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

Question

To run a process in the background in bash is fairly easy.

$ echo \"Hello I\'m a background task\" &
[1] 2076
Hello I\'m a background task
[1]+  Done             


        

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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

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

Answer 6

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.