I have a bash script (this_script.sh) that invokes multiple instances of another TCL script.
set -m
for vars in $( cat vars.txt );
do
exec tclsh8.5 the_script.tcl "$vars" &
done
while [ 1 ]; do fg 2> /dev/null; [ $? == 1 ] && break; done
The multi threading portion was taken from Aleksandr's answer on: Forking / Multi-Threaded Processes | Bash.
The script works perfectly (still trying to figure out the last line). However, this line is always displaed: exec tclsh8.5 the_script.tcl "$vars"
How do I hide that line? I tried running the script as :
bash this_script.sh > /dev/null
But this hides the output of the invoked tcl scripts too (I need the output of the TCL scripts).
I tried adding the /dev/null to the end of the statement within the for statement, but that too did not work either. Basically, I am trying to hide the command but not the output.
You should use $! to get the PID of the background process just started, accumulate those in a variable, and then wait for each of those in turn in a second for loop.
set -m
pids=""
for vars in $( cat vars.txt ); do
tclsh8.5 the_script.tcl "$vars" &
pids="$pids $!"
done
for pid in $pids; do
wait $pid
# Ought to look at $? for failures, but there's no point in not reaping them all
done
Related
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
I know I can run my bash script in the background by using bash script.sh & disown or alternatively, by using nohup. However, I want to run my script in the background by default, so when I run bash script.sh or after making it executable, by running ./script.sh it should run in the background by default. How can I achieve this?
Self-contained solution:
#!/bin/sh
# Re-spawn as a background process, if we haven't already.
if [[ "$1" != "-n" ]]; then
nohup "$0" -n &
exit $?
fi
# Rest of the script follows. This is just an example.
for i in {0..10}; do
sleep 2
echo $i
done
The if statement checks if the -n flag has been passed. If not, it calls itself with nohup (to disassociate the calling terminal so closing it doesn't close the script) and & (to put the process in the background and return to the prompt). The parent then exits to leave the background version to run. The background version is explicitly called with the -n flag, so wont cause an infinite loop (which is hell to debug!).
The for loop is just an example. Use tail -f nohup.out to see the script's progress.
Note that I pieced this answer together with this and this but neither were succinct or complete enough to be a duplicate.
Simply write a wrapper that calls your actual script with nohup actualScript.sh &.
Wrapper script wrapper.sh
#! /bin/bash
nohup ./actualScript.sh &
Actual script in actualScript.sh
#! /bin/bash
for i in {0..10}
do
sleep 10 #script is running, test with ps -eaf|grep actualScript
echo $i
done
tail -f 10 nohup.out
0
1
2
3
4
...
Adding to Heath Raftery's answer, what worked for me is a variation of what he suggested such as this:
if [[ "$1" != "-n" ]]; then
$0 -n & disown
exit $?
fi
For following bash statement:
tail -Fn0 /tmp/report | while [ 1 ]; do echo "pre"; exit; echo "past"; done
I got "pre", but didn't quit to the bash prompt, then if I input something into /tmp/report, I could quit from this script and get into bash prompt.
I think that's reasonable. the 'exit' make the 'while' statement quit, but the 'tail' still alive. If something input into /tmp/report, the 'tail' will output to pipe, then 'tail' will detect the pipe is close, then 'tail' quits.
Am I right? If not, would anyone provide a correct interpretation?
Is it possible to add anything into 'while' statement to quit from the whole pipe statement immediately? I know I could save the pid of tail into a temporary file, then read this file in the 'while', then kill the tail. Is there a simpler way?
Let me enlarge my question. If use this tail|while in a script file, is it possible to fulfill following items simultaneously?
a. If Ctrl-C is inputed or signal the main shell process, the main shell and various subshells and background processes spawned by the main shell will quit
b. I could quit from tail|while only at a trigger case, and preserve other subprocesses keep running
c. It's better not use temporary file or pipe file.
You're correct. The while loop is executing in a subshell because its input is redirected, and exit just exits from that subshell.
If you're running bash 4.x, you may be able to achieve what you want with a coprocess.
coproc TAIL { tail -Fn0 /tmp/report.txt ;}
while [ 1 ]
do
echo "pre"
break
echo "past"
done <&${TAIL[0]}
kill $TAIL_PID
http://www.gnu.org/software/bash/manual/html_node/Coprocesses.html
With older versions, you can use a background process writing to a named pipe:
pipe=/tmp/tail.$$
mkfifo $pipe
tail -Fn0 /tmp/report.txt >$pipe &
TAIL_PID=$!
while [ 1 ]
do
echo "pre"
break
echo "past"
done <$pipe
kill $TAIL_PID
rm $pipe
You can (unreliably) get away with killing the process group:
tail -Fn0 /tmp/report | while :
do
echo "pre"
sh -c 'PGID=$( ps -o pgid= $$ | tr -d \ ); kill -TERM -$PGID'
echo "past"
done
This may send the signal to more processes than you want. If you run the above command in an interactive terminal you should be okay, but in a script it is entirely possible (indeed likely) the the process group will include the script running the command. To avoid sending the signal, it would be wise to enable monitoring and run the pipeline in the background to ensure that a new process group is formed for the pipeline:
#!/bin/sh
# In Posix shells that support the User Portability Utilities option
# this includes bash & ksh), executing "set -m" turns on job control.
# Background processes run in a separate process group. If the shell
# is interactive, a line containing their exit status is printed to
# stderr upon their completion.
set -m
tail -Fn0 /tmp/report | while :
do
echo "pre"
sh -c 'PGID=$( ps -o pgid= $$ | tr -d \ ); kill -TERM -$PGID'
echo "past"
done &
wait
Note that I've replaced the while [ 1 ] with while : because while [ 1 ] is poor style. (It behaves exactly the same as while [ 0 ]).
Suppose I have test.sh as below. The intent is to run some background task(s) by this script, that continuously updates some file. If the background task is terminated for some reason, it should be started again.
#!/bin/sh
if [ -f pidfile ] && kill -0 $(cat pidfile); then
cat somewhere
exit
fi
while true; do
echo "something" >> somewhere
sleep 1
done &
echo $! > pidfile
and want to call it like ./test.sh | otherprogram, e. g. ./test.sh | cat.
The pipe is not being closed as the background process still exists and might produce some output. How can I tell the pipe to close at the end of test.sh? Is there a better way than checking for existence of pidfile before calling the pipe command?
As a variant I tried using #!/bin/bash and disown at the end of test.sh, but it is still waiting for the pipe to be closed.
What I actually try to achieve: I have a "status" script which collects the output of various scripts (uptime, free, date, get-xy-from-dbus, etc.), similar to this test.sh here. The output of the script is passed to my window manager, which displays it. It's also used in my GNU screen bottom line.
Since some of the scripts that are used might take some time to create output, I want to detach them from output collection. So I put them in a while true; do script; sleep 1; done loop, which is started if it is not running yet.
The problem here is now that I don't know how to tell the calling script to "really" detach the daemon process.
See if this serves your purpose:
(I am assuming that you are not interested in any stderr of commands in while loop. You would adjust the code, if you are. :-) )
#!/bin/bash
if [ -f pidfile ] && kill -0 $(cat pidfile); then
cat somewhere
exit
fi
while true; do
echo "something" >> somewhere
sleep 1
done >/dev/null 2>&1 &
echo $! > pidfile
If you want to explicitly close a file descriptor, like for example 1 which is standard output, you can do it with:
exec 1<&-
This is valid for POSIX shells, see: here
When you put the while loop in an explicit subshell and run the subshell in the background it will give the desired behaviour.
(while true; do
echo "something" >> somewhere
sleep 1
done)&
How can you suppress the Terminated message that comes up after you kill a
process in a bash script?
I tried set +bm, but that doesn't work.
I know another solution involves calling exec 2> /dev/null, but is that
reliable? How do I reset it back so that I can continue to see stderr?
In order to silence the message, you must be redirecting stderr at the time the message is generated. Because the kill command sends a signal and doesn't wait for the target process to respond, redirecting stderr of the kill command does you no good. The bash builtin wait was made specifically for this purpose.
Here is very simple example that kills the most recent background command. (Learn more about $! here.)
kill $!
wait $! 2>/dev/null
Because both kill and wait accept multiple pids, you can also do batch kills. Here is an example that kills all background processes (of the current process/script of course).
kill $(jobs -rp)
wait $(jobs -rp) 2>/dev/null
I was led here from bash: silently kill background function process.
The short answer is that you can't. Bash always prints the status of foreground jobs. The monitoring flag only applies for background jobs, and only for interactive shells, not scripts.
see notify_of_job_status() in jobs.c.
As you say, you can redirect so standard error is pointing to /dev/null but then you miss any other error messages. You can make it temporary by doing the redirection in a subshell which runs the script. This leaves the original environment alone.
(script 2> /dev/null)
which will lose all error messages, but just from that script, not from anything else run in that shell.
You can save and restore standard error, by redirecting a new filedescriptor to point there:
exec 3>&2 # 3 is now a copy of 2
exec 2> /dev/null # 2 now points to /dev/null
script # run script with redirected stderr
exec 2>&3 # restore stderr to saved
exec 3>&- # close saved version
But I wouldn't recommend this -- the only upside from the first one is that it saves a sub-shell invocation, while being more complicated and, possibly even altering the behavior of the script, if the script alters file descriptors.
EDIT:
For more appropriate answer check answer given by Mark Edgar
Solution: use SIGINT (works only in non-interactive shells)
Demo:
cat > silent.sh <<"EOF"
sleep 100 &
kill -INT $!
sleep 1
EOF
sh silent.sh
http://thread.gmane.org/gmane.comp.shells.bash.bugs/15798
Maybe detach the process from the current shell process by calling disown?
The Terminated is logged by the default signal handler of bash 3.x and 4.x. Just trap the TERM signal at the very first of child process:
#!/bin/sh
## assume script name is test.sh
foo() {
trap 'exit 0' TERM ## here is the key
while true; do sleep 1; done
}
echo before child
ps aux | grep 'test\.s[h]\|slee[p]'
foo &
pid=$!
sleep 1 # wait trap is done
echo before kill
ps aux | grep 'test\.s[h]\|slee[p]'
kill $pid ## no need to redirect stdin/stderr
sleep 1 # wait kill is done
echo after kill
ps aux | grep 'test\.s[h]\|slee[p]'
Is this what we are all looking for?
Not wanted:
$ sleep 3 &
[1] 234
<pressing enter a few times....>
$
$
[1]+ Done sleep 3
$
Wanted:
$ (set +m; sleep 3 &)
<again, pressing enter several times....>
$
$
$
$
$
As you can see, no job end message. Works for me in bash scripts as well, also for killed background processes.
'set +m' disables job control (see 'help set') for the current shell. So if you enter your command in a subshell (as done here in brackets) you will not influence the job control settings of the current shell. Only disadvantage is that you need to get the pid of your background process back to the current shell if you want to check whether it has terminated, or evaluate the return code.
This also works for killall (for those who prefer it):
killall -s SIGINT (yourprogram)
suppresses the message... I was running mpg123 in background mode.
It could only silently be killed by sending a ctrl-c (SIGINT) instead of a SIGTERM (default).
disown did exactly the right thing for me -- the exec 3>&2 is risky for a lot of reasons -- set +bm didn't seem to work inside a script, only at the command prompt
Had success with adding 'jobs 2>&1 >/dev/null' to the script, not certain if it will help anyone else's script, but here is a sample.
while true; do echo $RANDOM; done | while read line
do
echo Random is $line the last jobid is $(jobs -lp)
jobs 2>&1 >/dev/null
sleep 3
done
Another way to disable job notifications is to place your command to be backgrounded in a sh -c 'cmd &' construct.
#!/bin/bash
# ...
pid="`sh -c 'sleep 30 & echo ${!}' | head -1`"
kill "$pid"
# ...
# or put several cmds in sh -c '...' construct
sh -c '
sleep 30 &
pid="${!}"
sleep 5
kill "${pid}"
'
I found that putting the kill command in a function and then backgrounding the function suppresses the termination output
function killCmd() {
kill $1
}
killCmd $somePID &
Simple:
{ kill $! } 2>/dev/null
Advantage? can use any signal
ex:
{ kill -9 $PID } 2>/dev/null