How do we avoid Cron jobs interruption? - parallel-processing

I am new at using Cron Jobs on Google Cloud: I was wondering if it is possible to launch a job on an instance and have it run continuously without interruption even after I shut down my local (Laptop). Is it possible to have a job running without any ssh connection?

The CronJobs are a possibility, but they are not meant to be used in your scenario, but when you want to run a command with a certain frequency over the time.
A Bash Builtin command that suits better your needs is disown. First, run your process/script in the background (using &, or stopping it with ^Z and then restarting with bg):
$ long_operation_command &
[1] 1156
Note that at this point the process is still linked to the session and in case it is closed it will be killed.
You can the process attached to the session check running jobs in the background:
$ jobs
[1]+ Running long_operation_command
Therefore you can run disown in order to detach the processes from the session:
$ disown
You can confirm this checking the result of your script or command logging in again or checking with top the process still running.
Check also this because it could be interesting, i.e. the difference between nohup foo, foo & and $ foo & disown
P.S.
The direct answer to your question is yes, the cronjobs run even if you shutdown your laptop/shutdown the session.

Related

How to run a script in background in Linux. - shell scripting

A script that keeps updating the log file. data like system time and date, users currently logged in etc for every interval of time say 5 minutes. THE SCRIPT MUST RUN EVEN AFTER THE TERMINAL HAS BEEN CLOSED.
Actually, no.
First of, you don't need sh:
$ ./newscript.sh &
This is enough. This will start a background process. But your terminal is still controlling it. To achieve the behavior you want, do this:
$ disown %1
This will disown the job with the jobspec 1 (which is like an id), which was the one you started beforehand. Now you can close the terminal.
Hurrah!! I would like to answer my question since i have got the solution.
For example, I'm running a script newscript.sh I want to run this in background and continue someother job in the terminal or i can close the terminal.
[yourname # username ~]$ sh newscript.sh &
and hit enter. You will get a PID and your job will be attached to the background.
To kill the same process, use the PID
For eg.,
kill 1205212
Thank you.

How to make a shell script wait for another with out using sleep

I want to know how to make a shell script wait till other script finishes its execution with out the help of sleep command.
suppose i have scripts run.sh and kill.sh, where run.sh will make all the processes up(means to start running the image on the box) whereas kill.sh contains just the kill commands to kill all the running processes.
Whenever i have run the run.sh, it will make all the processes up and it will end. Then what happens here is all the running processes becoming orphan(handled by init). Whenever we run kill.sh, some of the processes are becoming zombies.
Means, Orphan processes becoming zombies.
To avoid this, i want to make the run.sh wait till the end of kill.sh script.
So, How to make a shell script wait for another script ? Please provide the comments.
Thanks in Advance
You can use wait to let the first script finish without giving an explicit sleep.
#!/bin/bash
./first_script.sh
wait
./second_script.sh

Script which launches another application will bring it down on exit

I have a script which does launch another application using nohup my_app &, but when the initial script dies the launched process also goes down. As per my understanding since since it has been ran with nohup that should not happen. The original script also called with nohup.
What went wrong there?
A very reliable script that has been used successfully for years, and has always terminated after invoking a nohup uses this construct:
nohup ${BinDir}/${Watcher} >${DataDir}/${Watcher}.nohup.out 2>&1 &
Perhaps the problem is that output is not being managed?
nohup does not mean that a (child) process is still running when the (parent) process is killed. nohup is used f.e. when you're connecting over ssh to a server and there starting a process. If you log out, the process will terminate (logging out sents the signal SIGHUP to the process causing the process to terminate), using nohup avoid this behaviour and you're process is still running when you logged out.
If you need a program which runs in the background even it's parent process has terminated try using daemons.
It depends what my-app does - it might set its own signal mask. You probably know that nohup ignores the hang-up signal SIGHUP, and this is inherited by the target program. If that target program does its own signal handling then it might be setting SIGHUP to, for example SIG_DFT - the default action (which is to die).
To check, run strace -f -o out or truss -f -o out on the command. This will give you all the kernel calls in the file called 'out'. You should be able to spot the signal mask being changed if it is.

Are forked processes (bash) subject to server timeout disconnection?

If I am working on a remote server (ssh) and I fork a process using bash & operator, will that process be killed if I am booted off the server due to server time-out? I'm pretty sure the answer is yes, but would love to know if there are any juicy details.
It might depend, but generally when you log out with your "connection program" (e.g. ssh in your case although it could have been rlogin or telnet as well), the shell and children (I think?) will receive a SIGHUP signal (hangup) which will make them terminate when you log out. There are two common ways to avoid this, running the program you want to keep running through nohup or screen. If the server have some other time limitation on running processes you will have to look into that.
bash will send a HUP signal to all background jobs. You can stop this from happening by starting the job with nohup (which should have a man page). If it's too late for nohup, you can use disown to stop the shell from sending a HUP to a job. disown is a builtin, so help disown will tell you everything you need to know.

Why do unix background processes sometimes die when I exit my shell?

I wanted to know why i am seeing a different behaviour in the background process in Bash shell
Case 1: Logged in to Unix server using Putty(SSH)
By default it uses csh shell
I changed to bash shell
typed sleep 2000 &
press enter
It gave me the job number. Now i killed my session by clicking the x in the putty window
Now open another session and tried to lookup the process..the process died.
Case 2:Case 1: Logged in to Unix server using Putty(SSH)
By default it uses csh shell
I changed to bash shell
vi mysleep.sh
sleep 2000 & Saved mysleep.sh
./mysleep.sh
Diff here is..instead of executing the sleep command directly i am storing the sleep command in a file and executing the file.
Now i killed my session by clicking the x in the putty window
Now open another session and tried to lookup the process..the process is still there
Not sure why this is happening. I thought i need to do disown in bash to run the process even after logging out.
One diff i see in the parent process id..In the second case..the parent process id for the sleep 2000 becomes 1. Looks like as soon as process for mysleep.sh died the kernel assigned the parent process to 1.
The difference here is indeed the intervening process.
When you close the terminal window, a HUP signal (related to "nohup" as an0nymo0usc0ward mentioned) is sent to the processes running in it. The default action on receiving HUP is to die - from the signal(3) manpage,
No Name Default Action Description
1 SIGHUP terminate process terminal line hangup
In your first example, the sleep process directly receives this HUP signal and dies because it isn't set to do anything else. (Some processes catch HUP and use it to perform some action, e.g. reread some configuration files)
In the second example, the shell process running your shell script has already died, so the sleep process never gets the signal. In UNIX, every process must have a parent process due to the internals of how the wait(2) family of calls works and indeed processes in general. So when the parent process dies, the kernel gives it to init (pid 1, as you note) as a foster child.
Orphan process (on wikipedia) has some more information available about it, also see Zombie process for some additional technical details.
Already running process?
^z
bg
disown %<jobid>
New process/script (on local machine's console)?
nohup script.sh &
New process/script (on remote machine's console)?
Depending on your need,
there are two options [ there will be more ;-) ]
ssh remotehost 'nohup /path/to/script.sh </dev/null > nohup.out 2>&1 &'
OR
use 'screen'
Try "nohup cmd args..."
Steven's answer is correct, but I'd like to highlight the tricky part here again:
=> Using a bash script that just executes sleep in the background
The effect of this is that the "script" exits almost immediately (since it's done all its commands). However, it did create a child process (sleep) during its lifetime. The effect of this is that:
The "script" cannot be the parent anymore, and sleep is orphaned to init (which shows nicely in a pstree)
The bash shell where you started the script from has no underlying jobs anymore
Note that this stuff all happens when you executed the script, and has nothing to do with any ssh logout/putty closing.
When you then finally close your putty session, bash receives a "SIGHUP", but doesn't forward it to any other process (since there are no jobs left)
In the other case, bash did still have a job left, which it then sent the SIGHUP to, causing it to end (as you noticed)
Hope this helps

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