Batch user's environment variables not getting set in su -c - python

I'm trying to set up my /etc/rc.local to automatically start up a process on reboot as another user. For some reason, the .bash_rc for this user does not seem to be getting initialized.
Here's the command I added to /etc/rc.local :
sudo su -l batchuser -c "/home/batchuser/app/run_prod.sh &"
this didn't work, so I also tried this:
sudo su -l batchuser -c ". /home/batchuser/.profile; /home/batchuser/app/run_prod.sh &"
run_prod.sh just starts up a python script. The python script fails because it references modules which are in a python path which gets initialized in the .bash_rc
EDIT: it works when I do this
sudo su -l batchuser -c "export PYTHONPATH=/my/python/path; /home/batchuser/app/run_prod.sh &"
Why does this work and not the statement above? How come the .bashrc is not getting initialized?

I have run into this same problem. I can't fully explain the behavior, but I ended up doing this type of thing:
sudo $PYTHONPATH=$PYTHONPATH the_command
or more specifically for your case,
sudo $PYTHONPATH=$PYTHONPATH su -l batchuser -c /home/batchuser/app/run_prod.sh &"
Does that work for you? If it does, you may find it doesn't return immediately like you expect it to. You may need to move the & outside the quotes so it applies to the sudo command.

Related

Fabric run command - start sh script on the remote & leaving the remote script running [duplicate]

Through Fabric, I am trying to start a celerycam process using the below nohup command. Unfortunately, nothing is happening. Manually using the same command, I could start the process but not through Fabric. Any advice on how can I solve this?
def start_celerycam():
'''Start celerycam daemon'''
with cd(env.project_dir):
virtualenv('nohup bash -c "python manage.py celerycam --logfile=%scelerycam.log --pidfile=%scelerycam.pid &> %scelerycam.nohup &> %scelerycam.err" &' % (env.celery_log_dir,env.celery_log_dir,env.celery_log_dir,env.celery_log_dir))
I'm using Erich Heine's suggestion to use 'dtach' and it's working pretty well for me:
def runbg(cmd, sockname="dtach"):
return run('dtach -n `mktemp -u /tmp/%s.XXXX` %s' % (sockname, cmd))
This was found here.
As I have experimented, the solution is a combination of two factors:
run process as a daemon: nohup ./command &> /dev/null &
use pty=False for fabric run
So, your function should look like this:
def background_run(command):
command = 'nohup %s &> /dev/null &' % command
run(command, pty=False)
And you can launch it with:
execute(background_run, your_command)
This is an instance of this issue. Background processes will be killed when the command ends. Unfortunately on CentOS 6 doesn't support pty-less sudo commands.
The final entry in the issue mentions using sudo('set -m; service servicename start'). This turns on Job Control and therefore background processes are put in their own process group. As a result they are not terminated when the command ends.
For even more information see this link.
you just need to run
run("(nohup yourcommand >& /dev/null < /dev/null &) && sleep 1")
DTACH is the way to go. It's a software you need to install like a lite version of screen.
This is a better version of the "dtach"-method found above, it will install dtach if necessary. It's to be found here where you can also learn how to get the output of the process which is running in the background:
from fabric.api import run
from fabric.api import sudo
from fabric.contrib.files import exists
def run_bg(cmd, before=None, sockname="dtach", use_sudo=False):
"""Run a command in the background using dtach
:param cmd: The command to run
:param output_file: The file to send all of the output to.
:param before: The command to run before the dtach. E.g. exporting
environment variable
:param sockname: The socket name to use for the temp file
:param use_sudo: Whether or not to use sudo
"""
if not exists("/usr/bin/dtach"):
sudo("apt-get install dtach")
if before:
cmd = "{}; dtach -n `mktemp -u /tmp/{}.XXXX` {}".format(
before, sockname, cmd)
else:
cmd = "dtach -n `mktemp -u /tmp/{}.XXXX` {}".format(sockname, cmd)
if use_sudo:
return sudo(cmd)
else:
return run(cmd)
May this help you, like it helped me to run omxplayer via fabric on a remote rasberry pi!
You can use :
run('nohup /home/ubuntu/spider/bin/python3 /home/ubuntu/spider/Desktop/baidu_index/baidu_index.py > /home/ubuntu/spider/Desktop/baidu_index/baidu_index.py.log 2>&1 &', pty=False)
nohup did not work for me and I did not have tmux or dtach installed on all the boxes I wanted to use this on so I ended up using screen like so:
run("screen -d -m bash -c '{}'".format(command), pty=False)
This tells screen to start a bash shell in a detached terminal that runs your command
You could be running into this issue
Try adding 'pty=False' to the sudo command (I assume virtualenv is calling sudo or run somewhere?)
This worked for me:
sudo('python %s/manage.py celerycam --detach --pidfile=celerycam.pid' % siteDir)
Edit: I had to make sure the pid file was removed first so this was the full code:
# Create new celerycam
sudo('rm celerycam.pid', warn_only=True)
sudo('python %s/manage.py celerycam --detach --pidfile=celerycam.pid' % siteDir)
I was able to circumvent this issue by running nohup ... & over ssh in a separate local shell script. In fabfile.py:
#task
def startup():
local('./do-stuff-in-background.sh {0}'.format(env.host))
and in do-stuff-in-background.sh:
#!/bin/sh
set -e
set -o nounset
HOST=$1
ssh $HOST -T << HERE
nohup df -h 1>>~/df.log 2>>~/df.err &
HERE
Of course, you could also pass in the command and standard output / error log files as arguments to make this script more generally useful.
(In my case, I didn't have admin rights to install dtach, and neither screen -d -m nor pty=False / sleep 1 worked properly for me. YMMV, especially as I have no idea why this works...)

Fabric sudo() not respecting env.password

I'm trying to prefill env.password using --initial-password-prompt, but remote is throwing back some strangeness. Let's say that I'm trying to cat a root-owned file as testuser, with 600 permissions on the file. I'm calling sudo('cat /home/testuser/test.txt'), and getting this back:
[testuser#testserver] sudo: cat /home/testuser/test.txt
[testuser#testserver] out: cat: /home/testuser/test.txt: Permission denied
[testuser#testserver] out:
Fatal error: sudo() received nonzero return code 1 while executing!
Requested: cat /home/testuser/test.txt
Executed: sudo -S -p 'sudo password:' -u "testuser" /bin/bash -l -c "cat /home/testuser/test.txt"
Is that piping the prompt right back into the input? I tried using sudo() with pty=False to see if it was an issue with the pseudoterminal, but to no avail.
Here's the weird part: calling run('sudo cat /home/testuser/test.txt') and invoking fab without --initial-password-prompt passes back a password prompt from remote, and on entering the password, everything works fine.
Naturally, running ssh -t testuser#testserver 'sudo cat /home/user/test.txt' prompts for a password and returns the contents of the file correctly. Do I have an issue with my server's shell config, or is the issue with how I'm using sudo()?
Down the line, I'm likely to set up a deploy user with no-password sudo and restricted commands. That'll probably moot the issue, but I'd like to figure this one out if possible. I'm running an Ubuntu 14.10 VPS, in case that's relevant.
Oh, my mistake. I had foolishly set env.sudo_user to my deploy user testuser, thinking that it was specifying the invoking user on remote. In fact, it was specifying the target user, and I was attempting to sudo into myself. Whoops.

How to customize virtualenv shell prompt

How do you define a custom prompt to use when activating a Python virtual environment?
I have a bash script for activating a virtualenv I use when calling specific Fabric commands. I want the shell prompt to say something like "(fab)" so I can easily distinguish it from other shells I have open. Following this example, I've tried:
#!/bin/bash
script_dir=`dirname $0`
cd $script_dir
/bin/bash -c ". .env/bin/activate; PS1='(fab) '; exec /bin/bash -i"
but there's no change to the prompt. What am I doing wrong?
The prompt is set in the virtualenv's activate script (located in the bin folder under the virtualenv). If you only want to change the prompt some times, you could set an environment variable before calling activate (make sure to clear it in the corresponding deactivate file). If you simply want the prompt to be different all the time, you can do that right in activate at the line that looks like
set "PROMPT=(virtualenvname) %PROMPT%"
If you're using virtualenvwrapper, you could do all of this in the postactivate and postdeactivate scripts as well.
I couldn't find any way to do this via a script executed as a child process. Calling a separate bash process seems to forget any previously set PS1. However, it turned out to be trivial if I just sourced the script:
#!/bin/bash
script_dir=`dirname $0`
cd $script_dir
. .env/bin/activate
PS1="(fab) "
It appears the
exec /bin/bash -i
is resetting the PS1 variable. When I run
export PS1="foo "; bash
it resets it too. Curiously, when I look into the bash sources (shell.c and variables.c) it appears to use
set_if_not ("PS1", primary_prompt);
to init it. But I'm not exactly sure what happens between this and main(). Giving up.
I tried on cygwin and on linux (RedHat CentOS) as well. I found solution for both.
CYGWIN
After some investigation I found that the problem is that PS1 is set by /etc/bash.bashrc which overrides the PS1 env.var. So You need to disable to run this file using:
/bin/bash -c ". .env/bin/activate; PS1='(fab) ' exec /bin/bash -i --norc"
or
/bin/bash -c ". .env/bin/activate; export PS1='(fab) '; exec /bin/bash -i --norc"
LINUX
It works much simpler:
/bin/bash -c ". .env/bin/activate; PS1='(fab) ' exec /bin/bash -i"
or
/bin/bash -c ". .env/bin/activate; export PS1='(fab) '; exec /bin/bash -i"
If the script You are calling does not export the variables (and I suppose it does not) and the set variables does not appears in the environment then You could try something like this:
/bin/bash -c "PS1='(fab) ' exec /bin/bash --rcfile .env/bin/activate; "
I hope I could help!

Fabric: Run command as root automatically

I have some freebsd servers and don't have sudo. But I want to run some command automatically with root just like the following command:
def autodeploy(url):
with cd('/tmp'):
if not exists('releasetar.sh'):
put('/tmp/releasetar.sh', 'releasetar.sh', mode=0644)
run("wget '{}'".format(url))
run('su - -m -c "cd /tmp && bash /tmp/releasetar.sh"')
the su with -c option worked to linux but didn't worked on freebsd. How can I solved this problem ? I'm wish your solution can both worked on linux and freebsd. Thank you for your answer~~
If you're using fabric you can just provide the -u argument from the command line to specify which user you want to run the task as
fab -u root <task name>
For more options from the command line check out http://docs.fabfile.org/en/1.7/usage/fab.html#command-line-options
You can also set your username programmatically
from fabric.api import run, settings
with settings(user="root"):
run("some-command")
Install sudo from ports (/usr/ports/security/sudo).

ubuntu ec2 - run python script at startup with arguments

I have a python script i'd like to start on startup on an ubuntu ec2 instance but im running into troubles.
The script runs in a loop and takes care or exiting when its ready so i shouldn't need to start or stop it after its running.
I've read and tried a lot of approaches with various degrees of success and honestly im confused about whats the best approach. I've tried putting a shell script that starts the python script in /etc/init.d, making it executable and doing update-rc.d to try to get it to run but its failed at every stage.
here's the contents of the script ive tried:
#!/bin/bash
cd ~/Dropbox/Render\ Farm\ 1/appleseed/bin
while :
do
python ./watchfolder18.py -t ./appleseed.cli -u ec2 ../../data/
done
i then did
sudo chmod +x /etc/init.d/script_name
sudo sudo update-rc.d /etc/init.d/script_name defaults
This doesn't seem to run on startup and i cant see why, if i run the command manually it works as expected.
I also tried adding a line to rc.local to start the script but that doesn't seem to work either
Can anybody share what they have found is the simplest way to run a python script in the background with arguments on startup of an ec2 instance.
UPDATE: ----------------------
I've since moved this code to a file called /home/ubuntu/bin/watch_folder_start
#!/bin/bash
cd /home/ubuntu/Dropbox/Render\ Farm\ 1/appleseed/bin
while :
do
python ./watchfolder18.py -t ./appleseed.cli -u ec2 ../../data/
done
and changed my rc.local file to this:
nohup /home/ubuntu/bin/watch_folder_start &
exit 0
Which works when i manually run rc.local but wont fire on startup, i did chmod +x rc.local but that didn't change anything,
Your /etc/init.d/script_name is missing the plumbing that update-rc.d and so on use, and won't properly handle stop, start, and other init-variety commands, so...
For initial experimentation, take advantage of the /etc/init.d/rc.local script (which should be linked to by default from /etc/rc2/S99rc.local). The gets you out of having to worry about the init.d conventions and just add things to /etc/rc.local before the exit 0 at its end.
Additionally, that ~ isn't going to be defined, you'll need to use a full pathname - and furthermore the script will run as root. We'll address how to avoid this if desired in a bit. In any of these, you'll need to replace "whoeveryouare" with something more useful. Also be warned that you may need to prefix the python command with a su command and some arguments to get the process to run with the user id you might need.
You might try (in /etc/rc.local):
( if cd '/home/whoeveryouare/Dropbox/Render Farm 1/appleseed/bin' ; then
while : ; do
# This loop should respawn watchfolder18.py if it dies, but
# ideally one should fix watchfolder18.py and remove this loop.
python ./watchfolder18.py -t ./appleseed.cli -u ec2 ../../data/
done
else
echo warning: could not find watchfolder 1>&2
fi
) &
You could also put all that in a script and just call it from /etc/rc.local.
The first pass is roughly what you had, but if we assume that watchfolder18.py will arrange to avoid dying we can cut it down to:
( cd '/home/whoeveryouare/Dropbox/Render Farm 1/appleseed/bin' \
&& exec python ./watchfolder18.py -t ./appleseed.cli -u ec2 ../../data/ ) &
These aren't all that pretty, but it should let you get your daemon sorted out so you can debug it and so on, then come back to making a proper /etc/init.d or /etc/init script later. Something like this might work in /etc/init/watchfolder.conf, but I'm not yet facile enough to claim this is anything other than a rough stab at it:
# watchfolder - spawner for watchfolder18.py
description "watchfolder program"
start on runlevel [2345]
stop on runlevel [!2345]
script
if cd '/home/whoeveryouare/Dropbox/Render Farm 1/appleseed/bin' ; then
exec python ./watchfolder18.py -t ./appleseed.cli -u ec2 ../../data/0
fi
end script
I found that the best solution in the end was to use 'upstart' and create a file in etc/init called myfile.conf that contained the following
description "watch folder service"
author "Jonathan Topf"
start on startup
stop on shutdown
# Automatically Respawn:
respawn
respawn limit 99 5
script
HOST=`hostname`
chdir /home/ubuntu/Dropbox/Render\ Farm\ 1/appleseed/bin
exec /usr/bin/python ./watchfolder.py -t ./appleseed.cli -u $HOST ../../data/ >> /home/ubuntu/bin/ec2_server.log 2>&1
echo "watch_folder started"
end script
More info on using the upstart system here
http://upstart.ubuntu.com/
https://help.ubuntu.com/community/UbuntuBootupHowto
http://blog.joshsoftware.com/2012/02/14/upstart-scripts-in-ubuntu/

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