CODEWITHSUNDEEP
awk
achille
achille

Reputation: 315

print part of a field with awk

I run:

  ss -atp | grep -vi state | awk '{ print $2"   "$3"   "$4"  "$5"   "$6 }'

output:

0   0   192.168.1.14:49254  92.222.106.156:http   users:(("firefox-esr",pid=696,fd=95))

From the last column, I want to strip everything but firefox-esr (in this case); more precisely I want to only fetch what's between "".

I have tried:

ss -atp | grep -vi state | awk '{ sub(/users\:\(\("/,"",$6); print $2"    "$3"   "$4"   "$5"    "$6 }'
0    0   192.168.1.14:49254   92.222.106.156:http    firefox-esr",pid=696,fd=95))

There is still the last part to strip; the problem is that the pid and fd are not a constant value and keep changing.

Upvotes: 0

Views: 453

Answers (2)

Daweo
Daweo

Reputation: 36370

You might harness gensub reference ability for that. For simplicity let file.txt content be

users:(("firefox-esr",pid=696,fd=95))

then

awk '{print gensub(/.*"(.+)".*/,"\\1",1,$1)}' file.txt

outputs:

firefox-esr

Keep in mind that gensub do not alter string it gets as 4th argument, but return new string, so I print it.

Upvotes: 1

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626728

You can use

awk '{ gsub(/^[^\"]*\"|\".*/, "", $6); print $2"   "$3"   "$4"  "$5"   "$6 }'

Here, gsub(/^[^\"]*\"|\".*/, "", $6) will take Field 6 as input, and remove all chars from start till the first " including it (see the ^[^\"]*\" part) and then the next " and all text after it (using \".*).

See this online awk demo:

s='0 0   0   192.168.1.14:49254  92.222.106.156:http   users:(("firefox-esr",pid=696,fd=95))'
awk '{gsub(/^[^\"]*\"|\".*/, "",$6); print $2"   "$3"   "$4"  "$5"   "$6 }' <<< "$s"
# => 0   0   192.168.1.14:49254  92.222.106.156:http   firefox-esr

Upvotes: 1

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