Getting None type error from a valid regex query

Question

I'm trying to extract the value of os (Linux 3.11 and newer) from a program's output. I came up with this:

import re

p0f = '''
--- p0f 3.08b by Michal Zalewski <[email protected]> ---

[+] Closed 3 file descriptors.
[+] Loaded 324 signatures from '/etc/p0f/p0f.fp'.
[+] Will read pcap data from file 'temp.pcap'.
[+] Default packet filtering configured [+VLAN].
[+] Processing capture data.

.-[ 10.0.7.20/37462 -> 216.58.209.229/443 (syn) ]-
|
| client   = 10.0.7.20/37462
| os       = Linux 3.11 and newer
| dist     = 0
| params   = none
| raw_sig  = 4:64+0:0:1460:mss*20,7:mss,sok,ts,nop,ws:df,id+:0
|
`----

.-[ 10.0.7.20/37462 -> 216.58.209.229/443 (mtu) ]-
|
| client   = 10.0.7.20/37462
| link     = Ethernet or modem
| raw_mtu  = 1500
|
`----


All done. Processed 1 packets.
'''


print p0f
os = re.match(r"os\\s*= (.*)", p0f).group(1)
print os

According to this Regex101, my regex should be spot on. But I'm getting an error NoneType has no 'group'.

docs.python.org/2/library/re.html#search-vs-match

Konstantin
– Konstantin

2015-06-17 13:39:00 +00:00
Commented Jun 17, 2015 at 13:39 — Konstantin
– Konstantin, Commented Jun 17, 2015 at 13:39

Martijn Pieters · Accepted Answer · 2015-06-17 13:38:39Z

5

You have two problems:

You are using re.match() where you should be using re.search(). re.match() only matches against the start of the string. See search() vs. match() in the module documentation.
You doubled the \\ backslash on the \s metacharacter, but are using a r'..' raw string literal.

This works:

re.search(r"os\s*= (.*)", p0f)

Demo:

>>> import re
>>> re.search(r"os\s*= (.*)", p0f).group(1)
'Linux 3.11 and newer'

answered Jun 17, 2015 at 13:38

Martijn Pieters

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Comments

user1907906 · Accepted Answer · 2015-06-17 13:38:41Z

1

If you are using r, don't escape the \. This works:

re.search(r"os\s*= (.*)", p0f).group(1)

answered Jun 17, 2015 at 13:38

user1907906

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