ssh and bring up many processes in parallel + solaris - shell

I have many processes and each of them take a lot of time to come up (5-10 min).
I am running my script in abc#abc1 and ssh to xyz#xyz1 to bring up the daemons.
There in the other machine(xyz#xyz1) I want to bring up 10 processes in parallel (call there startup scripts).
Then after 10 min I will check there status are they up or down.
I am doing this because I want the execution time of (my) script to be minimum.
How to do this using shell script with minimum amount of time ?
Thanks

Something like this should get your processes started:
for cmd in bin/proc1 bin/proc2 bin/procn; do
logfile=var/${cmd#bin/}.out
ssh xyz#xyz1 "bash -c '$cmd > $logfile 2>&1 &' && echo 'started $cmd in the background. See $logfile for its output.'"
done

Related

How to run Bash Script on startup and keep monitoring the results on the terminal

Due to some issues I wont elaborate here to not waste time, I made a bash script which will ping google every 10 minutes and if there is a response it will keep the loop running and if not then the PC will restart. After a lot of hurdle I have been able to make the script and also make it start on bootup. However the issue is that i want to see the results on the terminal, meaning I want to keep monitoring it but the terminal does not open on bootup. But it does open if I run it as ./net.sh.
The script is running on startup, that much I know because I use another script to open an application and it works flawlessly.
My system information
NAME="Linux Mint"
VERSION="18.3 (Sylvia)"
ID=linuxmint
ID_LIKE=ubuntu
PRETTY_NAME="Linux Mint 18.3"
VERSION_ID="18.3"
HOME_URL="http://www.linuxmint.com/"
SUPPORT_URL="http://forums.linuxmint.com/"
BUG_REPORT_URL="http://bugs.launchpad.net/linuxmint/"
VERSION_CODENAME=sylvia
UBUNTU_CODENAME=xenial
The contents of my net.sh bash script are
#! /bin/bash
xfce4-terminal &
sleep 30
while true
do
ping -c1 google.com
if [ $? == 0 ]; then
echo "Ping Sucessful. The Device will Continue Operating"
sleep 600
else
systemctl reboot
fi
done
I have put the scripts in /usr/bin and inserted the scripts for startup at boot in /etc/rc.local
So I did some further research and with help from reddit I realized that the reason I couldnt get it to show on terminal was because the script was starting on bootup but I needed it to start after user login. So I added the script on startup application (which can be found searching on start menu if thats whats it called). But it was still giving issues so I divided the script in two parts.
I put the net.sh script on startup and directed that script to open my main script which i named net_loop.sh
This is how the net.sh script looks
#! /bin/bash
sleep 20
xfce4-terminal -e usr/bin/net_loop.sh
And the net_loop.sh
#! /bin/bash
while true
do
ping -c1 google.com
if [ $? == 0 ]; then
echo "Ping Sucessful. The Device will Continue Operating"
sleep 600
else
systemctl reboot
fi
done
The results are the results of the net_loop.sh script are open in another terminal.
Note: I used help from this thread
If minute interval is usable why not use "cron" to start your?
$> crontab –e
or
$> sudo crontab –e

checking per ssh if a specific program is still running, in parallel

I have several machines where I have a program running. Every 30 seconds or so I want to check if those programs are still running. I use the following command to do that.
ssh ${USER}#${HOSTS[i]} "bash -c 'if [[ -z \"\$(pgrep -u ${USER} program)\" ]]; then exit 1; else exit 0; fi'"
Now running this on >100 machines takes a long time and I want to speed that up by checking in parallel. I am aware of '&' and 'parallel', but I am unsure how to retreive the return value (task completed or not).
The following lets all connections complete before starting any in the next batch, and thus can potentially wait for more than 30 seconds -- but should give you a good idea of how to do what you're looking for:
hosts=( host1 host2 host3 )
user=someuser
script="script you want to run on each remote host"
last_time=$(( SECONDS - 30 ))
while (( ( SECONDS - last_time ) >= 30 )) || \
sleep $(( 30 - (SECONDS - last_time) )); do
last_time=$SECONDS
declare -A pids=( )
for host in "${hosts[#]}"; do
ssh "${user}#${host}" "$script" & pids[$!]="$host"
done
for pid in "${!pids[#]}"; do
wait "$pid" || {
echo "Failure monitoring host ${pids[$pid]} at time $SECONDS" >&2
}
done
done
Now, bigger picture: Don't do that.
Almost every operating system has a process supervision framework. Ubuntu has Upstart; Fedora and CentOS 7 have systemd; MacOS X has launchd; runit, daemontools, and others can be installed anywhere (and are very, very easy to use -- look at the run scripts at http://smarden.org/runit/runscripts.html for examples).
Using these tools are the Right Way to monitor a process and ensure that it restarts whenever it exits: Unlike this (very high-overhead) solution they have almost no overhead at all, since they rely on the operating system notifying a process's parent when that process exits, rather than doing the work of polling for a process (and that only after all the overhead of connecting via SSH, negotiating a pair of session keys, starting a shell to run your script, etc, etc, etc).
Yes, this may be a small private project. Still, you're making extra complexity (and thus, extra bugs) for yourself -- and if you learn to use the tools to do this right, you'll know how to do things right when you have something that isn't a small private project.

Running processes simultaneously, Bash

I would like to run n processes (in my case simulations) simultaneously, using bash.
Right now this is what I'm running:
for file in $ini/SAN*.ini;
do
echo "Running $file...";
temp=$(basename $file .ini)
mosrun -G opp_run -r 0 -u Cmdenv -n ..:../../src -l ../../src/inet SAN.ini > $outputs/$temp.out;
done
Problem is, the loop only progresses to the next iteration after the simulation is done. Any suggestions? Thanks!
You should be able to run your command in the background by adding a & after it.
Should make them run in parallell, although in the background.
(Small side note: the processes will continue to run even if you abort the script, so you might want to add a trap to kill the processes if you hit for eg. ctrl-c when script is running. Look at bash manual.)

Kill process in bash that runs more than specified time?

I have a shutdown script for Oracle in /etc/init.d dir
on "stop" command it does:
su oracle -c "lsnrctl stop >/dev/null"
su oracle -c "sqlplus sys/passwd as sysdba #/usr/local/PLATEX/scripts/orastop.sql >/dev/null"
..
The problem is when lsnrctl or sqlplus are unresponsive - in this case this "stop" script just never ends and server cant shutdown. The only way - is to "kill - 9 " that.
I'd like to rewrite script so that after 5min (for example) if command is not finished - it should be terminated.
How I can achieve this? Could you give me an example?
I'm under Linux RHEL 5.1 + bash.
If able to use 3rd-party tools, I'd leverage one of the 3rd-party, pre-written helpers you can call from your script (doalarm and timeout are both mentioned by the BashFAQ entry on the subject).
If writing such a thing myself without using such tools, I'd probably do something like the following:
function try_proper_shutdown() {
su oracle -c "lsnrctl stop >/dev/null"
su oracle -c "sqlplus sys/passwd as sysdba #/usr/local/PLATEX/scripts/orastop.sql >/dev/null"
}
function resort_to_harsh_shutdown() {
for progname in ora_this ora_that ; do
killall -9 $progname
done
# also need to do a bunch of cleanup with ipcs/ipcrm here
}
# here's where we start the proper shutdown approach in the background
try_proper_shutdown &
child_pid=$!
# rather than keeping a counter, we check against the actual clock each cycle
# this prevents the script from running too long if it gets delayed somewhere
# other than sleep (or if the sleep commands don't actually sleep only the
# requested time -- they don't guarantee that they will).
end_time=$(( $(date '+%s') + (60 * 5) ))
while (( $(date '+%s') < end_time )); do
if kill -0 $child_pid 2>/dev/null; then
exit 0
fi
sleep 1
done
# okay, we timed out; stop the background process that's trying to shut down nicely
# (note that alone, this won't necessarily kill its children, just the subshell we
# forked off) and then make things happen.
kill $child_pid
resort_to_harsh_shutdown
wow, that's a complex solution. here's something easier. You can track the PID and kill it later.
my command & #where my command is the command you want to run and the & sign backgrounds it.
PID=$! #PID = last run command.
sleep 120 && doProperShutdown || kill $PID #sleep for 120 seconds and kill the process properly, if that fails, then kill it manually.. this can be backgrounded too.

Why can't I use job control in a bash script?

In this answer to another question, I was told that
in scripts you don't have job control
(and trying to turn it on is stupid)
This is the first time I've heard this, and I've pored over the bash.info section on Job Control (chapter 7), finding no mention of either of these assertions. [Update: The man page is a little better, mentioning 'typical' use, default settings, and terminal I/O, but no real reason why job control is particularly ill-advised for scripts.]
So why doesn't script-based job-control work, and what makes it a bad practice (aka 'stupid')?
Edit: The script in question starts a background process, starts a second background process, then attempts to put the first process back into the foreground so that it has normal terminal I/O (as if run directly), which can then be redirected from outside the script. Can't do that to a background process.
As noted by the accepted answer to the other question, there exist other scripts that solve that particular problem without attempting job control. Fine. And the lambasted script uses a hard-coded job number — Obviously bad. But I'm trying to understand whether job control is a fundamentally doomed approach. It still seems like maybe it could work...
What he meant is that job control is by default turned off in non-interactive mode (i.e. in a script.)
From the bash man page:
JOB CONTROL
Job control refers to the ability to selectively stop (suspend)
the execution of processes and continue (resume) their execution at a
later point.
A user typically employs this facility via an interactive interface
supplied jointly by the system’s terminal driver and bash.
and
set [--abefhkmnptuvxBCHP] [-o option] [arg ...]
...
-m Monitor mode. Job control is enabled. This option is on by
default for interactive shells on systems that support it (see
JOB CONTROL above). Background processes run in a separate
process group and a line containing their exit status is
printed upon their completion.
When he said "is stupid" he meant that not only:
is job control meant mostly for facilitating interactive control (whereas a script can work directly with the pid's), but also
I quote his original answer, ... relies on the fact that you didn't start any other jobs previously in the script which is a bad assumption to make. Which is quite correct.
UPDATE
In answer to your comment: yes, nobody will stop you from using job control in your bash script -- there is no hard case for forcefully disabling set -m (i.e. yes, job control from the script will work if you want it to.) Remember that in the end, especially in scripting, there always are more than one way to skin a cat, but some ways are more portable, more reliable, make it simpler to handle error cases, parse the output, etc.
You particular circumstances may or may not warrant a way different from what lhunath (and other users) deem "best practices".
Job control with bg and fg is useful only in interactive shells. But & in conjunction with wait is useful in scripts too.
On multiprocessor systems spawning background jobs can greatly improve the script's performance, e.g. in build scripts where you want to start at least one compiler per CPU, or process images using ImageMagick tools parallely etc.
The following example runs up to 8 parallel gcc's to compile all source files in an array:
#!bash
...
for ((i = 0, end=${#sourcefiles[#]}; i < end;)); do
for ((cpu_num = 0; cpu_num < 8; cpu_num++, i++)); do
if ((i < end)); then gcc ${sourcefiles[$i]} & fi
done
wait
done
There is nothing "stupid" about this. But you'll require the wait command, which waits for all background jobs before the script continues. The PID of the last background job is stored in the $! variable, so you may also wait ${!}. Note also the nice command.
Sometimes such code is useful in makefiles:
buildall:
for cpp_file in *.cpp; do gcc -c $$cpp_file & done; wait
This gives much finer control than make -j.
Note that & is a line terminator like ; (write command& not command&;).
Hope this helps.
Job control is useful only when you are running an interactive shell, i.e., you know that stdin and stdout are connected to a terminal device (/dev/pts/* on Linux). Then, it makes sense to have something on foreground, something else on background, etc.
Scripts, on the other hand, doesn't have such guarantee. Scripts can be made executable, and run without any terminal attached. It doesn't make sense to have foreground or background processes in this case.
You can, however, run other commands non-interactively on the background (appending "&" to the command line) and capture their PIDs with $!. Then you use kill to kill or suspend them (simulating Ctrl-C or Ctrl-Z on the terminal, it the shell was interactive). You can also use wait (instead of fg) to wait for the background process to finish.
It could be useful to turn on job control in a script to set traps on
SIGCHLD. The JOB CONTROL section in the manual says:
The shell learns immediately whenever a job changes state. Normally,
bash waits until it is about to print a prompt before reporting
changes in a job's status so as to not interrupt any other output. If
the -b option to the set builtin command is enabled, bash reports
such changes immediately. Any trap on SIGCHLD is executed for each
child that exits.
(emphasis is mine)
Take the following script, as an example:
dualbus#debian:~$ cat children.bash
#!/bin/bash
set -m
count=0 limit=3
trap 'counter && { job & }' CHLD
job() {
local amount=$((RANDOM % 8))
echo "sleeping $amount seconds"
sleep "$amount"
}
counter() {
((count++ < limit))
}
counter && { job & }
wait
dualbus#debian:~$ chmod +x children.bash
dualbus#debian:~$ ./children.bash
sleeping 6 seconds
sleeping 0 seconds
sleeping 7 seconds
Note: CHLD trapping seems to be broken as of bash 4.3
In bash 4.3, you could use 'wait -n' to achieve the same thing,
though:
dualbus#debian:~$ cat waitn.bash
#!/home/dualbus/local/bin/bash
count=0 limit=3
trap 'kill "$pid"; exit' INT
job() {
local amount=$((RANDOM % 8))
echo "sleeping $amount seconds"
sleep "$amount"
}
for ((i=0; i<limit; i++)); do
((i>0)) && wait -n; job & pid=$!
done
dualbus#debian:~$ chmod +x waitn.bash
dualbus#debian:~$ ./waitn.bash
sleeping 3 seconds
sleeping 0 seconds
sleeping 5 seconds
You could argue that there are other ways to do this in a more
portable way, that is, without CHLD or wait -n:
dualbus#debian:~$ cat portable.sh
#!/bin/sh
count=0 limit=3
trap 'counter && { brand; job & }; wait' USR1
unset RANDOM; rseed=123459876$$
brand() {
[ "$rseed" -eq 0 ] && rseed=123459876
h=$((rseed / 127773))
l=$((rseed % 127773))
rseed=$((16807 * l - 2836 * h))
RANDOM=$((rseed & 32767))
}
job() {
amount=$((RANDOM % 8))
echo "sleeping $amount seconds"
sleep "$amount"
kill -USR1 "$$"
}
counter() {
[ "$count" -lt "$limit" ]; ret=$?
count=$((count+1))
return "$ret"
}
counter && { brand; job & }
wait
dualbus#debian:~$ chmod +x portable.sh
dualbus#debian:~$ ./portable.sh
sleeping 2 seconds
sleeping 5 seconds
sleeping 6 seconds
So, in conclusion, set -m is not that useful in scripts, since
the only interesting feature it brings to scripts is being able to
work with SIGCHLD. And there are other ways to achieve the same thing
either shorter (wait -n) or more portable (sending signals yourself).
Bash does support job control, as you say. In shell script writing, there is often an assumption that you can't rely on the fact that you have bash, but that you have the vanilla Bourne shell (sh), which historically did not have job control.
I'm hard-pressed these days to imagine a system in which you are honestly restricted to the real Bourne shell. Most systems' /bin/sh will be linked to bash. Still, it's possible. One thing you can do is instead of specifying
#!/bin/sh
You can do:
#!/bin/bash
That, and your documentation, would make it clear your script needs bash.
Possibly o/t but I quite often use nohup when ssh into a server on a long-running job so that if I get logged out the job still completes.
I wonder if people are confusing stopping and starting from a master interactive shell and spawning background processes? The wait command allows you to spawn a lot of things and then wait for them all to complete, and like I said I use nohup all the time. It's more complex than this and very underused - sh supports this mode too. Have a look at the manual.
You've also got
kill -STOP pid
I quite often do that if I want to suspend the currently running sudo, as in:
kill -STOP $$
But woe betide you if you've jumped out to the shell from an editor - it will all just sit there.
I tend to use mnemonic -KILL etc. because there's a danger of typing
kill - 9 pid # note the space
and in the old days you could sometimes bring the machine down because it would kill init!
jobs DO work in bash scripts
BUT, you ... NEED to watch for the spawned staff
like:
ls -1 /usr/share/doc/ | while read -r doc ; do ... done
jobs will have different context on each side of the |
bypassing this may be using for instead of while:
for `ls -1 /usr/share/doc` ; do ... done
this should demonstrate how to use jobs in a script ...
with the mention that my commented note is ... REAL (dunno why that behaviour)
#!/bin/bash
for i in `seq 7` ; do ( sleep 100 ) & done
jobs
while [ `jobs | wc -l` -ne 0 ] ; do
for jobnr in `jobs | awk '{print $1}' | cut -d\[ -f2- |cut -d\] -f1` ; do
kill %$jobnr
done
#this is REALLY ODD ... but while won't exit without this ... dunno why
jobs >/dev/null 2>/dev/null
done
sleep 1
jobs

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