grep regular expression

[mootroot]

Hi all,
Im having a problem with my regular expression to determine IP addresses in a file. At the mo, im using

grep '[0-9]*[0-9]*[0-9][.][0-9]*[0-9]*[0-9][.][0-9]*[0-9]*[0-9]'

which returns ip addresses fine, however its also returning ip addresses which have something attached on the end (or before), such as:

127.0.0.1.random

i tried using grep -w, and also

grep '\<[0-9]*[0-9]*[0-9][.][0-9]*[0-9]*[0-9][.][0-9]*[0-9]*[0-9]\>'

but they dont seem to work, can anybody help?
Cheers
Mootroot

[bigtomrodney]

Grep will return the entire string that contains the search terms. Are the returned products the strings contained in the source data? Also you seem to be sarching for only to the subnet stage of the IP address - 127.0.0 for example.

[vsemaska]

Specify that nothing can be before or after the string with the anchors '^' - beginning of line and '$' - end of line. Your string should look like:

grep '^[0-9]*[0-9]*[0-9][.][0-9]*[0-9]*[0-9][.][0-9]*[0-9]*[0-9]$'

You can reduce your string down to this and it'll work:

grep '^[0-9]*\.[0-9]*\.[0-9]*\.[0-9]*$'

Vic

[mootroot]

thanks for your replies.

i didnt spot that id only put 3 octets!

I dont think i can use the beginning of line and end of line thing, cos the IP address will be somewhere in the middle. What i have is a file with a load of lines, where a line could be like one of the following:

hello a 10.10.10.10
test b 192.168.2.1.test.test

but i only want to return the first line not the one with test.test. Then i will use awk to obtain just the ip.

What i would really like is an expression where i could grep for each line with an IP address (which is by itself) no matter where it is in the line

Thanks again

[vsemaska]

OK, then I'll assume there's spaces and/or TABs delimiting the IP address. I'll also take into account that the IP address may be either at the beginning or end of the line. This command should work:

grep -E '(^|[[:space:]])[0-9]*\.[0-9]*\.[0-9]*\.[0-9]*([[:space:]]|$)'

The -E option means to use extended regular expressions. This is needed to search to beginning of line or whitespace ''(^|[[:space:]])" or whitespace or end-of-line "([[:space:]]|$)".

Vic

[mootroot]

thanks mate, that seems to have done the job

[ghostdog74]

with GNU awk,

awk --re-interval '{
        for(i=1;i<=NF;i++) {             
             if ( $i ~ /^([[:digit:]]{1,3}\.){3}[[:digit:]]{1,3}$/ ) {
                print $0
             }
        }
      }' "file"