I have a script called first.sh and this script calls another script using "./second.sh". In second.sh there are commands to play songs. For example, the content of second.sh could be:
play song1.mp3
play song2.mp3
...
I want to stop the script second.sh at certain times during the day, the problem is that using killall (and similar commands) do not help because the name of the script "second.sh" does not appear among the list of commands when I use "ps aux", I only see "play song1.mp3" and then "play song2.mp3" once song2 starts playing.
What can I do to stop second.sh using a command in the terminal? Or at least tie all the commands in it to a single process so I can kill that particular process?
Any help is appreciated, I've tried many ideas I found online but nothing seems to work.
Because you said :
at certain times during the day,
I would recommend crontab.
Use crontab -e and append the below line
0 12 * * * kill -9 `ps aux | awk '/play/{print $2}'`
This kills the parent shell that invoked play
The syntax for the crontab file is
m h dom mon dow command
where:
m - minute
h - hours
dom - day of month
mon - month
dow - day of week
command - the command that you wish to execute.
Edit
Or you could do something like this :
0 12 * * * killall -sSIGSTOP play
0 16 * * * killall -sSIGCONT play
which will pause all the play processes from 12 hours till 16 hours.
Requirement
You need to have the cron daemon up and running on your system.
You can save the pgid of the process explicitly, and then use the signals SIGSTOP and SIGCONT to start and stop the process group.
first.sh
#!/bin/bash
nohup ./second.sh > /dev/null 2>&1 &
echo $$ > /tmp/play.pid ### save process group id
second.sh
#!/bin/bash
play ...
play ...
third.sh
#!/bin/bash
case $1 in
(start)
kill -CONT -$(cat /tmp/play.pid)
;;
(stop)
kill -STOP -$(cat /tmp/play.pid)
;;
esac
Now you can launch and control the play as follows:
./first.sh
./third.sh stop
./third.sh start
You just need to stop second.sh and it will kill all its child processes automatically.
killall second.sh
Related
I want to search all the jobs which are running more that one hour. kill them. Then sleep for 60 mins. Again search if any job is running more than 60? loop the process.
If you want to find the PIDs for the processes running for more than 60 minutes on your linux box you can use a very simple and basic bash script like the one bellow:
#!/bin/sh
MIN=60
SEC=$((MIN*60))
ps -eo etimes=,pid= | while read sec pid; do
if [ ${sec} -gt ${SEC} ]; then
echo ${pid}
#kill -9 ${pid} # remove the # at the beginning of the line to actually kill those processes
fi
done
This will display the PIDs of the running processes, one per line
Assuming you name this script 60min.sh, you can run it every 60 minute using a cron job:
0 * * * * /bin/bash /path_to/60min.sh
This cron job will run your 60min.sh script every 60 minutes (or every hour)
Please keep in mind that you might accidentally kill system processes and your system might become unstable or unusable so you will have to reboot.
If you run different processes using a specific linux user I would recommend you to search the processes beloging to that user only and not to user root.
I need something like $command & stop This should execute a command and suspend it. The application later resumes back the command for complete results.
I understand that job can be suspended with stop signal to the corresponding pid.
$kill -SIGSTOP 12753
When we execute a command, we barely know its pid. There is extra command involved to take a pid and do the required. I want to avoid the extra command and a time interval.
Basically The application is for a measure of network performance. Trigger all the commands put them in halt mode. The halted commands are resumed back as per the kind of traffic needed.
The process ID of the most recently started background command is available in the shell parameter $!:
$ command & kill -SIGSTOP $!
(Check the documentation for your shell's implementation of kill for the correct format.)
Try killall with the --signal option where you can specify the name of the process.
linux:~ # killall
Usage: killall [OPTION]... [--] NAME...
killall -l, --list
killall -V, --version
-e,--exact require exact match for very long names
-I,--ignore-case case insensitive process name match
-g,--process-group kill process group instead of process
-i,--interactive ask for confirmation before killing
-l,--list list all known signal names
-q,--quiet don't print complaints
-r,--regexp interpret NAME as an extended regular expression
-s,--signal SIGNAL send this signal instead of SIGTERM
-u,--user USER kill only process(es) running as USER
-v,--verbose report if the signal was successfully sent
-V,--version display version information
-w,--wait wait for processes to die
Verified by starting md5sum in a shell session:
linux$ md5sum
and in another session, ran:
killall -s SIGSTOP md5sum
yielding the following in the md5sum session:
[1]+ Stopped md5sum
Kindly confirm if you want to halt your command or run in background(append '&' to your command)?
If your application is expected to start halted command later, then why dont you start your command(to be halted) in that application itself.
This helps :
sleep 5 & kill -SIGSTOP $!
In above, have executed sleep(demo command) for 5 seconds in background.
Next have send to kill for stopping it using its PID obtained by $!.
Demo & kludge using timeout, (for some reason timeout intereprets a '0s' duration as "run forever"), to stop yes before it outputs anything:
# run 'yes' command, let it print 5 numbered lines, but stop it immediately
timeout -s SIGSTOP .000000001s yes | head -n 5 | cat -n
Output (to STDERR):
[1]+ Stopped timeout -s SIGSTOP .000000001s yes | head -n 5 | cat -n
Now restart it:
fg > /dev/null
Output:
1 y
2 y
3 y
4 y
5 y
Technique for users stuck with v8.12 or earlier coreutils, (pre-2011), wherein timeout lacks sub-second intervals. Requires waiting a second.
Wrap the command string in a shell invocation, preceded by a 1s wait -- so timeout waits 1 second, and simultaneously, so does the command string. Total wait time 1 second:
timeout -s SIGSTOP 1s sh -c "sleep 1s; yes | head -n 5 | cat -n"
Output is the same as before, fg is the same too.
Finesse, if waiting even 1 second before sleeping is too much, it can be run in the background like so:
timeout -s SIGSTOP 1s sh -c "sleep 1s; yes | head -n 5 | cat -n" &
Output (process number will vary):
[1] 14601
Then after a second, the output will be the same as the previous two timeout examples.
Assuming you are using the same command, find the command name in ps output, you can launch it in one terminal then open a new terminal
ps -ely
after retrieving the command name:
command & kill -SIGSTOP $(pidof command_name)
pidof needs the exact command name to be able to find the pid.
then to resume it:
kill -SIGCONT $(pidof command_name)
if the command name is not constant, but there is a pattern, you can create a script like this, you can call it pof.sh:
ps -ely | grep $1 | tr -s ' ' | cut -d" " -f3
command & kill -SIGSTOP $(bash pof.sh pattern)
One drawback with this script, is that in case many lines match the pattern it will returns all of theirs pids, if this is a problem, you can put the output in an array and go on from there.
I want to run the script every 30 minutes with cron but I have problem with my code.
In every 30 min I have to kill old script and run it again. I have somethink like this, but it is not working:
cd /var/www/scripts
pkill -f bot
now="$(date +%Y%m%d%H%M%S)"
screen -S bot
node mybot.js >> logi/logi_$now.txt
You may not use screen for running things in background in a script. Use ampersand (&) to background a process and nohup so it won't be killed when cron script exits. Also remember a subprocess PID in a file.
Something like this:
kill -- "$(cat mybot.pid)"
now="$(date +%Y%m%d%H%M%S)"
nohup node mybot.js >> "logi/logi_$now.txt" &
echo $! > mybot.pid
use crontab :
crontab -l
*/30 * * * * /path/to/your/command
save and run
The line
node mybot.js >> logi/logi_$now.txt
is never reached, as screen -S <session name> will start a screen session and therefore a new shell and connect to it. The rest of the script would only execute once that 'inner' session terminates.
screen is more for interactive use. Calling it in a script like this is rather strange. I guess you want to have node mybot.js >> logi/logi_$now.txt running in the background, so that your script can terminate while node keeps running. See Redirecting stdout & stderr from background process and Node.js as a background service for options how to do that.
(OSX 10.7) An application we use let us assign scripts to be called when certain activities occur within the application. I have assigned a bash script and it's being called, the problem is that what I need to do is to execute a few commands, wait 30 seconds, and then execute some more commands. If I have my bash script do a "sleep 30" the entire application freezes for that 30 seconds while waiting for my script to finish.
I tried putting the 30 second wait (and the second set of commands) into a separate script and calling "./secondScript &" but the application still sits there for 30 seconds doing nothing. I assume the application is waiting for the script and all child processes to terminate.
I've tried these variations for calling the second script from within the main script, they all have the same problem:
nohup ./secondScript &
( ( ./secondScript & ) & )
( ./secondScript & )
nohup script -q /dev/null secondScript &
I do not have the ability to change the application and tell it to launch my script and not wait for it to complete.
How can I launch a process (I would prefer the process to be in a scripting language) such that the new process is not a child of the current process?
Thanks,
Chris
p.s. I tried the "disown" command and it didn't help either. My main script looks like this:
[initial commands]
echo Launching second script
./secondScript &
echo Looking for jobs
jobs
echo Sleeping for 1 second
sleep 1
echo Calling disown
disown
echo Looking again for jobs
jobs
echo Main script complete
and what I get for output is this:
Launching second script
Looking for jobs
[1]+ Running ./secondScript &
Sleeping for 1 second
Calling disown
Looking again for jobs
Main script complete
and at this point the calling application sits there for 45 seconds, waiting for secondScript to finish.
p.p.s
If, at the top of the main script, I execute "ps" the only thing it returns is the process ID of the interactive bash session I have open in a separate terminal window.
The value of $SHELL is /bin/bash
If I execute "ps -p $$" it correctly tells me
PID TTY TIME CMD
26884 ?? 0:00.00 mainScript
If I execute "lsof -p $$" it gives me all kinds of results (I didn't paste all the columns here assuming they aren't relevant):
FD TYPE NAME
cwd DIR /private/tmp/blahblahblah
txt REG /bin/bash
txt REG /usr/lib/dyld
txt REG /private/var/db/dyld/dyld_shared_cache_x86_64
0 PIPE
1 PIPE -> 0xffff8041ea2d10
2 PIPE -> 0xffff 8017d21cb
3r DIR /private/tmp/blahblah
4r REG /Volumes/DATA/blahblah
255r REG /Volumes/DATA/blahblah
The typical way of doing this in Unix is to double fork. In bash, you can do this with
( sleep 30 & )
(..) creates a child process, and & creates a grandchild process. When the child process dies, the grandchild process is inherited by init.
If this doesn't work, then your application is not waiting for child processes.
Other things it may be waiting for include the session and open lock files:
To create a new session, Linux has a setsid. On OS X, you might be able to do it through script, which incidentally also creates a new session:
# Linux:
setsid sleep 30
# OS X:
nohup script -q -c 'sleep 30' /dev/null &
To find a list of inherited file descriptors, you can use lsof -p yourpid, which will output something like:
sleep 22479 user 0u CHR 136,32 0t0 35 /dev/pts/32
sleep 22479 user 1u CHR 136,32 0t0 35 /dev/pts/32
sleep 22479 user 2u CHR 136,32 0t0 35 /dev/pts/32
sleep 22479 user 5w REG 252,0 0 1048806 /tmp/lockfile
In this case, in addition to the standard FDs 0, 1 and 2, you also have a fd 5 open with a lock file that the parent can be waiting for.
To close fd 5, you can use exec 5>&-. If you think the lock file might be stdin/stdout/stderr themselves, you can use nohup to redirect them to something else.
Another way is to abandon the child
#!/bin/bash
yourprocess &
disown
As far as I understand, the application replaces the normal bash shell because it is still waiting for a process to finish even if init should have taken care of this child process.
It could be that the "application" intercepts the orphan handling which is normally done by init.
In that case, only a parallel process with some IPC can offer a solution (see my other answer)
I think it depends on how your parent process tries to detect if your child process has been finished.
In my case (my parent process was gnu make), I succeed by closing stdout and stderr (slightly based on the answer of that other guy) like this:
sleep 30 >&- 2>&- &
You might also close stdin
sleep 30 <&- >&- 2>&- &
or additionally disown your child process (not for Mac)
sleep 30 <&- >&- 2>&- & disown
Currently tested only in bash on kubuntu 14.04 and Mac OSX.
If all else fails:
Create a named pipe
start the "slow" script independent from the "application", make sure executes it's task in an endless loop, starting with reading from the pipe. It will become read-blocked when it tries to read..
from the application, start your other script. When it needs to invoke the "slow" script, just write some data to the pipe. The slow script will start independently so your script won't wait for the "slow" script to finish.
So, to answer the question:
bash - how can I launch a new process that is NOT a child of the original process?
Simple: don't launch it but let an independent entity launch it during boot...like init or on the fly with the command at or batch
Here I have a shell
└─bash(13882)
Where I start a process like this:
$ (urxvt -e ssh somehost&)
I get a process tree (this output snipped from pstree -p):
├─urxvt(14181)───ssh(14182)
where the process is parented beneath pid 1 (systemd in my case).
However, had I instead done this (note where the & is) :
$ (urxvt -e ssh somehost)&
then the process would be a child of the shell:
└─bash(13882)───urxvt(14181)───ssh(14182)
In both cases the shell prompt is immediately returned and I can exit
without terminating the process tree that I started above.
For the latter case the process tree is reparented beneath pid 1 when
the shell exits, so it ends up the same as the first example.
├─urxvt(14181)───ssh(14182)
Either way, the result is a process tree that outlives the shell. The
only difference is the initial parenting of that process tree.
For reference, you can also use
nohup urxvt -e ssh somehost &
urxvt -e ssh somehost & disown $!
Both give the same process tree as the second example above.
└─bash(13882)───urxvt(14181)───ssh(14182)
When the shell is terminated the process tree is, like before, reparented
to pid 1.
nohup additionally redirects the process' standard output to a file
nohup.out so, if that is a useful trait, it may be a more useful choice.
Otherwise, with the first form above, you immediately have a completely
detached process tree.
I have a bash script that looks like this:
#!/bin/bash
wget LINK1 >/dev/null 2>&1
wget LINK2 >/dev/null 2>&1
wget LINK3 >/dev/null 2>&1
wget LINK4 >/dev/null 2>&1
# ..
# ..
wget LINK4000 >/dev/null 2>&1
But processing each line until the command is finished then moving to the next one is very time consuming, I want to process for instance 20 lines at once then when they're finished another 20 lines are processed.
I thought of wget LINK1 >/dev/null 2>&1 & to send the command to the background and carry on, but there are 4000 lines here this means I will have performance issues, not to mention being limited in how many processes I should start at the same time so this is not a good idea.
One solution that I'm thinking of right now is checking whether one of the commands is still running or not, for instance after 20 lines I can add this loop:
while [ $(ps -ef | grep KEYWORD | grep -v grep | wc -l) -gt 0 ]; do
sleep 1
done
Of course in this case I will need to append & to the end of the line! But I'm feeling this is not the right way to do it.
So how do I actually group each 20 lines together and wait for them to finish before going to the next 20 lines, this script is dynamically generated so I can do whatever math I want on it while it's being generated, but it DOES NOT have to use wget, it was just an example so any solution that is wget specific is not gonna do me any good.
Use the wait built-in:
process1 &
process2 &
process3 &
process4 &
wait
process5 &
process6 &
process7 &
process8 &
wait
For the above example, 4 processes process1 ... process4 would be started in the background, and the shell would wait until those are completed before starting the next set.
From the GNU manual:
wait [jobspec or pid ...]
Wait until the child process specified by each process ID pid or job specification jobspec exits and return the exit status of the last
command waited for. If a job spec is given, all processes in the job
are waited for. If no arguments are given, all currently active child
processes are waited for, and the return status is zero. If neither
jobspec nor pid specifies an active child process of the shell, the
return status is 127.
See parallel. Its syntax is similar to xargs, but it runs the commands in parallel.
In fact, xargs can run commands in parallel for you. There is a special -P max_procs command-line option for that. See man xargs.
You can run 20 processes and use the command:
wait
Your script will wait and continue when all your background jobs are finished.