CODEWITHSUNDEEP
awksed
xxjoe
xxjoe

Reputation: 11

bash awk with ps cannot print the results correctly

I want to get the related pids which match certain pattern, and then show the results with the pids and the matching part of the commands

samples results from ps (note that the target column is not necessary $9. eg $9 in line 1 and $10 in line 2):

root     23775     1  0 18:40 ?        00:00:01 /bin/bash /opt1/scripts/datamon/check_usage.sh -t 10

root     23777     1  0 18:40 ?        00:00:01 /bin/bash /opt1/servers/rt/extract_data.sh /opt1/scripts/datamon/results.data

what I am doing is:

pat1="/opt1/scripts/"

ps -ef | grep 'datamon' | awk -v RS='\n' -v pattern="$pat1" '{for (i=1;i<=NF;i++){if($i == /^pattern/) {print $2" "$i}}}'

the results obtained via the above commands are (pids are correct but not the command):

23775 0

23777 0

but actually what i want is:

23775 /opt1/scripts/datamon/check_usage.sh

23777 /opt1/scripts/datamon/print_results.sh

question:

1 why I got "0" instead of the matching commands?

2 how can I print the correct results above?

Upvotes: 1

Views: 178

Answers (3)

RavinderSingh13
RavinderSingh13

Reputation: 133468

With your shown samples please try following awk code. We need NOT to traverse through all fields, since its output of ps command so values will be searched in 9th and 10th fields, please try following.

pat1="/opt1/scripts/"
awk -v pattern="$pat1" 'index($9,pattern) || index($10,pattern){print $2,$9}' Input_file

Upvotes: 3

sseLtaH
sseLtaH

Reputation: 11217

Using GNU sed

$ pat1="/opt1/scripts/"
$ ps -ef | sed -En "/datamon/{s~[^ ]* +([^ ]* +).*($pat1[^ ]*).*?~\1\2~p}"
23775     /opt1/scripts/datamon/check_usage.sh
23777     /opt1/scripts/datamon/results.data

Upvotes: 1

The fourth bird
The fourth bird

Reputation: 163287

In the second part you are doing a string compare == instead of using ~ for a regex match.

If you print /^pattern/ it will give 0.

So if($i == /^pattern/) becomes if(0 == 0) and is true for field 4, which has 0 in the data and is printed.

You can print field 9 instead of $i as that field contains the data that you are interested in.

pat1="/opt1/scripts/"

awk -v pattern="$pat1" '
{
for (i=1;i<=NF;i++){
  if($i ~ pattern) {
    print $2" "$9
  }
}
}' file

Output

23775 /opt1/scripts/datamon/check_usage.sh
23777 /opt1/servers/rt/extract_data.sh

You can also print the field where the match is with print $2" "$i but then the output will be:

23775 /opt1/scripts/datamon/check_usage.sh
23777 /opt1/scripts/datamon/results.data

Upvotes: 3

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