Prevent grep returning an error when input doesn't match

Question

I want to write in a bash script a piece of code that checks if a program is already running. I have the following in order to search whether bar is running

Answer 1

Try to make so:

ps auxw | grep -v grep | cat

cat returns always 0 and ignores exit code of grep

Answer 2

Short answer

Write

ps -ef | grep bar | { grep -v grep || test $? = 1; }

if you are using set -e.

If you use bash's pipefail option (set -o pipefail), remember to apply the exception handling (||test) to every grep in the pipeline:

ps -ef | { grep bar || test $? = 1; } | { grep -v grep || test $? = 1; }

In shell scripts I suggest you to use the ”catch-1-grep“ (c1grep) utility function:

c1grep() { grep "$@" || test $? = 1; }

---

Explained

grep's exit status is either 0, 1 or 2: [1]

• 0 means a line is selected
• 1 means no lines were selected
• 2 means an error occurred

grep can also return other codes if it's interrupted by a signal (e.g. 130 for SIGINT).

Since we only want to ignore exit status 1, we use test to suppress that specific exit status.

• If grep returns 0, test is not run.
• If grep returns 1, test is run and returns 0.
• If grep returns any other value, test is run and returns 1.

In the last case, the script will exit immediately due to set -e or set -o pipefail. However, if you don't care about grep errors at all, you can of course write

ps -ef | grep bar | { grep -v grep || true; }

as suggested by Sean.

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[additional] usage in shell scripts

In shell scripts, if you are using grep a lot, I suggest you to define an utility function:

# "catch exit status 1" grep wrapper
c1grep() { grep "$@" || test $? = 1; }

This way your pipe will get short & simple again, without losing the features of set -e and set -o pipefail:

ps -ef | c1grep bar | c1grep -v grep

FYI:

• I called it c1grep to emphasize it's simply catching exit status 1, nothing else.
• I could have called the function grep instead (grep() { env grep "$@" ...; }), but I prefer a less confusing and more explicit name, c1grep.

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[additional] ps + grep

So if you want to know how to avoid grep -v grep or even the | grep part of ps|grep at all, take a look at some of the other answers; but this is somewhat off-topic imho.

---

[1] grep manpage

Answer 3

Sure:

ps -ef | grep bar | { grep -v grep || true; }

Or even:

ps -ef | grep bar | grep -v grep | cat

Answer 4

Why ask ps to provide massive amounts of output with -ef if you only are going to throw away 99% of it? ps and especially the GNU version is a swiss army knife of handy functionality. Try this:

ps -C bar -o pid= 1>/dev/null

I specify -o pid= here just because, but in fact it's pointless since we throw away all of stdout anyway. It would be useful if you wanted to know the actual running PID, though.

ps automatically will return with a non-zero exist status if -C fails to match anything and with zero if it matches. So you could simply say this

ps -C bar 1>/dev/null && echo bar running || echo bar not running

Or

if ps -C bar 1>/dev/null ; then
    echo bar running
else
    echo bar not running
fi

Isn't that simpler? No need for grep, not twice or even once.

Answer 5

foo=`ps -ef | grep bar | grep -v grep` || true

Answer 6

A good trick to avoid grep -v grep is this:

ps -ef | grep '[b]ar'

That regular expression only matches the string "bar". However in the ps output, the string "bar" does not appear with the grep process.

---

In the days before I learned about pgrep, I wrote this function to automate the above command:

psg () { 
    local -a patterns=()
    (( $# == 0 )) && set -- $USER
    for arg do
        patterns+=("-e" "[${arg:0:1}]${arg:1}")
    done
    ps -ef | grep "${patterns[@]}"
}

Then,

psg foo bar

turns into

ps -ef | grep -e '[f]oo' -e '[b]ar'