CODEWITHSUNDEEP
unixawksed
Anand Abhay
Anand Abhay

Reputation: 359

Grep Number before pattern

I have file with following content 

4 packets transmitted, 2 received, 50% packet loss, time 2002ms
rtt min/avg/max/mdev = 0.904/1.108/1.235/0.148 ms

Now I am trying to find the packet loss percentage and received value 

loss=`more file.txt  | grep "% packet loss" | awk -v RS=[0-9]+ '{print RT+0;exit}'`

But it always returns me the 4 ( first number in the line). But I am expecting loss to be 50% and received to be 2

Upvotes: 1

Views: 273

Answers (2)

Ed Morton
Ed Morton

Reputation: 203532

Is this all you're trying to do?

$ awk '/% packet loss/{print $6, $4}' file
50% 2

and if you need them separately:

$ arr=( $(awk '/% packet loss/{print $6, $4}' file) )
$ echo "${arr[0]}"
50%
$ echo "${arr[1]}"
2

Upvotes: 0

Inian
Inian

Reputation: 85590

Why would need to make things hard using RS and using cat .. | grep when awk can just solve the problem with just regex match. Just do

awk 'match($0, /.*([0-9]) received, ([0-9%]+)/, arr){ print "received="arr[1]; print "loss="arr[2]  }' file.txt

Storing it in a shell variable, e.g. on bash supporting process-substitution

read -r received loss < <(awk 'match($0, /.*([0-9]) received, ([0-9%]+)/, arr){ printf "%d %s", arr[1]+0, arr[2] ; }' file)

Upvotes: 3

Related Questions