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I am currently starting a background process within a subshell and was wondering how can I assign its pid number to a variable outside the subshell scope?
I have tried many different ways but MYPID always stays set to 0.
MYPID = 0;
({ sleep 2 & }; $MYPID=$!)
({ sleep 2 & }; echo $!) > $MYPID
The only way it returns a value is with:
$MYPID=$({ sleep 2 & }; echo $!)
However this discards the background process instruction and it will only return a result after 2 seconds.
bash shouldn't print the job status when non-interactive.
If that's indeed for an interactive bash, you can do:
{ pid=$(sleep 20 >&3 3>&- & echo "$!"); } 3>&1
We want sleep's stdout to go to where it was before, not the pipe that feeds the $pid variable. So we save the outer stdout in the file descriptor 3 (3>&1) and restore it for sleep inside the command substitution. So pid=$(...) returns as soon as echo terminates because there's nothing left with an open file descriptor to the pipe that feeds $pid.
However note that because it's started from a subshell (here in a command substitution), that sleep will not run in a separate process group. So it's not the same as running sleep 20 & with regards to I/O to the terminal for instance.
Maybe better would be to use a shell that supports spawning disowned background jobs like zsh where you can do:
sleep 20 &! pid=$!
With bash, you can approximate it with:
{ sleep 20 2>&3 3>&- & } 3>&2 2> /dev/null; pid=$!; disown "$pid"
bash outputs the [1] 21578 to stderr. So again, we save stderr before redirecting to /dev/null, and restore it for the sleep command. That way, the [1] 21578 goes to /dev/null but sleep's stderr goes as usual.
If you're going to redirect everything to /dev/null anyway, you can simply do:
{ apt-get update & } > /dev/null 2>&1; pid=$!; disown "$pid"
To redirect only stdout:
{ apt-get-update 2>&3 3>&- & } 3>&2 > /dev/null 2>&1; pid=$!; disown "$pid"
apt-get update instead sleep I usually would send that to > /dev/null however since it is sending ’2>&3 the output still showing?
– zanona
Sep 24 '14 at 14:49
apt-get update unfortunately when there is an error it won’t be logged since it is simple discarding everything. I was wondering about a way to keep the errors and discard the rest, however, I am not sure if that is possible. But I just noted that the process output [1] 23929 is sent to stderr so I believe this is a tricky thing :) Thanks once again for your help.
– zanona
Sep 24 '14 at 15:24
second line can't work
({ sleep 2 & }; $MYPID=$!)
the ( ... ) fork a subshell, MYPID is set, then subshell exit and it's value is lost.
third line didn't work either
({ sleep 2 & }; echo $!) > $MYPID
you just echo $! to a file named 0 on local dir.
The only answer I can provide is via a temporary file (as var can't be push back)
( { sleep 100 & } ; echo $! > u ) > /dev/null &> /dev/null
MYPID=$(<u)
you can replace u with any file.
mktemp or > /tmp/file Also what would be the best procedure after getting the PID? Should I remove the temporary file?
– zanona
Sep 24 '14 at 14:33
sleep 2 & mypid=$!? – Stéphane Chazelas Sep 24 '14 at 13:56