当输入不匹配时,防止 grep 返回错误

我想在 bash 脚本中编写一段代码来检查程序是否已经在运行。 我有以下为了搜索是否酒吧正在运行

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

那个

 grep -v grep

部分是确保在 ps 结果中不考虑“ grep bar”

当 bar 不运行时,foo 正确地为空

 set -e

如果某个命令返回错误,则终止该脚本的标志。 结果表明,当 bar 不运行时,“ grep-v grep”与任何内容都不匹配,grep 返回一个错误。我尝试使用-q 或-s,但没有用。

有什么办法吗? 谢谢

40229 次浏览

Sure:

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

Or even:

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

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'

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.

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.


[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.

[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

Try to make so:

ps auxw | grep -v grep | cat

cat returns always 0 and ignores exit code of grep