Thus, using -c "alias" after the --login suppresses the processing of /etc/zshrc and ~/.zshrc. If you want those executed even so, you need to use something like:
zsh --login -c "[ -f /etc/zshrc ] && . /etc/zshrc; [ -f ~/.zshrc ] && . ~/.zshrc; alias"
Using -x to debug login processing is often informative.
It's nice that modern shells provide a command line option to induce login processing. I still have a program (which I don't use any more) that runs a login shell the old-fashioned way, by adding a - before the shell name in argv[0]. Thus, running -ksh would trigger login processing; the login program would run the login shell with the - at the start.
Related
According to bash man page, it says -i is for interactive mode of shell.
I tried example code to find out what -i option does.
interactive.sh is script that needs user input, which means interactive script.
The default bash option is non-interactive mode.
But the interactive.sh runs without any problem with non-interactive mode.
It also runs well with interactive mode. It confuses me.
What is the exact usage of -i option in bash?
What is the difference between interactive and non-interactive mode in shell?
$cat interactive.sh
#!/bin/bash
echo 'Your name ?'
read name
echo "Your name is $name"
$ bash interactive.sh
Your name ?
ABC
Your name is ABC
$ bash -i interactive.sh
Your name ?
DEF
Your name is DEF
With bash -i script you are running interactive non-login shell.q
What is the exact usage of -i option in bash?
From man bash:
-i If the -i option is present, the shell is interactive.
What is the difference between interactive and non-interactive mode in shell?
There are some differences. Look at man bash | grep -i -C5 interactive | less:
An interactive shell is one started without non-option arguments (unless -s is specified) and without the -c option whose standard input and er‐
ror are both connected to terminals (as determined by isatty(3)), or one started with the -i option. PS1 is set and $- includes i if bash is
interactive, allowing a shell script or a startup file to test this state.
When an interactive login shell exits, or a non-interactive login shell executes the exit builtin command, bash reads and executes commands from
the file ~/.bash_logout, if it exists.
When an interactive shell that is not a login shell is started, bash reads and executes commands from ~/.bashrc, if that file exists. This may
be inhibited by using the --norc option. The --rcfile file option will force bash to read and execute commands from file instead of ~/.bashrc.
[...]
When bash is interactive, in the absence of any traps, it ignores SIGTERM (so that kill 0 does not kill an interactive shell), and SIGINT is
caught and handled (so that the wait builtin is interruptible). In all cases, bash ignores SIGQUIT. If job control is in effect, bash ignores
SIGTTIN, SIGTTOU, and SIGTSTP.
etc. For example bash -i -c 'echo $PS1'.
I want to execute my shell with the environment I provide as the command line arguments. For that I have a script that ends with exec zsh -d -f after setting all the variables I want, which gives me a new shell with all variables set.
e.g:
export MY_SESSION="$1"
cd $2
export PS1="$3; "
exec zsh -d -f
My issue is that, I also want to execute bindkey -e on the new shell before it is made available. How do I do that?
I managed to do this with expect. Here is what I did.
#!/usr/bin/env expect
spawn /bin/zsh -d -f
send "source ./bin/activate\r"
send "export PS1='*| '\r"
send "bindkey -e\r"
interact
This successfully sends the bindkey -e on the produced zsh shell.
This should be very simple.
I recently noticed that when I type 'bash' into Terminal on Mac it shows this:
Jays-MacBook-Pro: ~ $ bash
bash: parse_git_branch: command not found
When before it didn't. Can someone explain why and how to resolve.
It is likely that you configured BASH to run parse_git_branch and print the result as part of PS1 (or alike). You can check this by: "echo $PS1" and "echo $PROMPT_COMMAND".
However, parse_git_branch is not a builtin function of bash. Below is how I configured my PS1. You may want to copy my git_branch_4_ps1 as your parse_git_branch
PS1='\n' # begin with a newline
PS1=$PS1'\[\e[38;5;101m\]\! \t ' # time and command history number
PS1=$PS1'\[\e[38;5;106m\]\u#\h ' # user#host
PS1=$PS1'\[\e[7;35m\]${MY_WARN}\[\e[0m\] ' # warning message if there is any
PS1=$PS1'\[\e[38;5;10m\]${MY_EXTRA} ' # extra info if there is any
PS1=$PS1'\[\e[0;36m\]$(git_branch_4_ps1) ' # git_branch_4_ps1 defined below
PS1=$PS1'\[\e[38;5;33m\]\w' # working directory
PS1=$PS1'\n\[\e[32m\]\$ ' # "$"/"#" sign on a new line
PS1=$PS1'\[\e[0m\]' # restore to default color
function git_branch_4_ps1 { # get git branch of pwd
local branch="$(git branch 2>/dev/null | grep "\*" | colrm 1 2)"
if [ -n "$branch" ]; then
echo "(git: $branch)"
fi
}
If your parse_git_branch is defined in ~/.bash_profile, it will not be loaded when you open a non-login shell (e.g. by running bash).
The differences between login and non-login shells are described here: Difference between Login Shell and Non-Login Shell? For our purposes, the main difference is that login shells (e.g. that when you first open Terminal) automatically source ~/.bash_profile upon startup, whereas non-login shells (e.g. that when you run bash from within Terminal) do not.
To fix this error, simply source your ~/.bash_profile after running bash:
user#host:~ $ bash
bash: parse_git_branch: command not found
user#host:~ $ source .bash_profile
Alternatively, place the function in ~/.bashrc instead, which will be automatically sourced by non-login shells (as covered in the earlier link).
Instead of having
parse_git_branch
call in PS1 definition alone you may use
parse_git_branch 2>/dev/null
to send stderr to /dev/null. This will silence the error you don't want to see.
have you export your $PS1 ?
You can check by run command:
printenv
else you should export it by run:
export -n PS1
after you will can run sudo or sudo su without problem
The key to this is to NOT export PS1. If it's exported, then any non-login shell also takes PS1. Since .bash_profile is automatically source'd by the login shell, the PS1 variable only affects the login shell.
I am trying to run a command that has been aliased in my ~/.bashrc from Perl using the system command. It works well running the command only once, but when I run it twice the second invocation is run as a background job and then suspended (the same as pressing <CTRL-Z>) and I have to type fg to complete the command. For example
use strict;
use warnings;
system ('bash -ic "my_cmd"');
system ('bash -ic "my_cmd"');
The second call never completes. The output is [1]+ Stopped a.pl.
Note:
The same result is obtained when replacing my_cmd with any other command, for example ls.
It seems not to depend of the contents of my ~/.bashrc file. I tried to remove everything from it, and the problem still persisted.
I am using Ubuntu 14.04 and Perl version 5.18.2.
Update
For debugging I reduced my ~/.bashrc to
echo "Entering ~/.bashrc .."
alias my_cmd="ls"
alias
and my ~/.bash_profile to
if [ -f ~/.bashrc ]; then
echo "Entering ~/.bash_profile .."
. ~/.bashrc
fi
Now running:
system ('bash -lc "my_cmd"');
system ('bash -lc "my_cmd"');
gives
Entering ~/.bash_profile ..
Entering ~/.bashrc ..
alias my_cmd='ls'
bash: my_cmd: command not found
Entering ~/.bash_profile ..
Entering ~/.bashrc ..
alias my_cmd='ls'
bash: my_cmd: command not found
and running
system ('bash -ic "my_cmd"');
system ('bash -ic "my_cmd"');
gives
Entering ~/.bashrc ..
alias my_cmd='ls'
a.pl p.sh
[1]+ Stopped a.pl
Rather than using the -i switch for an interactive shell, I think you should use the -l (or --login) switch, which causes bash to act as if it had been invoked as a login shell.
Using the -l switch doesn't load ~/.bashrc by default. According to man bash, in a login shell, /etc/profile/ is loaded, followed by the first file found from ~/.bash_profile/, ~/.bash_login or ~/.profile/. On my system, I have the following in ~/.bash_profile, so ~/.bashrc is loaded:
# Source .bashrc
if [ -f ~/.bashrc ]; then
. ~/.bashrc
fi
Now that your ~/.bashrc is being loaded, you need to enable the expansion of aliases, which is off in a non-interactive shell. To do this, you can add the following line before setting your aliases:
shopt -s expand_aliases
A process randomly stopping - aside from ctrl-z is usually when it needs STDIN, but doesn't have it attached.
Try it with - for example passwd &. It'll background and go straight into 'stopped' state. This may well be what's happening with your bash command. -i means interactive shell, explicitly, and you're trying to do something noninteractive with it.
That's almost certainly not the best approach, you probably want to do something different. bash --login might be closer to what you're after.
Tom Fenech's answer worked for me in Ubuntu 16.04.1 LTS with a small addition. At the top of my ~/.bashrc file, I commented out the following section so that if the shell is not interactive (e.g., a login shell), ~/.bashrc is still read. On some other versions of Linux I don't see this section.
# If not running interactively, don't do anything
case $- in
*i*) ;;
*) return;;
esac
I have a host on which I don't have sudo. Its been setup with ksh, I'm too used to bash and chsh doesn't work. So I put in a /bin/bash as the first line in the .profile on the system.
So the result is, when I login to this system, it automatically gets me into bash. However, when I exit the shell, not suprisingly I land up in ksh.
Any tricks to avoid this?
Use exec to replace the current process (shell) with the new process (shell).
I recommend two steps:
if [ $SHELL != /bin/bash ]
then SHELL=/bin/bash exec /bin/bash --login
fi
Or, you can compress that to:
[ $SHELL != /bin/bash ] && SHELL=/bin/bash exec /bin/bash --login
You can then put the rest of your Bash profile after this. Note that probably you don't put a shebang on the first line - that will confuse things. Also, while testing, make sure you have a second connection (window) open so that you can adjust problems. It is annoying to get locked out by an erroneous profile.
You may write a script named myexit like this:
kill -1 $(ps | sed 1d | awk '{print $1}')
It sends the signal hang up (SIGHUP) to process attached to this terminal.
And would not affect any process started up by nohup.