script1.sh
script2.sh #launch another script
sudo reboot #reboot computer
script2.sh
# disown myself from script1
pkill script2.sh # hence preventing the reboot
# continue doing other stuff, even after the parent is dead
If possible, how is this done?
You don't need to disown the child script:
$ cat parent.sh
#!/usr/bin/env bash
echo 'Parent start'
./child.sh
echo 'Parent end'
$ cat child.sh
#!/usr/bin/env bash
echo 'Child start'
kill $PPID
sleep 1
echo 'Child end'
$ ./parent.sh
Parent start
Child start
Terminated
$ Child end
So the parent is terminated by the child, and one second later the child runs its own last line. The child has simply been reparented to the init process. If you up the timeout you can see the entry in the process table:
$ ps a | grep '[.]sh'
9752 pts/1 S 0:00 bash ./child.sh
Related
Creating a bash script with this command:
cat <<"END"> z
#! /bin/bash
sleep 20 && exit 1 &
ret=$!
ps $ret | grep $ret
END
and then running it gives:
7230 pts/39 S+ 0:00 /bin/bash ./z
I was expecting to see sleep 20 ... which is the child process. If I remove the && exit 1 it does return the child process.
Whats the reason? How can I get the child process id in above statement?
You already get the right information about the child process. Only in your case, ps doesn't know or want to show a proper COMMAND name for your chained sub-process you start in the background - what probably confused you.
Looks like this is the case with the chained commands (.. && ..., thus it has nothing to do with exit 1 could be also echo 5 etc.) where the process group leader name is showed as cmd name instead.
From the (ps man page)
`cmd | COMMAND`: simple name of executable
# Process state codes
`S`: interruptible sleep (waiting for an event to complete)
`+`: is in the foreground process group
See the S+ in your ps | grep output.
So, you can adapt your script a bit to confirm that you actually capture(d) the right information about the child process, like so:
cat <<"END"> z
#! /bin/bash
sleep 20 && exit 1 &
ret=$!
echo $ret
jobs -l
# display parent and child process info
# -j Jobs format
ps -j $$ $ret
END
Output of echo $ret:
30274
Output of jobs -l:
[1]+ 30274 Running sleep 20 && exit 1 &
Output of ps -j $$ $ret:
PID PGID SID TTY STAT TIME COMMAND
30273 30273 21804 pts/0 S+ 0:00 /bin/bash ./z
30274 30273 21804 pts/0 S+ 0:00 /bin/bash ./z
Note that both the parent and child have the same PGID, whereas the pid 30274 of the child process displayed by jobs -l and ps ... matches.
Further, if you change sleep 20 && exit 1 & as bash -c 'sleep 20 && exit 1' & you would get a proper command name for the child this time, as follows (cf. output order above):
30384
[1]+ 30384 Running bash -c 'sleep 20 && exit 1' &
PID PGID SID TTY STAT TIME COMMAND
30383 30383 21804 pts/0 S+ 0:00 /bin/bash ./z
30384 30383 21804 pts/0 S+ 0:00 bash -c sleep 20 && exit 1
Last but not least, in your original version instead of ps $ret | grep $ret you could also try
pstree -s $ret
From pstree man page
-s: Show parent processes of the specified process.
Which will provide you with an output similar to that one below, which would also confirm that you get the right process info for sleep 20 && exit 1 &:
systemd───systemd───gnome-terminal-───bash───bash───sleep
What you see is not parent pid, but sub-shell pid
When you run :
sleep 20 && exit 1 &
The processes tree is like :
current-shell ---> sub-shell ---> 'sleep 20 && exit 1'
When you run :
sleep 20 &
The processes tree is like :
current-shell ---> 'sleep 20'
Reason why you see pid for 'sleep 20'
Whats the reason?
The reason is that some entity has to do &&. It can't be sleep, because sleep only sleeps, and after sleep terminates (so there is no longer sleep to make any decision), some "entity" needs to compare the exit status of sleep and decide and then execute exit 1. That "entity" is the shell, that has to be "above" sleep to do the action. So the "real" background process is the shell, and sleep is it's child process.
In case of only sleep 20 & there is an optimization in bash that the parent shell in case bash sees there is only a single command to do. So bash scans the whole command command bla bla & and sees there is only one command to do. Because of that bash does only call to exec instead of the standard fork+exec and becomes sleep itself instead of running a child process. Because of the exec the subshell becomes sleep, so you see it in process name. It's a resource optimization done bash.
What is the shortest way to sleep a bash script at a certain location until another script wakes it up to continue it's job?
Mayby using flock -u .. or blocking read on a pipe ?
Say scriptA sleeps and waits for being waken up by scriptB.
One way is, in A, before you sleep, write the pid to some file say scriptA.pid then falling in sleep.
When B is running, at the right moment, you can read the scriptA.pid file, to get the pid of A, then do pkill -P pidofA sleep thus, the sleep sub-process will be killed, and A will continue its execution.
I'm a fan of named pipes (fifo). scriptA.sh:
pipe='/tmp/mypipe'
mkfifo "$pipe"
echo "$0 going to sleep..."
# Should block
read < "$pipe"
echo "$0 continuing"
scriptB.sh
pipe='/tmp/mypipe'
mkfifo "$pipe"
echo "$0 waking other process"
# might block
echo > "$pipe"
echo "$0 exiting"
You will get a mkfifo: /tmp/mypipe: File exists from the second mkfifo, if that bothers you then test for existence first (-e "$pipe"). This does not tidy-up (rm) the fifo, not sure where that should go because timing of the application is critical to where you put that.
You could use the inter process signals: the kill command should be used to send a signal to a process using its pid.
The SIGSTOP signal stops the execution of the process.
The SIGCONT signal resumes the process execution.
The example script below:
stores the pid of the process in a file.
the script sends to its own process the SIGSTOP signal ($$ is the pid of the current bash process).
Hopefully, another process will resume the execution.
Give a try to this:
#!/bin/bash --
printf "%s" $$ > /tmp/aScript.pid
kill -STOP $$ # STOP the execution here
# execution continues here when the SIGCONT signal is received
printf "script %s: received the SIGCONT signal\n" $$
Test in a terminal:
$ ./aScript.sh &
[1] 26444
$ kill -CONT $(cat /tmp/aScript.pid)
script 26444: received the SIGCONT signal
1st method
The running script can stop itself -
$: cat flagfile
#!/usr/bin/bash
echo $$ > /tmp/flagfile.pid
kill -STOP $$
date
$: ./flagfile &
[1] 24679
$: ps -fu $LOGNAME | grep 'flagfile$'
P2759474 24679 24521 0 13:29 pts/0 00:00:00 /usr/bin/bash ./flagfile
[1]+ Stopped ./flagfile
Then any other script can restart it.
$: kill -CONT $(</tmp/flagfile.pid)
$: Wed Dec 12 13:36:01 CST 2018
That last line gave me back a prompt before the background process managed to output the date. :)
2nd method
If a delay is ok, you can have a trap break it out.
This isn't totally stopping the script, but you can set the delay and make it as freindly as you have leeway to wait for it to wake up.
$: cat flagfile
#!/usr/bin/bash
trap 'loop=0' USR1
loop=1
delay=2
echo $$ > /tmp/flagfile.pid
while (( loop )); do sleep $delay; done
date
$: ./flagfile &
[1] 25018
$: ps -fu $LOGNAME | grep 'flagfile$'
P2759474 25018 24521 0 13:42 pts/0 00:00:00 /usr/bin/bash ./flagfile
Wait as long as you like....
$: kill -USR1 $(</tmp/flagfile.pid)
$: Wed Dec 12 13:42:43 CST 2018
[1]+ Done ./flagfile
i want to start programs in a chain, like start a bash script startScript.sh and then in the startScript.sh chain load certain programs.
startScript.sh
./program1 &
#PID=(echo $!)
PID=$!
echo "Wait for 10 seconds here"
if ps -p $PID > /dev/null; then echo "continue"; ./program2; else echo "PID is not running"; exit; fi
echo "Wait for 10 seconds here"
#if program1 is running and program2 is running, then ./program3, else exit.
I tried to find the running pid of program1 with $!, but the problem is that program1 is itself a shell script and they invokes further shell scripts in which I am not interested. Hence $! never gives me the pid of ./program1 but something irrelevant. HOw can I get the PID of program1?
Also how can I get the PID of program1 as well as program2 to see if they are running and then start program3.
The special parameter $! in shell (from man bash):
Expands to the process ID of the job most recently placed into the background, whether executed as an asynchronous command or using the bg builtin (see JOB CONTROL below).
It's important to notice that job control is shell-local (in your case local to the bash interpreter executing your startScript.sh).
That means the pid from below:
#!/bin/bash
./program.sh
pid=$!
will always contain the PID of the program.sh script (i.e. the process executing it, defined with a hashbang, in your case probably another bash process), regardless of the background processes spawned by that child process (remember that $! is local to your script; the $! in the child will be undefined, until a background process is spawned there).
What maybe sidetracked you initially is the line:
PID=(echo $!)
that didn't properly set the PID. The PID was set to an array containing two words (elements): echo and <pid>. You want this:
pid=$!
(and maybe to use lowercase name for the non-global variable pid).
A simple demo:
$ cat script1.sh
#!/bin/bash
./script2.sh &
echo "in script1: $!"
$ cat script2.sh
#!/bin/bash
echo "in script2: $!"
sleep 5 &
echo "in script2 after sleep: $!"
$ ./script1.sh
in script1: 19537
in script2:
in script2 after sleep: 19541
Notice how the $! in initially undefined in script2.sh - if $! were to return the last PID globally, it would be set to PID of script2.sh.
(or How to kill the child process)?
inotifywait -mqr --format '%w %f %e' $feedDir | while read dir file event
do
#something
done &
echo $! #5431
ps eg:
>$ ps
PID TTY TIME CMD
2867 pts/3 00:00:02 bash
5430 pts/3 00:00:00 inotifywait
5431 pts/3 00:00:00 bash
5454 pts/3 00:00:00 ps
It seems if I kill 5431 then 5430 (inotifywait) will be left running, but if I kill 5430 then both processes die. I don't suppose I can reliably assume that the pid of inotifywait will always be 1 less than $!?
When we run a pipe, each command is executed in a separated process. The interpreter waits for the last one but if we use ampersand (&).
cmd1 | cmd2 &
The pid of processes will be probably close, but we cannot assume it reliably. In the case where the last command is a bash reserved word as while, it creates a dedicated bash (that's why your 'dir', 'file' variables won't exist after the done keyword). Example:
ps # shows one bash process
echo "azerty" | while read line; do ps; done # shows one more bash
When the first command exits, the second one will terminate because the read on the pipe return EOF.
When the second command exits, the first command will be terminated by the signal SIGPIPE (write on a pipe with no reader) when it tries to write to the pipe. But if the command waits indefinitely... it is not terminated.
echo "$!" prints the pid of the last command executed in background. In your case, the bash process that is executing the while loop.
You can find the pid of "inotifywait" with the following syntax. But it's uggly:
(inotifywait ... & echo "$!">inotifywait.pid) | \
while read dir file event
do
#something
done &
cat inotifywait.pid # prints pid of inotifywait
If you don't want the pid, but just be sure the process will be terminated, you can use the -t option of inotifywait:
(while true; do inotifywait -t 10 ...; done)| \
while read dir file event
do
#something
done &
kill "$!" # kill the while loop
None of this solution are nice. What is your real achievement? Maybe we can find a more elegant solution.
If your goal is to make sure all of the children can be killed or interrupted elegantly. If you're using BusyBox's Ash, you don't have process substitution. If you don't want to use an fd either, check out this solution.
#!/bin/sh
pid=$$
terminate() {
pkill -9 -P "$pid"
}
trap terminate SIGHUP SIGINT SIGQUIT SIGTERM
# do your stuff here, note: should be run in the background {{{
inotifywait -mqr --format '%w %f %e' $feedDir | while read dir file event
do
#something
done &
# }}}
# Either pkill -9 -P "$pid" here
wait
# or pkill -9 -P "$pid" here
Or in another shell:
kill <pid ($$)>
I am trying to write a wrapper which will execute a script as a session leader.
I am confused by the behaviour of the linux command setsid. Consider this script, called test.sh:
#!/bin/bash
SID=$(ps -p $$ --no-headers -o sid)
if [ $# -ge 1 -a $$ -ne $SID ] ; then
setsid bash test.sh
echo pid=$$ ppid=$PPID sid=$SID parent
else
sleep 2
echo pid=$$ ppid=$PPID sid=$SID child
sleep 2
fi
The output differs depending on whether it is executed or sourced:
$ bash
$ SID=$(ps -p $$ --no-headers -o sid)
$ echo pid=$$ ppid=$PPID sid=$SID
pid=9213 ppid=9104 sid= 9104
$ ./test.sh 1 ; sleep 5
pid=9326 ppid=9324 sid= 9326 child
pid=9324 ppid=9213 sid= 9104 parent
$ . ./test.sh 1 ; sleep 5
pid=9213 ppid=9104 sid= 9104 parent
pid=9336 ppid=1 sid= 9336 child
$ echo $BASH_VERSION
4.2.8(1)-release
$ exit
exit
So, it seems to me that setsid returns immediately when the script is sourced, but it waits for its child when the script is executed.
Why would the presence of a controlling tty have anything to do with setsid? Thanks!
Edit: For clarification I added pid/ppid/sid reporting to all relevant commands.
The source code of the setsid utility is actually very straightforward. You'll note that it only fork()s if it sees that its process ID and process-group ID are equal (i.e., if it sees that it's a process group leader) — and that it never wait()s for its child process: if it fork()s, then the parent process just returns immediately. If it doesn't fork(), then it gives the appearance of wait()ing for a child, but really what happens is just that it is the child, and it's Bash that's wait()ing (just as it always does). (Of course, when it really does fork(), Bash can't wait() for the child it creates, because processes wait() for their children, not their grandchildren.)
So the behavior that you're seeing is a direct consequence of a different behavior:
when you run . ./test.sh or source ./test.sh or whatnot — or for that matter, when you just run setsid directly from the Bash prompt — Bash will launch setsid with a new process-group-ID for job control purposes, so setsid will have the same process-ID as its process-group-ID (that is, it's a process group leader), so it will fork() and won't wait().
when you run ./test.sh or bash test.sh or whatnot and it launches setsid, setsid will be part of the same process group as the script that's running it, so its process-ID and process-group-ID will be different, so it won't fork(), so it'll give the appearance of waiting (without actually wait()ing).
The behavior I observe is what I expect, though different from yours. Can you use set -x to make sure you're seeing things right?
$ ./test.sh 1
child
parent
$ . test.sh 1
child
$ uname -r
3.1.10
$ echo $BASH_VERSION
4.2.20(1)-release
When running ./test.sh 1, the script's parent — the interactive shell — is the session leader, so $$ != $SID and the conditional is true.
When running . test.sh 1, the interactive shell is executing the script in-process, and is its own session leader, so $$ == $SID and the conditional is false, thus never executing the inner child script.
I do not see any problem with your script as is. I added extra statements in your code to see what is happening:
#!/bin/bash
ps -H -o pid,ppid,sid,cmd
echo '$$' is $$
SID=`ps -p $$ --no-headers -o sid`
if [ $# -ge 1 -a $$ -ne $SID ] ; then
setsid bash test.sh
echo pid=$$ ppid=$PPID sid=$SID parent
else
sleep 2
echo pid=$$ ppid=$PPID sid=$SID child
sleep 2
fi
The case that concerns you is:
./test.sh 1
And trust me run this modified script and you will see exactly what is happening. If the shell which is not a session leader runs the script then it simply goes to else block. Am I missing something?
I see now what you mean: When you do ./test.sh 1 with your script as is then parent waits for the child to complete. child blocks the parent. But if you start the child in background then you will notice that parent completes before child. So just make this change in your script:
setsid bash test.sh &