Perl regex in a variable string

Question

I'm new to perl with regex.

I'm trying to have a string of oid 1.3.6.1.2.1.4.22.1.2.*.192.168.1.1, but I'm not sure how to do it. I tried the below, but it is getting error which is saying not able to recognize the oid.

my $matchanyoid = "/(\d+)$/";
my $dot1dTpFdbAddress = '1.3.6.1.2.1.4.22.1.2.',$matchanyoid,'\.',$srcip;

choroba · Accepted Answer · 2013-07-05 01:17:59Z

1

Comma is not a concatenation operator, dot is:

my $dot1dTpFdbAddress = '1.3.6.1.2.1.4.22.1.2.' . $matchanyoid . '\.' . $srcip;

If you are trying to build a regular expression, note that the first several dots are not backslashed, so they can match anything. To avoid lots of backslashes, you can use the \Q ... \E construct:

my $matchanyoid = '(\d+)';
my $srcip = 12;
my $regex = qr/\Q1.3.6.1.2.1.4.22.1.2.\E$matchanyoid\.$srcip/;
print '1.3.6.1.2.1.4.22.1.2.123.12' =~ $regex;

edited Jul 5, 2013 at 1:17

answered Jul 5, 2013 at 1:12

choroba

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3 Comments

kyusan93 Over a year ago

I got the error: ERROR: The base OBJECT IDENTIFIER "(?^:1\.3\.6\.1\.2\.1\.4\.22\.1\.2\.\/(d\+)\\.192\.1168\.1\.32)" is expected in dotted decimal notation.

choroba Over a year ago

@kyusan93: Then you are probably not telling us all the important details. I updated the answer, it works for me.

kyusan93 Over a year ago

I'm trying to use Net::SNMP to get the mac address via oid, but I only know the ip address and the 1.3.6.1.2.1.4.22.1.2, but between the 1.3.6.1.2.1.4.22.1.2 and the IP, there is some other * numbers and I do not how to use the perl regex to do the *.

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