From the command line, typing cat waits for user input.
But in the following script, wait ignores the background process.
#!/bin/bash
cat &
wait
echo "After wait"
This script immediately blasts right past the wait command. How can I make wait actually wait for the cat command to finish? I've tried waiting for the specific PID or job number, but the effect is the same.
That's because cat is exiting right away, because stdin is not inherited. Try this instead:
cat <&0 &
Related
I have below scripts ready with me -
1.sh:
echo "Good"
sleep 10
echo "Morning"
2.sh:
echo "Whats"
sleep 30
echo "Up"
script1.sh:
sh1.sh &
sh2.sh &
script2.sh:
echo "Hello world"
Requirement:
Execute script1.sh and do not wait for its completion or failure i.e., let the script run in background As soon as script1.sh is triggered the very next second execute the script2.sh.
./script1.sh
./script2.sh
Challenge:
./script2.sh keeps on waiting for completion of . ./script1.sh.
Like ./script2.sh I have lot of scripts to be run one after another but they should never wait for completion of ./script1.sh
Thanks,
B.J.
Just as youdid in 1.sh, you should append & after script1.sh:
#! /bin/bash
./script1.sh &
./script2.sh
exit 0
This will create a background process of script1.sh and continues in the main thread with script2.sh.
Usually, it a good practice not to leave background processes (unless they are long running servers, daemons, etc.). Better to make the parent script wait for all the children. Otherwise, you might have lot of orphan processes, which may use resources and have unintended consequences (e.g., open files, logging, ...)
Consider
#! /bin/bash
script1.sh &
script2.sh
script3.sh
wait # wait for any backgrounded processs
One immediate advantage is that killing the main script will also kill running script1 and script2. If for some reason the main script exit before all background childs are terminated, they can not be easily stopped (other then killing them by PID).
Also, using ps/pstree will show system status in clear way
I am writing a bash script to execute 2 commands at a time on 2 different terminal & original terminal wait for both 2 terminal to finish & then continue with remaining script.
I am able to open a different terminal with required command, however the original terminal seems not waiting for the 2nd one to complete & auto close before proceeding with remaining of the script.
#!/bin/bash
read -p "Hello"
read -p "Press enter to start sql installation"
for i in 1
do
xterm -hold -e mysql_secure_installation &
done
echo "completed installation"
Use the Bash wait command to cause the calling script to wait for a background process to complete. Your for loop implies that you may be launching multiple background processes in parallel even though in your question there's only one. Without any options wait will wait for all of them.
I wonder why you're launching the processes in xterm instead of directly.
I have a bash script with a loop that calls a hard calculation routine every iteration. I use the results from every calculation as input to the next. I need make bash stop the script reading until every calculation is finished.
for i in $(cat calculation-list.txt)
do
./calculation
(other commands)
done
I know the sleep program, and i used to use it, but now the time of the calculations varies greatly.
Thanks for any help you can give.
P.s>
The "./calculation" is another program, and a subprocess is opened. Then the script passes instantly to next step, but I get an error in the calculation because the last is not finished yet.
If your calculation daemon will work with a precreated empty logfile, then the inotify-tools package might serve:
touch $logfile
inotifywait -qqe close $logfile & ipid=$!
./calculation
wait $ipid
(edit: stripped a stray semicolon)
if it closes the file just once.
If it's doing an open/write/close loop, perhaps you can mod the daemon process to wrap some other filesystem event around the execution? `
#!/bin/sh
# Uglier, but handles logfile being closed multiple times before exit:
# Have the ./calculation start this shell script, perhaps by substituting
# this for the program it's starting
trap 'echo >closed-on-calculation-exit' 0 1 2 3 15
./real-calculation-daemon-program
Well, guys, I've solved my problem with a different approach. When the calculation is finished a logfile is created. I wrote then a simple until loop with a sleep command. Although this is very ugly, it works for me and it's enough.
for i in $(cat calculation-list.txt)
do
(calculations routine)
until [[ -f $logfile ]]; do
sleep 60
done
(other commands)
done
Easy. Get the process ID (PID) via some awk magic and then use wait too wait for that PID to end. Here are the details on wait from the advanced Bash scripting guide:
Suspend script execution until all jobs running in background have
terminated, or until the job number or process ID specified as an
option terminates. Returns the exit status of waited-for command.
You may use the wait command to prevent a script from exiting before a
background job finishes executing (this would create a dreaded orphan
process).
And using it within your code should work like this:
for i in $(cat calculation-list.txt)
do
./calculation >/dev/null 2>&1 & CALCULATION_PID=(`jobs -l | awk '{print $2}'`);
wait ${CALCULATION_PID}
(other commands)
done
In my bash script I am writing, I am trying to start a process (sleep) in the background and then suspend it. Finally, the process with be finished. For some reason through, when I send the kill command with the stop signal, it just keeps running as if it received nothing. I can do this from the command line, but the bash script is not working as intended.
sleep 15&
pid=$!
kill -s STOP $pid
jobs
kill -s CONT $pid
You can make it work by enabling 'monitor mode' in your script: set -m
Please see why-cant-i-use-job-control-in-a-bash-script for further information
My question is very similar to this one except that my background process was started from a script. I could be doing something wrong but when I try this simple example:
#!/bin/bash
set -mb # enable job control and notification
sleep 5 &
I never receive notification when the sleep background command finishes. However, if I execute the same directly in the terminal,
$ set -mb
$ sleep 5 &
$
[1]+ Done sleep 5
I see the output that I expect.
I'm using bash on cygwin. I'm guessing that it might have something to do with where the output is directed, but trying various output redirection, I'm not getting any closer.
EDIT: So I have more information about the why (thx to jkramer), but still looking for the how. How do I get "push" notification that a background process started from a script has terminated? Saving a PID to a file and polling is not what I'm looking for.
if you run it as source file . then it can notify. e.g.
cat foo.sh
#!/bin/bash
set -mb # enable job control and notification
sleep 5 &
. foo.sh
[1]+ Done sleep 5
The job control of your shell only affects processes controlled by your terminal, that is tasks started directly from the shell.
When the parent process (your script) dies, the init process automatically becomes the parent of the child process (your sleep command), effectively killing all output. Try this example:
[jkramer/sgi5k:~]# cat foo.sh
#!/bin/bash
sleep 20 &
echo $!
[jkramer/sgi5k:~]# bash foo.sh
19638
[jkramer/sgi5k:~]# ps -ef | grep 19638
jkramer 19638 1 0 23:08 pts/3 00:00:00 sleep 20
jkramer 19643 19500 0 23:08 pts/3 00:00:00 grep 19638
As you can see, the parent process of sleep after the script terminates is 1, the init process. If you need to get notified about the termination of your child process, you can save the content of the $! variable (PID of the last created process) in a file or somewhere and use the `wait´ command to wait for its termination.