Throw away standard output and error

Question

I have redircted my ouput using /dev/null in bash script but it is still throwing an error. Code is following

ps -p $proc | fgrep $proc> /dev/null
if [ $? -ne '0' ] ; then
......
fi    

below is error

error: list of process IDs must follow -p

Usage:
 ps [options]

 Try 'ps --help <simple|list|output|threads|misc|all>'
  or 'ps --help <s|l|o|t|m|a>'
 for additional help text.

For more details see ps(1).
Usage: fgrep [OPTION]... PATTERN [FILE]...
Try 'fgrep --help' for more information.

How can I suppress this error without affecting $? output?

Solution 1

You can use command grouping:

{ ps -p "$proc" | fgrep "$proc";} >/dev/null 2>&1

or wrap pipe in subshell:

(ps -p "$proc" | fgrep "$proc") >/dev/null 2>&1

Solution 2

&> /dev/null throws away both stderr and stdout. Same answer as the others, just a few characters shorter.

Comments

[Admin]

Not really an answer to your question but I think you'd better check if $proc is empty before using it.

[Admin]

@Aashu; What are you trying to do in the first place? - If you ps for a PID using -p the grep seems unnecessary. - You can do (for example): ps -p "$pid" >/dev/null 2>&1 || echo error, or: if ps -p "$pid" >/dev/null 2>&1 ; then echo okay ; else echo error ; fi.

[Admin]

Are you testing for existence of the PID $proc? If so then use test -n "$proc" && kill -0 "$proc"

[Aashu]

Still getting same error.

[Bazi]

edited the answer @Aashu, try it now

[Janis]

You don't need a subshell if all you want is redirecting I/O; you can use command grouping: { ps -p "$proc" | fgrep "$proc" ;} >/dev/null 2>&1.

[Aloha]

I never knew subshells exist. Now I know.

[cuonglm]

@Janis: Good point, forgot about it. Updated my answer. Thanks.