Timeline for Daemonize a process in shell?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3, 2018 at 5:19 | comment | added | Bruno Bronosky | @StoneThrow but doing this test echo "outer tty: $(tty)"; ls -la $(dirname $(tty)); bash -c 'echo "inner tty: $(tty)"; ls -la $(dirname $(tty));' will show you that it is the exact same TTY for both, but the demonstrated behavior of getting a TTY of ? still happens even though you're never "logging out from the shell" nor closing the TTY. | |
| Dec 3, 2018 at 5:12 | comment | added | Bruno Bronosky | @StoneThrow "without logging out from the shell from which the process was spawned" If you use the bash -c wrapper, there is no TTY associated with the process. It really is just like he demonstrated it. | |
| May 28, 2018 at 0:38 | vote | accept | Tim | ||
| Aug 23, 2017 at 22:22 | comment | added | StoneThrow | I'm just curious if you know a way to dissociate the process from the tty without logging out from the shell from which the process was spawned? | |
| May 23, 2017 at 12:40 | history | edited | CommunityBot | replaced http://stackoverflow.com/ with https://stackoverflow.com/ | |
| Apr 21, 2016 at 20:50 | comment | added | Gene Pauly | Take a look at daemonize. Besides a nice tool, it has pretty good explanations on what a daemon is. | |
| Apr 21, 2016 at 16:52 | review | Late answers | |||
| Apr 21, 2016 at 17:04 | |||||
| Apr 21, 2016 at 16:37 | review | First posts | |||
| Apr 21, 2016 at 17:20 | |||||
| Apr 21, 2016 at 16:37 | history | answered | Gene Pauly | CC BY-SA 3.0 |