Shell script running binaries in multiple terminals

Question

I'm pretty new to shell scripts but I have some binaries I'd like to run in a loop over a set of different input files. I need them to be run in sequence: repeat after current one complete.

I got as far as trying to run two process in separate terminals and saving the pid but when I go to kill the process, it looks like the pids have changed.

#!/bin/bash
gnome-terminal -e "vi" & MAIN_PID=$!
gnome-terminal -e "gedit" & SIM_PID=$!
echo "MAIN Process ID is " $MAIN_PID
echo "SIM Process ID is " $SIM_PID
sudo kill $MAIN_PID

Terminal Output:

MAIN Process ID is  8532
SIM Process ID is  8542
# Option “-e” is deprecated and might be removed in a later version of gnome-terminal.
# Use “-- ” to terminate the options and put the command line to execute after it.
kill: (8532): No such process
user:~/Desktop$ ps -A | grep vi
 1026 ?        00:00:01 dconf-service
 8541 pts/3    00:00:00 vi
user:~/Desktop$ ps -A | grep gedit
 8550 pts/4    00:00:00 gedit

If there's an easier way to get what I want started, I'd like to know as well. I was planning to start these processes and check some condition to know when they have completed, kill the process and repeat with different set of inputs.

Answer 1

When you run gnome-terminal -e "vi" & MAIN_PID=$!, you are not capturing the PID of vi in MAIN_PID; you are capturing the PID of that instance of gnome-terminal. If you want to find the PID of a process owned by gnome_terminal, you could do some magic on the output of pstree:

pstree -p $MAIN_PID | grep -o 'vim([0-9]*) | grep -o '[0-9]*'

To check whether a process still exists, you can use kill -0 which will not send any signals, and either return with an exit code of 0 if the process exists, or 1 if it does not.