Is there a way to determine (in bash) how much time is remaining on a process that is running for a specified time?
For example, some time after executing
caffeinate -s -t 8000 &
is there a command or technique for determining when my system will be allowed to sleep?
Bash won't know that caffeinate has a timer attached to it; for all it knows, -t refers to the number of times you'll place an Amazon order of Red Bull before the process exits five minutes later.
If you know this, however, you can detect when the command was started and do the math yourself.
$ sleep 45 &
[1] 16065
$ ps -o cmd,etime
CMD ELAPSED
sleep 45 00:03
ps -o cmd,etime 00:00
/bin/bash 6-21:11:11
On OS X, this will be ps -o command,etime; see the FreeBSD ps man page or Linux ps docs for details and other switches/options.
Related
The perf-stat command in linux runs until crtl-c is pressed. I am trying to use this command in script to profile a loop. The recommended solution to simulate sending crtl-c is to issue a kill command with -2 or -SIGINT flag.
However this does not work for me. I am on RHEL.
The script more or less looks as follows:
for i in {1..12}
do
pid=$1
perf stat -e dTLB-loads -p $pid > perf.out&
perf_pid=$!
sleep 10
kill -SIGINT $perf_pid
done
Even after the kill the perf process is still active. All the ctrl-c's are executed at the end when the script gets over.
Reading the man page for perf, I came across the --control option which seems is the proper approach to profile a portion of running command.
However, this option is not available on RHEL .
I was able find workaround by using the -INT option for kill mentioned here. For some reason -2 or -SIGINT doesn't work on RHEL.
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.)
I'm debugging a shell script
so I add set -x at the beginning
a code snippet are as below
tcpdump -i $interface host $ip_addr and 'port 80' -w flash/flash.pcap &
sudo -u esolve firefox /tor_capture/flash.html &
sleep $capture_time
but I noticed that the execution sequence is as below
++ sleep 5
++ sudo -u esolve firefox /tor_capture/flash.html
++ tcpdump -i eth0 host 138.96.192.56 and 'port 80' -w flash/flash.pcap
tcpdump: listening on eth0, link-type EN10MB (Ethernet), capture size 65535 bytes
so the execution sequence is reversed compared to the command sequence in the script
what is wrong with this and how to deal with it?
thanks!
Since those lines are being backgrounded, I think the output from set -x comes from the subshell that is spawned to run the program, and the main shell gets to the sleep command before the subshells have proceeded to the point that they generate the output. That would explain why the sleep command shows up first. With regards to the other two, I would think you might occasionally get them in the other order, as well, since there's no synchronization between the two - depending on how many CPUs you have, how busy the system is, etc., the timing between the subshells is pseudo-non-deterministic...
Do you need the first 2 lines to run as background processes?
If not, remove the & at the end and try again.
I have many processes and each of them take a lot of time to come up (5-10 min).
I am running my script in abc#abc1 and ssh to xyz#xyz1 to bring up the daemons.
There in the other machine(xyz#xyz1) I want to bring up 10 processes in parallel (call there startup scripts).
Then after 10 min I will check there status are they up or down.
I am doing this because I want the execution time of (my) script to be minimum.
How to do this using shell script with minimum amount of time ?
Thanks
Something like this should get your processes started:
for cmd in bin/proc1 bin/proc2 bin/procn; do
logfile=var/${cmd#bin/}.out
ssh xyz#xyz1 "bash -c '$cmd > $logfile 2>&1 &' && echo 'started $cmd in the background. See $logfile for its output.'"
done
I'm not a scripting expert and I was wondering what was an acceptable way to run a script for at most x milliseconds (and yet finish before x milliseconds if the script is done before the timeout).
I solved that problem using Bash in a way that I think is very hacky and I wonder if there's a better way to do it.
Basically I've got one shell script called sleep_kill.sh that takes a PID as the first argument and a timeout as its second argument and that does this:
sleep $2
kill -9 $1 2> /dev/null 1> /dev/null
So if the PID corresponds to a script that finishes before timing out, nothing is going to be killed (I take it that the OS shall not have the time to be reusing this PID for another [unrelated] process seen that it's 'cycling' through all the process IDs before starting to reuse them).
Anyway, then I call my script that may "hang" or timeout:
command_that_may_hang.sh
PID=$!
sleep_kill.sh $PID .3
wait $PID > /dev/null 2>&1
And I'll be waiting at most 300 ms for command_that_may_hang.sh. Yet if command_that_may_hang.sh took only 10 ms to execute, I won't be "stuck" for 300 ms.
It would be great if some shell expert could explain the drawbacks of this approach and what should be done instead.
Have a look at this script: http://www.pixelbeat.org/scripts/timeout
Note timeouts of less that one second are pretty much nonsensical on most systems due to scheduling delays etc. Note also that newer coreutils has the timeout command included and it has a resolution of 1 second.