Running bash commands in the background without printing job and process ids

隐身守侯 提交于 2019-11-28 04:27:45

Not related to completion, but you could supress that output by putting the call in a subshell:

(echo "Hello I'm a background task" &)
shellter

In some newer versions of bash and in ksh93 you can surround it with a sub-shell or process group (i.e. { ... }).

/home/shellter $ { echo "Hello I'm a background task" & } 2>/dev/null
Hello I'm a background task
/home/shellter $

Building off of @shellter's answer, this worked for me:

tyler@Tyler-Linux:~$ { echo "Hello I'm a background task" & disown; } 2>/dev/null; sleep .1;
Hello I'm a background task
tyler@Tyler-Linux:~$

I don't know the reasoning behind this, but I remembered from an old post that disown prevents bash from outputting the process ids.

Building on the above answer, if you need to allow stderr to come through from the command:

f() { echo "Hello I'm a background task" >&2; }
{ f 2>&3 &} 3>&2 2>/dev/null

Based on this answer, I came up with the more concise and correct:

silent_background() {
    { 2>&3 "$@"& } 3>&2 2>/dev/null
    disown &>/dev/null  # Prevent whine if job has already completed
}
silent_background date

Try:

user@host:~$ read < <( echo "Hello I'm a background task" & echo $! )
user@host:~$ echo $REPLY
28677

And you have hidden both the output and the PID. Note that you can still retrieve the PID from $REPLY

Sorry for the response to an old post, but I figure this is useful to others, and it's the first response on Google.

I was having an issue with this method (subshells) and using 'wait'. However, as I was running it inside a function, I was able to do this:

function a {
    echo "I'm background task $1"
    sleep 5
}

function b {
    for i in {1..10}; do
        a $i &
    done
    wait
} 2>/dev/null

And when I run it:

$ b
I'm background task 1
I'm background task 3
I'm background task 2
I'm background task 4
I'm background task 6
I'm background task 7
I'm background task 5
I'm background task 9
I'm background task 8
I'm background task 10

And there's a delay of 5 seconds before I get my prompt back.

The subshell solution works, but I also wanted to be able to wait on the background jobs (and not have the "Done" message at the end). $! from a subshell is not "waitable" in the current interactive shell. The only solution that worked for me was to use my own wait function, which is very simple:

myWait() {
  while true; do
    sleep 1; STOP=1
    for p in $*; do
      ps -p $p >/dev/null && STOP=0 && break
    done
    ((STOP==1)) && return 0
  done
}

i=0
((i++)); p[$i]=$(do_whatever1 & echo $!)
((i++)); p[$i]=$(do_whatever2 & echo $!)
..
myWait ${p[*]}

Easy enough.

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