CODEWITHSUNDEEP
linuxshellwildcard
czchlong
czchlong

Reputation: 2574

Linux wildcard not behaving correctly

I will be running a list of processes that will have the following naming conventions:

a_feed
b_feed
c_feed
...

I have written a bash shell script that will allow me to filter out the processes with these naming patterns. Please look at the one line in my shell script below:

ps -ef | grep -i *_feed | grep -v grep | awk '{print $2, " ", $8, " ", $10}'

For some reason, grep -i *_feed is unable to find any process that conforms to the pattern *_feed.

Does anyone have any ideas why?

Thanks.

Upvotes: 0

Views: 681

Answers (4)

Simon C
Simon C

Reputation: 2017

I usually save the output of the list of processes in a shell variable and then search for the matching lines as a new command. This avoids needing a grep -v to remove the running grep command.

I also match the lines inside awk so that no grep is needed at all. I think this is easier to read and understand:

p="$(ps -ef)"
echo "$p" | awk '/_feed/ {print $2, " ", $8, " ", $10}'

Upvotes: 0

Kent
Kent

Reputation: 195029

from grep man page:

-G, --basic-regexp Interpret PATTERN as a basic regular expression (BRE, see below). This is the default.

so, by default the pattern would be regular expression. in your example, you could use grep -i ".*_feed"

Upvotes: 0

Karoly Horvath
Karoly Horvath

Reputation: 96258

* need something in front of it.

Also, if you have a file with the pattern *_feed in your working directory bash will do wildcard expansion.

Use:

grep -i '.*_feed'

Upvotes: 1

Maxim Krizhanovsky
Maxim Krizhanovsky

Reputation: 26699

grep users regular expression, in which * means matches 0 or more times, and not any character. You should replace it with grep -i .*_feed

Upvotes: 3

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