pgrep -f exits with 1

Question

RUNNING_APPS=$(pgrep -f "somePattern")
echo $?

#results in
1

How can I make my command pass with exit code 0?

What are you trying to do exactly? Please [edit] your question and give us some context. Why are you checking the exit status of the command? — terdon, Aug 11 '20 at 13:32
I've got script with `set -e` flag which fails after this command. Echo is added just to illustrate behavior. — makozaki, Aug 11 '20 at 13:34
Please ***edit*** your question to add this then. And give us enough context so we can offer an alternative: this is how `pgrep` behaves so if you _must_ use `set -e`, then you cannot use `prep` like this. Make sure to mention what operating system you are running. Also, a point on style: avoid using CAPS for variable names in shell scripts, that is bad practice since global environment variables are capitalized and this can cause variable name collisions with unexpected consequences. — terdon, Aug 11 '20 at 13:36
`pgrep -f whatever || true` should return 0 and not trip the -e action. — Paul_Pedant, Aug 11 '20 at 14:27

terdon · Accepted Answer · 2020-08-11T14:51:04.973

3

On my Arch system, with the pgrep from procps-ng, I see this in man pgrep:

EXIT STATUS
       0      One  or  more processes matched the criteria. For
              pkill the process must also  have  been  success‐
              fully signalled.
       1      No  processes  matched  or  none of them could be
              signalled.
       2      Syntax error in the command line.
       3      Fatal error: out of memory etc.

So this is just the way it is: pgrep will exit with 1 if everything worked fine but there were no processes matching the search string. This means you will need to use a different tool. Perhaps something like Kusalananda suggested in the comments and ilkkachu posted as an answer:

running_apps=$(pgrep -f "somePattern" || exit 0)

But a better approach, IMO, would be to change your script. And instead of using set -e, have it exit manually at the important steps. Then, you can use something like this:

running_apps=$(pgrep -fc "somePattern")
if [ "$running_apps" = 0 ]; then
    echo "none found"
else
    echo "$running_apps running apps"
fi

edited Aug 11 '20 at 14:51

answered Aug 11 '20 at 13:41

terdon

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Or just use `pgrep ... || exit 0` in the command substitution. Or don't use `set -e`. – Kusalananda Aug 11 '20 at 13:52
`pgrep -fc` is just a pseudo-code? I get `invalid option -- 'c'` – makozaki Aug 11 '20 at 13:54
@MarcinKozakiewicz that's why I asked you to tell us your operating system. It looks like you are using a different version of `pgrep`. What OS are you on? – terdon Aug 11 '20 at 13:57
1

`... | grep -v grep` returns with an error if there are no lines without `grep` in them. Just go with `|| true` to suppress the error exit. – ilkkachu Aug 11 '20 at 14:33
@ilkkachu is that ever possible given that the command is part of `ps aux | grep one | grep two`? That said, yes the `|| exit 0` trick is indeed a better approach. – terdon Aug 11 '20 at 14:41
@terdon, well, if you go `ps aux | grep xterm | grep -v xterm` and there's no `xterm` running on the system, then the first `grep` only matches the line for itself (or not even that), and the second doesn't select (print) anything, so it exits `1`. – ilkkachu Aug 11 '20 at 14:46
@ilkkachu duh, of course. Thanks. – terdon Aug 11 '20 at 14:50

score 3 · Answer 2 · answered Aug 11 '20 at 14:36

With set -e, commands that on the left side of the AND (&&) or OR (||) operators don't cause the shell to exit, so you can suppress the error by adding || true.

So this should output 0 regardless of the processes found (and not exit before outputting it):

set -e
RUNNING_APPS=$(pgrep -f "somePattern" || true)
echo $?

pgrep -f exits with 1

2 Answers2