BASH get error code in parent terminal from child terminal? - bash

In a bash script I start a new terminal with a command that gives an error. However I don't seem to be able to grab that error code:
#! /bin/bash
gnome-terminal -x bash -c "cat dksdamfasdlm"
echo $?
Output:
0
So I get the error code of the gnome-terminal command instead of the cat one. One suggestion I got, was to make a file with the code and read that from the parent bash. The problem is that I still seem to not be able and read the error code even that way:
#! /bin/bash
gnome-terminal -x bash -c "cat dksdamfasdlm; echo $?; sleep 2"
Output (on new terminal):
cat: dksdamfasdlm: No such file or directory
0
Why is that? Some suggestion on how to solve this? I just want to somehow grab the error in the new terminal from the parent bash.

It seems GNOME Terminal exits immediately after starting, which is obvious if you run for example gnome-terminal -x sleep 10. Since it doesn't wait for the command to finish, there's no way the return code will be that of the command. I could find no option in gnome-terminal --help-all to keep the process in the foreground.
Regarding your second question, you've double-quoted the command, so $? is expanded before running it. This should work:
gnome-terminal -x bash -c 'cat dksdamfasdlm; echo $?; sleep 2'
PS: The -x option is not documented in GNOME Terminal 3.8.4's gnome-terminal --help-all, various references don't help much, and there's no good explanation for why there's a -e option with identical semantics and different syntax.

Related

Saving the result of an echo command in a shell script?

I am attempting to store the result of an echo command as a variable to be used in a shell script. Debian 4.19.0-6-amd64
The command works in terminal: echo $HOSTNAME returns debian-base, the correct hostname.
I attempt to run it in a shell script, such as:
#!/usr/bin/bash
CURRENT_HOSTNAME=`echo $HOSTNAME`
echo $CURRENT_HOSTNAME
I have tried expansion:
CURRENT_HOSTNAME=$(echo $HOSTNAME)
And just to cover some more bases, I tried things like:
CURRENT_HOSTNAME=$HOSTNAME
# or
CURRENT_HOSTNAME="$HOSTNAME"
# also, in case a problem with reserved names:
test=$HOSTNAME
test="$HOSTNAME"
Works great in the terminal! Output is as follows:
root#debian-base:/scripts# echo $HOSTNAME
debian-base
root#debian-base:/scripts# TEST_HOSTNAME=$HOSTNAME
root#debian-base:/scripts# echo $TEST_HOSTNAME
debian-base
root#debian-base:/scripts# TEST_TWO_HOSTNAME=$(echo $HOSTNAME)
root#debian-base:/scripts# echo $TEST_TWO_HOSTNAME
debian-base
As soon as I run the script (as above):
root#debian-base:/scripts# sh test.sh
root#debian-base:/scripts#
What am I doing wrong?
You are using bash as your terminal. Bash has the variable $HOSTNAME set. You run your script with sh. sh does not have a $HOSTNAME.
Options:
bash test.sh
Or run it as a program:
chmod +x test.sh
./test.sh
But I think you need to change your first line to:
#!/bin/bash
As I don't think bash is installed in /usr/bin in most cases. But you need to try. To figure out where bash is installed use which bash
Another option is to use the hostname binary:
CURRENT_HOSTNAME=$(hostname)
echo $CURRENT_HOSTNAME
Which works in both bash and sh.
You can start sh by just running sh. You will see it has a bash-like terminal. You can try to do echo $HOSTNAME. It will not show, because it's not there. You can use set to see all the variables that are there (as sh does not have tab completion it's harder to figure out).

Allow user input in second command in bash pipe

I'm looking for how I might allow user input in a second command in a bash statement and I'm not sure how to go about it. I'd like to be able to provide a one-liner for someone to be able to install my application, but part of that application process requires asking some questions.
The current script setup looks like:
curl <url/to/bootstrap.sh> | bash
and then boostrap.sh does:
if [ $UID -ne 0 ]; then
echo "This script requires root to run. Restarting the script under root."
exec sudo $0 "$#"
exit $?
fi
git clone <url_to_repo> /usr/local/repo/
bash /usr/local/repo/.setup/install_system.sh
which in turn calls a python3 script that asks for input.
I know that the the curl in the first line is using stdin and so that might make what I'm asking impossible and that it has to be two lines to ever work:
wget <url/to/boostrap.sh>
bash bootstrap.sh
You can restructure your script to run this way:
bash -c "$(curl -s http://0.0.0.0//test.bash 2>/dev/null)"
foo
wololo:a
a
My test.bash is really just
#!/bin/bash
echo foo
python -c 'x = raw_input("wololo:");print(x)'`
To demonstrate that stdin can be read from in this way. Sure it creates a subshell to take care of curl but it allows you to keep reading from stdin as well.

How I can write a command in a terminal which is running a program and I open from a script

I already know how to open a terminal in a bash with gnome-terminal and execute a program:
gnome-terminal -e ./OpenBTSCLI
But I also need that once open that program in the new terminal, write another command inside.
When a I tried to use echo, the message appear in the terminal where I run the bash.
I tried: gnome-terminal -e "bash -c './OpenBTSCLI && echo message'" that I find online but its not working, it only do the first part.
Anyone have an idea of how to resolve this? Thank you
I think it does the second command as well, but the new terminal closes as soon as the command's finished, so you don't see it. I reversed the order of quotes and added a 1s sleep at the end to allow seeing the echo.
gnome-terminal -e 'bash -c "./OpenBTSCLI && echo message && sleep 1"'

How to exit multiple nested shells at once?

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.

How to change argv0 in bash so command shows up with different name in ps?

In a C program I can write argv[0] and the new name shows up in a ps listing.
How can I do this in bash?
You can do it when running a new program via exec -a <newname>.
Just for the record, even though it does not exactly answer the original poster's question, this is something trivial to do with zsh:
ARGV0=emacs nethack
I've had a chance to go through the source for bash and it does not look like there is any support for writing to argv[0].
I'm assuming you've got a shell script that you wish to execute such that the script process itself has a new argv[0]. For example (I've only tested this in bash, so i'm using that, but this may work elsewhere).
#!/bin/bash
echo "process $$ here, first arg was $1"
ps -p $$
The output will be something like this:
$ ./script arg1
process 70637 here, first arg was arg1
PID TTY TIME CMD
70637 ttys003 0:00.00 /bin/bash ./script arg1
So ps shows the shell, /bin/bash in this case. Now try your interactive shell's exec -a, but in a subshell so you don't blow away the interactive shell:
$ (exec -a MyScript ./script arg1)
process 70936 here, first arg was arg1
PID TTY TIME CMD
70936 ttys008 0:00.00 /bin/bash /path/to/script arg1
Woops, still showing /bin/bash. what happened? The exec -a probably did set argv[0], but then a new instance of bash started because the operating system read #!/bin/bash at the top of your script. Ok, what if we perform the exec'ing inside the script somehow? First, we need some way of detecting whether this is the "first" execution of the script, or the second, execed instance, otherwise the second instance will exec again, and on and on in an infinite loop. Next, we need the executable to not be a file with a #!/bin/bash line at the top, to prevent the OS from changing our desired argv[0]. Here's my attempt:
$ cat ./script
#!/bin/bash
__second_instance="__second_instance_$$"
[[ -z ${!__second_instance} ]] && {
declare -x "__second_instance_$$=true"
exec -a MyScript "$SHELL" "$0" "$#"
}
echo "process $$ here, first arg was $1"
ps -p $$
Thanks to this answer, I first test for the environment variable __second_instance_$$, based on the PID (which does not change through exec) so that it won't collide with other scripts using this technique. If it's empty, I assume this is the first instance, and I export that environment variable, then exec. But, importantly, I do not exec this script, but I exec the shell binary directly, with this script ($0) as an argument, passing along all the other arguments as well ($#). The environment variable is a bit of a hack.
Now the output is this:
$ ./script arg1
process 71143 here, first arg was arg1
PID TTY TIME CMD
71143 ttys008 0:00.01 MyScript ./script arg1
That's almost there. The argv[0] is MyScript like I want, but there's that extra arg ./script in there which is a consequence of executing the shell directly (rather than via the OS's #! processing). Unfortunately, I don't know how to get any better than this.
Update for Bash 5.0
Looks like Bash 5.0 adds support for writing to special variable BASH_ARGV0, so this should become far simpler to accomplish.
(see release announcement)
( exec -a foo bash -c 'echo $0' )
ps and others inspect two things, none of which is argv0: /proc/PID/comm (for the "process name") and /proc/PID/cmdline (for the command-line). Assigning to argv0 will not change what ps shows in the CMD column, but it will change what the process usually sees as its own name (in output messages, for example).
To change the CMD column, write to /proc/PID/comm:
echo -n mynewname >/proc/$$/comm; ps
You cannot write to or modify /proc/PID/cmdline in any way.
Process can set their own "title" by writing to the memory area in which argv & envp are located (note that this is different than setting BASH_ARGV0). This has the side effect of changing /proc/PID/cmdline as well, which is what some daemons do in order to prettify (hide?) their command lines. libbsd's setproctitle() does exactly that, but you cannot do that in Bash without support of external tools.
I will just add that this must be possible at runtime, at least in some environments. Assigning $0 in perl on linux does change what shows up in ps. I do not know how that is implemented, however. If I can find out, i'll update this.
edit:
Based on how perl does it, it is non-trivial. I doubt there is any bask built in way at runtime but don't know for sure. You can see how perl does sets the process name at runtime.
Copy the bash executable to a different name.
You can do this in the script itself...
cp /bin/bash ./new-name
PATH=$PATH:.
exec new-name $0
If you are trying to pretend you are not a shell script you can rename the script itself to something cool or even " " (a single space) so
exec new-name " "
Will execute bash your script and appears in the ps list as just new-name.
OK so calling a script " " is a very bad idea :)
Basically, to change the name
bash script
rename bash and rename the script.
If you are worried, as Mr McDoom. apparently is, about copying a binary to a new name (which is entirely safe) you could also create a symlink
ln -s /bin/bash ./MyFunkyName
./MyFunkyName
This way, the symlink is what appears in the ps list. (again use PATH=$PATH:. if you dont want the ./)

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