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1
 
[USER@SERVER ~] sleep 3 &
[1] 5232
[USER@SERVER ~] 
[1]+  Done                    sleep 3
[USER@SERVER ~]

How do I /dev/null these two messages?: 

[1] 5232
[1]+  Done                    sleep 3

p.s.: so I need the output of the process, but not the mentioned two lines!

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You only get those lines when you run it interactively. You don't if you run it from a script, for example. – derobert Jan 25 at 0:25

3 Answers

up vote 6 down vote accepted

It's not the program output, it's some useful shell information.

Anyway, those can be hided by using subshell and output redirection

( sleep 3 & ) > /dev/null 2>&1
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1  
"2>/dev/null" is the good thing, i need the output from the process – LanceBaynes Dec 10 '11 at 12:36
@LanceBaynes The process could also write something on stderr. – Chris Down Dec 10 '11 at 18:45
up vote 0 down vote

In bash or zsh, you can call disown %1 to tell the shell to forget about the job. Then the shell won't print any message about that job, nor will it show it when you run jobs or ever send a SIGHUP to it. In zsh, starting the job with &! instead of & is equivalent to calling disown on it immediately.

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up vote 0 down vote

Can't comment (yet) on @Gilles but it seems that & disown also works in bash:

sleep 3 & disown
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