在后台运行 bash 命令而不打印作业和进程 ID

在 bash 的后台运行一个进程是相当容易的。

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

但是输出是冗长的。在第一行输出作业 id 和后台任务的进程 id，然后输出命令，最后输出作业 id 及其状态和触发作业的命令。

有没有一种方法可以抑制运行后台任务的输出，使输出看起来与结尾没有与号的输出完全一样？例如:

$ echo "Hello I'm a background task" &
Hello I'm a background task

我这样问的原因是，我想运行一个后台进程作为制表符完成命令的一部分，因此该命令的输出必须是不间断的才有意义。

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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 $
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最佳答案

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

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

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.

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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
小开

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

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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.

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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.

小开

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

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