Linux Bash Regular Expressions, retrieving data from SNMPGet Output

Question

I've been working on getting a few simple monitoring tools running at home, and decided to be funny and retrieve the printer data along with everything else, however now that I've got the SNMP portion of it working quite well, I can't seem to be able to parse the data that my SNMPGET command retrieves properly in Linux, the current script I am using is as follows:

#!/usr/bin/env bash

# RegEx for Strings:    "(.+?)"| -?\d+ 
RegExStr='"(.+?)"| -?\d+'

# ***
# Brother HL-2150N Printer
# ***
# Order Data: Toner Naame, Toner Level, Drum Name, Drum Status, Total Pages Printer,  Display Status
Input=$(snmpget -v 1 -c public 192.168.16.112 SNMPv2-SMI::mib-2.43.11.1.1.6.1.1 SNMPv2-SMI::mib-2.43.11.1.1.8.1.1 SNMPv2-SMI::mib-2.43.11.1.1.6.1.2 SNMPv2-SMI::mib- 2.43.11.1.1.9.1.1 SNMPv2-SMI::mib-2.43.10.2.1.4.1.1 SNMPv2-SMI::mib-2.43.16.5.1.2.1.1 -m BROTHER-MIB)

Output1=( $(echo $Input | egrep -o $RegExStr) )

# Output
echo $Input
echo ${Output1[@]}

Which, oddly enough does not work. I'm fairly certain my regular expression ( "(.+?)" ) is correct, as I've tested it numerous times in various different syntax checkers and testers. It's supposed to select all the data that's between quotation marks ("").

Anyhow, the SNMPGET return is:

SNMPv2-SMI::mib-2.43.11.1.1.6.1.1 = STRING: "Black Toner Cartridge" SNMPv2-SMI::mib-2.43.11.1.1.8.1.1 = INTEGER: -2 SNMPv2-SMI::mib-2.43.11.1.1.6.1.2 = STRING: "Drum Unit" SNMPv2-SMI::mib-2.43.11.1.1.9.1.1 = INTEGER: -3 SNMPv2-SMI::mib-2.43.10.2.1.4.1.1 = Counter32: 13630 SNMPv2-SMI::mib-2.43.16.5.1.2.1.1 = STRING: "SLAAP "

I've tried various things myself, and using grep returns a blank string. to my understanding grep does not support every regular expression command by itself, so I started using egrep, while this returns SOMETHING, it is everything inside the original string divided by spaces, starting at the first quotation mark.

Is there anything I'm missing? I've looked around, and adjusted my methods a few times but never seemed to get a usable array in return.

Anyhow, I appreciate any help/pointers you'd be able to give me. I'd like to be able to get this running, even if just for fun and a good learning experience. Thank you in advance though! I'll be fidgeting on with it some more myself, but will check here every now and then.

@Jotne, At the end, I'm trying to retrieve all the data between brackets and the numerical values returned after INTEGER: and COUNTER32:. I'm sorry if that wasn't made clear up there. — Silvester van Harberden, Oct 03 '13 at 09:25
@anubhava, At the top of the bash script. I presume, I assign it a value over there. — Silvester van Harberden, Oct 03 '13 at 09:26
@anubhava, Ah! Sorry. A typo in my many edits, I've corrected it in my main script and it doesn't do a single thing still. — Silvester van Harberden, Oct 03 '13 at 09:32
If I got your idea right, you want to have the part ".+?" non-greedy; but does this really work in grep/egrep? I don't think so. — volferine, Oct 03 '13 at 09:45
Really? That'd be a bit bothersome. It's the only way I could think of when trying to select everything(including spaces) between the quotationmarks. Is there a replacement for that set of characters, or should I take a different approach entirely? — Silvester van Harberden, Oct 03 '13 at 09:51
if ungreedy-ness doesn't work couldn't you just use `"([^"]+)`? — JonM, Oct 03 '13 at 09:56
Correct (thank you!), however this returns me to another problem of mine, the array is built up out of the entire $Input string divided by spaces, rather than just what I am trying to request from the string. I'm guessing this is because grep returns lines, rather than full cutouts? — Silvester van Harberden, Oct 03 '13 at 10:00
might this answer help with multi/single line mode? http://stackoverflow.com/questions/3717772/regex-grep-for-multi-line-search-needed — JonM, Oct 03 '13 at 10:03
Ah, that does explain the use of some of the parameters yes. Thank you! — Silvester van Harberden, Oct 03 '13 at 10:48

score 1 · Accepted Answer · answered Oct 03 '13 at 10:00

1

From your output:

To get all strings:

grep -oP 'STRING: *"\K[^"]*'
Black Toner Cartridge
Drum Unit
SLAAP

To get all integers:

grep -oP '(INTEGER|Counter32): *\K[^ ]*'
-2
-3
13630

answered Oct 03 '13 at 10:00

anubhava

761,203
64
569
643

I've combined the two lines (and my original script) into: Output1=( $(echo $Input | grep -oP 'STRING: *"\K[^"]*|(INTEGER|Counter32): *\K[^ ]*') ) This seems to work like a charm! It still returns everything beyond a space as a new array piece, but I believe that's something Bash specific. Thank you for the help! I appreciate it. – Silvester van Harberden Oct 03 '13 at 10:43

score 0 · Answer 2 · answered Oct 03 '13 at 10:20

0

With awk you can do this:

awk 'NR%2==0' RS=\" <<< $Input
Black Toner Cartridge
Drum Unit
SLAAP

Or into a variable

Output1=$(awk 'NR%2==0' RS=\" <<< $Input)

answered Oct 03 '13 at 10:20

Jotne

40,548
12
51
55

This seems to work, although it returns it all into a single entry Input1(0). Not a huge problem, of course. But figured I'd let you know! Appreciate the alternative method however, as I'm still trying to learn how to work the Bash commands. :) – Silvester van Harberden Oct 03 '13 at 10:44

Linux Bash Regular Expressions, retrieving data from SNMPGet Output

2 Answers2