Creating a script that will execute a program and if it exits without crashing, will run it again

Question

I've already asked this question in the UNIx exchange but I was told to ask here also about AppleScripts if the standard shell does not work.

I require a script that will open an application. Wait for it to exit, if the application exited gracefully (I.E: did not crash) relaunch it. Otherwise, keep the crash report open and exit the script. I'd like this script to run until I quit it manually.

So far the following has been suggested:

while open -W /path/to/application.app
do
    :
done

But this opens the application and then either on crash or exit will automatically run it again. I assume this is because open regardless of waiting or not will terminate with a normal 0 exit code. Unless of course open crashes but in this case it will not.

Any assistance with this is appreciated.

Thanks.

Answer 1

To keep restarting the job on succesful exit You can use launchd on OSX. For this You have to prepare *.plist file in order to "tell" launchd how to behave. To restart application on succesfull exit please look at KeepAlive -> SuccessfulExit key (more info: https://developer.apple.com/library/mac/documentation/Darwin/Reference/ManPages/man5/launchd.plist.5.html). To start your job, You'll need to start it via launchctl command, for example:

launchctl load path/to/file.plist

To stop it:

launchctl unload path/to/file.plist

More informations about launchctl/launchd programming here:

http://alvinalexander.com/mac-os-x/launchd-examples-launchd-plist-file-examples-mac

http://launchd.info