I need to run a Unix application if a different application is running, from within a shell script. How do I do this?
If you know the name of the application and you are sure that only instantiation of it can be running at a given time, then the solution is pretty simple:
running="$(ps -aux | grep yourappname)"
if [ -n "$running" ] ; then
launch your secondary application
Obviously, you’d need to change ‘yourappname’ to the appropriate name of the app you seek, but that’s all you need to do.
If you could have more than one version of the process running or otherwise don’t have an easy way to identify it uniquely, then you have a greater challenge. For example, if you wanted to check for the Apache Web server (process name “httpd”) but wanted to make sure that you weren’t seeing a child process but the parent, then you’d also need to take into account the parent process ID (ppid) because the ppid of child processes is the parent process (which makes sense!): in the case of child Apache processes, their ppid’s are the process ID (pid) of the parent Apache process.
And that’s exactly how you do it. Use a flag like ‘-j’ to get the pid and ppid, use grep to pull out the matching processes by name, then check the ppid to identify which has the odd ppid out. Let me show you. Here’s a set of processes shown in this format:
$ ps -jax | grep http root 6706 1 0 Ss ?? 0:03.43 /usr/local/apache/bin/httpsd www 6707 6706 0 I ?? 0:11.91 /usr/local/apache/bin/httpsd www 6708 6706 0 S ?? 0:13.22 /usr/local/apache/bin/httpsd www 6709 6706 0 S ?? 0:13.46 /usr/local/apache/bin/httpsd www 6710 6706 0 I ?? 0:13.82 /usr/local/apache/bin/httpsd www 6711 6706 0 S ?? 0:12.36 /usr/local/apache/bin/httpsd www 19003 6706 0 S ?? 0:00.09 /usr/local/apache/bin/httpsd www 19046 6706 0 I ?? 0:00.01 /usr/local/apache/bin/httpsd www 19429 6706 0 S ?? 0:00.02 /usr/local/apache/bin/httpsd www 19432 6706 0 S ?? 0:00.04 /usr/local/apache/bin/httpsd www 19531 6706 0 I ?? 0:00.00 /usr/local/apache/bin/httpsd www 19612 6706 0 S ?? 0:00.01 /usr/local/apache/bin/httpsd www 19615 6706 0 S ?? 0:00.00 /usr/local/apache/bin/httpsd www 19616 6706 0 S ?? 0:00.00 /usr/local/apache/bin/httpsd www 19622 6706 0 S ?? 0:00.01 /usr/local/apache/bin/httpsd www 19624 6706 0 Z ?? 0:00.00 (httpsd)
Notice that the second column is the process ID and the third is the parent process id: as you can see, the parent process is 6706 in this instance.
You can identify this in your script (I’ve written about it here before) and then screen for a process by ID rather than by name (use the ‘-p process‘ flag to ps) then use the same basic test shown earlier to launch your secondary application, as needed.
Hope that helps you out!