how to program wait and continue in this bash script - bash

I have two shell scripts say A and B. I need to run A in the background and run B in the foreground till A finishes its execution in the background. I need to repeat this process for couple of runs, hence once A finishes, I need to suspend current iteration and move to next iteration.
Rough idea is like this:
for((i=0; i< 10; i++))
do
./A.sh &
for ((c=0; c< C_MAX; c++))
do
./B.sh
done
continue
done
how do I use 'wait' and 'continue' so that B runs as many times while A is in the background and the entire process moves to next iteration once A finishes

Use the PID of the current background process:
./A.sh &
while ps -p $! >/dev/null; do
./B.sh
done

I am just translating your rough idea into bash scripting.
The core idea to implement the wait-continue mechanism (while ps -p $A_PID >/dev/null; do...) is taken from #thiton who posted an answer earlier to your question.
for i in `seq 0 10`
do
./A.sh &
A_PID=$!
for i in `seq 0 $C_MAX`
do
./B.sh
done
while ps -p $A_PID >/dev/null; do
sleep 1
done
done

Related

Why is the second bash script not printing its iteration?

I have two bash scripts:
a.sh:
echo "running"
doit=true
if [ $doit = true ];then
./b.sh &
fi
some-long-operation-binary
echo "done"
b.sh:
for i in {0..50}; do
echo "counting";
sleep 1;
done
I get this output:
> ./a.sh
running
counting
Why do I only see the first "counting" from b.sh and then nothing anymore? (Currently some-long-operation-binary just sleep 5 for this example). I first thought that due to setting b.sh in the background, its STDOUT is lost, but why do I see the first output? More importantly: is b.sh still running and doing its thing (its iteration)?
For context:
b.sh is going to poll a service provided by some-long-operation-binary, which is only available after some time the latter has run, and when ready, would write its content to a file.
Apologies if this is just rubbish, it's a bit late...
You should add #!/bin/bash or the like to b.sh that uses a Bash-like expansion, to make sure Bash is actually running the script. Otherwise there may be (indeed) only one loop iteration happening.
When you start a background process, it is usually a good practice to kill it and wait for it, no matter which way the script exits.
#!/bin/bash
set -e -o pipefail
declare -i show_counter=1
counter() {
local -i i
for ((i = 0;; ++i)); do
echo "counting $((i))"
sleep 1
done
}
echo starting
if ((show_counter)); then
counter &
declare -i counter_pid="${!}"
trap 'kill "${counter_pid}"
wait -n "${counter_pid}" || :
echo terminating' EXIT
fi
sleep 10 # long-running process

Checking and killing hanged background processes in a bash script

Say I have this pseudocode in bash
#!/bin/bash
things
for i in {1..3}
do
nohup someScript[i] &
done
wait
for i in {4..6}
do
nohup someScript[i] &
done
wait
otherThings
and say this someScript[i] sometimes end up hanging.
Is there a way I can take the process IDs (with $!)
and check periodically if the process is taking more than a specified amount of time after which I want to kill the hanged processes with kill -9 ?
Unfortunately the answer from #Eugeniu did not work for me, timeout gave an error.
However I found useful doing this routine, I'll post it here so anyone can take advantage of it if in my same problem.
Create another script which goes like this
#!/bin/bash
#monitor.sh
pid=$1
counter=10
while ps -p $pid > /dev/null
do
if [[ $counter -eq 0 ]] ; then
kill -9 $pid
#if it's still there then kill it
fi
counter=$((counter-1))
sleep 1
done
then in the main work you just put
things
for i in {1..3}
do
nohup someScript[i] &
./monitor.sh $! &
done
wait
In this way for any of your someScript you will have a parallel process that checks if it's still there every chosen interval (until maximum time decided by the counter) and that actually quit itself if the associated process dies (or gets killed)
One possible approach:
#!/bin/bash
# things
mypids=()
for i in {1..3}; do
# launch the script with timeout (3600s)
timeout 3600 nohup someScript[i] &
mypids[i]=$! # store the PID
done
wait "${mypids[#]}"

How to stop a running script which calls a infinite loop

I'm writing a bash script for kicking up an uncertain program. The run time of the program is unknown. The script will also kick up a while loop for using linux commands or perf to record something in a 1 second manner.
./my_app &
$i=1
while true;
do
perf stat -a -A -e writeback:writeback_dirty_page sleep $i >> out
done
How can I stop the while loop while my_app is finished? Thank you.
Make your while loop conditional on the process id of the app existing:
./my_app &
app_pid=$!
i=1
while ps -p $app_pid >/dev/null 2>&1
do
perf stat -a -A -e writeback:writeback_dirty_page sleep $i >> out
done
Get the pid using
echo $!
then
kill
you can send kill signal from my_app to the process that spawn my_app
Here is the real example
test.sh
#!/bin/bash
./my_app.sh $$ &
while [ 1 ]
do
echo running....
sleep 2
done
my_app.sh
#!/bin/bash
sleep 10
kill -9 $1

How can I run multiple bash scripts in unison?

I'm learning Bash for a Unix class, and I'm trying to figure out how to run a script, then run a second script while the first is running and have the two interact. To clarify, the scripts look like this:
#!/bin/bash
num = 1
trap exit 0 SIGINT SIGTERM
trap "{ echo &num ; num++; }" SIGUSR1
while :
do
sleep 2
done
and the second one:
#!/bin/bash
if ps | grep "$1" > /dev/null
then
kill -SIGUSR1 $1
else
echo "Process doesn't exist"
fi
exit 0
In case the code isn't correct, the general idea is for the first script to loop until it recieves a SIGINT or SIGTERM, and echo and increment a number whenever it receives a SIGUSR1. The second script takes a pid as an argument and checks if it exists, and sends a SIGUSR1 to the given process. The problem is that when I run the first script, I can't do anything unless I move it to the background with ctrl-z, but when it's there it doesn't seem to respond to any signal except a kill signal. Any ideas on how to make this work?
You can use mycommand & to run a script in the background. Ctrl-Z stops the script, but you can then use bg to let it run in the background. In either case, you can use fg to bring it to the foreground again.
Also note that you can't have spaces around the = in assignments, and you can use let num++ to increment num. You should also singlequote the command in trap, to prevent "$num" from expanding.
All in all:
#!/bin/bash
num=1
trap exit 0 SIGINT SIGTERM
trap '{ echo $num ; let num++; }' SIGUSR1
while :
do
sleep 2
done
Finally, you can more easily check if a pid exists by just using kill -0 pid, or just attempting to sigusr1 it and check the result, to avoid grep "123" matching the substring of pid "1234" and such.
You need to make the first script run in the background. When you press Ctrl+Z it is suspended. Then you can type "bg" to make it run in the background (it will stop again if it tries to read from standard input, to allow you to switch back to it with the "fg" command).
Another way is to start script1 already in the background like this:
$ ./script1 &
The ampersand starts a job in the background and returns you to the prompt immediately.
Look in the bash man page under "JOB CONTROL" (here's a copy) for more information on how this works. The key commands to deal with jobs from an interactive shell is "jobs", "fg", and "bg".

How do I terminate all the subshell processes?

I have a bash script to test how a server performs under load.
num=1
if [ $# -gt 0 ]; then
num=$1
fi
for i in {1 .. $num}; do
(while true; do
{ time curl --silent 'http://localhost'; } 2>&1 | grep real
done) &
done
wait
When I hit Ctrl-C, the main process exits, but the background loops keep running. How do I make them all exit? Or is there a better way of spawning a configurable number of logic loops executing in parallel?
Here's a simpler solution -- just add the following line at the top of your script:
trap "kill 0" SIGINT
Killing 0 sends the signal to all processes in the current process group.
One way to kill subshells, but not self:
kill $(jobs -p)
Bit of a late answer, but for me solutions like kill 0 or kill $(jobs -p) go too far (kill all child processes).
If you just want to make sure one specific child-process (and its own children) are tidied up then a better solution is to kill by process group (PGID) using the sub-process' PID, like so:
set -m
./some_child_script.sh &
some_pid=$!
kill -- -${some_pid}
Firstly, the set -m command will enable job management (if it isn't already), this is important, as otherwise all commands, sub-shells etc. will be assigned to the same process group as your parent script (unlike when you run the commands manually in a terminal), and kill will just give a "no such process" error. This needs to be called before you run the background command you wish to manage as a group (or just call it at script start if you have several).
Secondly, note that the argument to kill is negative, this indicates that you want to kill an entire process group. By default the process group ID is the same as the first command in the group, so we can get it by simply adding a minus sign in front of the PID we fetched with $!. If you need to get the process group ID in a more complex case, you will need to use ps -o pgid= ${some_pid}, then add the minus sign to that.
Lastly, note the use of the explicit end of options --, this is important, as otherwise the process group argument will be treated as an option (signal number), and kill will complain it doesn't have enough arguments. You only need this if the process group argument is the first one you wish to terminate.
Here is a simplified example of a background timeout process, and how to cleanup as much as possible:
#!/bin/bash
# Use the overkill method in case we're terminated ourselves
trap 'kill $(jobs -p | xargs)' SIGINT SIGHUP SIGTERM EXIT
# Setup a simple timeout command (an echo)
set -m
{ sleep 3600; echo "Operation took longer than an hour"; } &
timeout_pid=$!
# Run our actual operation here
do_something
# Cancel our timeout
kill -- -${timeout_pid} >/dev/null 2>&1
wait -- -${timeout_pid} >/dev/null 2>&1
printf '' 2>&1
This should cleanly handle cancelling this simplistic timeout in all reasonable cases; the only case that can't be handled is the script being terminated immediately (kill -9), as it won't get a chance to cleanup.
I've also added a wait, followed by a no-op (printf ''), this is to suppress "terminated" messages that can be caused by the kill command, it's a bit of a hack, but is reliable enough in my experience.
You need to use job control, which, unfortunately, is a bit complicated. If these are the only background jobs that you expect will be running, you can run a command like this one:
jobs \
| perl -ne 'print "$1\n" if m/^\[(\d+)\][+-]? +Running/;' \
| while read -r ; do kill %"$REPLY" ; done
jobs prints a list of all active jobs (running jobs, plus recently finished or terminated jobs), in a format like this:
[1] Running sleep 10 &
[2] Running sleep 10 &
[3] Running sleep 10 &
[4] Running sleep 10 &
[5] Running sleep 10 &
[6] Running sleep 10 &
[7] Running sleep 10 &
[8] Running sleep 10 &
[9]- Running sleep 10 &
[10]+ Running sleep 10 &
(Those are jobs that I launched by running for i in {1..10} ; do sleep 10 & done.)
perl -ne ... is me using Perl to extract the job numbers of the running jobs; you can obviously use a different tool if you prefer. You may need to modify this script if your jobs has a different output format; but the above output is also on Cygwin, so it's very likely identical to yours.
read -r reads a "raw" line from standard input, and saves it into the variable $REPLY. kill %"$REPLY" will be something like kill %1, which "kills" (sends an interrupt signal to) job number 1. (Not to be confused with kill 1, which would kill process number 1.) Together, while read -r ; do kill %"$REPLY" ; done goes through each job number printed by the Perl script, and kills it.
By the way, your for i in {1 .. $num} won't do what you expect, since brace expansion is handled before parameter expansion, so what you have is equivalent to for i in "{1" .. "$num}". (And you can't have white-space inside the brace expansion, anyway.) Unfortunately, I don't know of a clean alternative; I think you have to do something like for i in $(bash -c "{1..$num}"), or else switch to an arithmetic for-loop or whatnot.
Also by the way, you don't need to wrap your while-loop in parentheses; & already causes the job to be run in a subshell.
Here's my eventual solution. I'm keeping track of the subshell process IDs using an array variable, and trapping the Ctrl-C signal to kill them.
declare -a subs #array of subshell pids
function kill_subs() {
for pid in ${subs[#]}; do
kill $pid
done
exit 0
}
num=1 if [ $# -gt 0 ]; then
num=$1 fi
for ((i=0;i < $num; i++)); do
while true; do
{ time curl --silent 'http://localhost'; } 2>&1 | grep real
done &
subs[$i]=$! #grab the pid of the subshell
done
trap kill_subs 1 2 15
wait
While these is not an answer, I just would like to point out something which invalidates the selected one; using jobs or kill 0 might have unexpected results; in my case it killed unintended processes which in my case is not an option.
It has been highlighted somehow in some of the answers but I am afraid not with enough stress or it has been not considered:
"Bit of a late answer, but for me solutions like kill 0 or kill $(jobs -p) go too far (kill all child processes)."
"If these are the only background jobs that you expect will be running, you can run a command like this one:"

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