parse host output with regex

Question

We have dig output which looks like

Trying "google.com"
Using domain server:
Name: 177.39.160.253
Address: 177.39.160.253#53
Aliases: 

;; ->>HEADER<<- opcode: QUERY, status: NOERROR, id: 52549
;; flags: qr rd ra; QUERY: 1, ANSWER: 1, AUTHORITY: 13, ADDITIONAL: 6

;; QUESTION SECTION:
;google.com.            IN  A

;; ANSWER SECTION:
google.com.     245 IN  A   216.58.202.46
google.com.     245 IN  A   216.58.202.47

;; AUTHORITY SECTION:
com.            16060   IN  NS  d.gtld-servers.net.
com.            16060   IN  NS  e.gtld-servers.net.
com.            16060   IN  NS  g.gtld-servers.net.
com.            16060   IN  NS  c.gtld-servers.net.
com.            16060   IN  NS  m.gtld-servers.net.
com.            16060   IN  NS  f.gtld-servers.net.
com.            16060   IN  NS  i.gtld-servers.net.
com.            16060   IN  NS  l.gtld-servers.net.
com.            16060   IN  NS  k.gtld-servers.net.
com.            16060   IN  NS  a.gtld-servers.net.
com.            16060   IN  NS  j.gtld-servers.net.
com.            16060   IN  NS  b.gtld-servers.net.
com.            16060   IN  NS  h.gtld-servers.net.

;; ADDITIONAL SECTION:
a.gtld-servers.net. 13518   IN  A   192.5.6.30
e.gtld-servers.net. 23554   IN  A   192.12.94.30
e.gtld-servers.net. 24306   IN  AAAA    2001:502:1ca1::30
g.gtld-servers.net. 13443   IN  AAAA    2001:503:eea3::30
k.gtld-servers.net. 2284    IN  A   192.52.178.30
k.gtld-servers.net. 10638   IN  AAAA    2001:503:d2d::30

Received 400 bytes from 177.39.160.253#53 in 306 ms

I need to write regex which will return 216.58.202.46 and 216.58.202.47

Right now I'm doing it using 2 regexes:

At first I'm taking content of the answer section with

;; ANSWER SECTION:[\s\S]*?\n\n

Then I'm parsing answer with

.*IN[\t *]A[\t *](?P<output>[\S]*)\n

Can I achieve same functionality with 1 regular expression? Thanks.

@WiktorStribiżew wow! Even with the regex101 explanation I still don't quite get all you did ... I'll need some time to figure out everything — Esteban
– Esteban, Commented Jun 28, 2017 at 8:18
@Esteban: If it works for OP, I will post with explanations. Илья, does it work for you? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jun 28, 2017 at 8:19
@WiktorStribiżew It works on his exampe for sure, only thing I changed on the regex was the ADDITIONAL SECTION to ANSWER SECTION, but all in all, it did what he wants to. Thanks for taking time to make the explanation — Esteban
– Esteban, Commented Jun 28, 2017 at 8:26
Yes, regex101.com/r/2Np9Mr/2. I did not pay attention to that bit. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jun 28, 2017 at 8:27

Graham · Accepted Answer · 2017-09-24 04:50:51Z

2

You may use a regex based on the \G operator that matches either the position at the start of a string, or the location after a successful match.

(?s)(?:\G(?!\A)|;; ANSWER SECTION:)(?:(?!\R{2}).)*? IN +A +\K\S+

See the regex demo.

Details:

(?s) - an inline DOTALL modifier that makes . match line break chars, too (now, it matches any chars)
(?:\G(?!\A)|;; ANSWER SECTION:) - either the location after the previous match (\G(?!\A), the start of string position is subtracted with the negative lookahead) or a literal string ;; ANSWER SECTION:
(?:(?!\R{2}).)*? - a tempered greedy token matching any char (.) that is not a starting point for a double line break sequence (\R{2}, \R matches any linebreak sequence), as few times as possible, as *? is a lazy quantifier
IN +A + - space, IN, 1+ spaces, A, 1+ spaces
\K - a match reset operator
\S+ - one or more non-whitespace chars.

edited Sep 24, 2017 at 4:50

Graham

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answered Jun 28, 2017 at 9:21

Wiktor Stribiżew

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