CODEWITHSUNDEEP
regexbashshellgrep
tommsen
tommsen

Reputation: 237

pgrep -f with multiple arguments

i try to find a specific process containing the term "someWord" and two other terms represented by $1 and $2

 7   regex="someWord.*$1.*$2"
 8   echo "$regex"
 9   [ `pgrep -f $regex` ] && return 1 || return 0

which returns

./test.sh foo bar
someWord.*foo bar.*
./test.sh: line 9: [: too many arguments

What happens to my regular expression? Doing that pgrep directly in the shell works fine.

Upvotes: 16

Views: 18236

Answers (3)

Zombo
Zombo

Reputation: 1

Good sir, perhaps this

[[ `pgrep -f "$regex"` ]] && return 1 || return 0

or this 

[ "`pgrep -f '$regex'`" ] && return 1 || return 0

Upvotes: 10

Anew
Anew

Reputation: 5492

If you really want to do something in case your command returns an error:

cmd="pgrep -f $regex"

if ! $cmd; then
  echo "cmd failed."
else
  echo "ok."
fi

Upvotes: 0

chepner
chepner

Reputation: 531055

First, there's no reason to wrap your pgrep command in anything. Just use its exit status:

pgrep -f "$regex" && return 1 || return 0.

If pgrep succeeds, you'll return 1; otherwise, you'll return 0. However, all you're doing is reversing the expected exit codes. What you probably want to do is simply let the pgrep be the last statement of your function; then the exit code of pgrep will be the exit code of your function.

something () {
   ...
   regex="someWord.*$1.*$2"
   echo "$regex"
   pgrep -f $regex
}

Upvotes: 2

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