On to my next learning exercise. As I parse a file I need to pull an IP
address out a line. Now I thought a regular expression would be the
ticket, but it's giving me a problem. The follow line is an example
string I need to pull one of two IP address out of: (they are not
always formed the same)

Received: from mmds-111-19-22-30.twm.ca.internet.net (HELO
?192.168.1.2?) (222.222.222.22)

I need that last IP address. Now the problem is that I can't always
count on it being enclosed in paren's. Although I can expect the right
paren to always be there.

So here's my regex exp:
sourceip = line.scan(/\b(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})\b/)
And as you might expect, it is pulling both IP addresses. Is there a
way I can adjust the expression to grab the second IP testing for the
")" or is there another method I can use? Short of dissecting the
entire string backwards and testing whether I have a number or a char,
decimal and at most 3 chars from it, etc?

tonyd
--
Posted via http://www.ruby-forum.com/.

Hi --

On Sat, 29 Mar 2008, Tony De wrote:

> On to my next learning exercise. As I parse a file I need to pull an IP
> address out a line. Now I thought a regular expression would be the
> ticket, but it's giving me a problem. The follow line is an example
> string I need to pull one of two IP address out of: (they are not
> always formed the same)
>
> Received: from mmds-111-19-22-30.twm.ca.internet.net (HELO
> ?192.168.1.2?) (222.222.222.22)
>
> I need that last IP address. Now the problem is that I can't always
> count on it being enclosed in paren's. Although I can expect the right
> paren to always be there.
>
> So here's my regex exp:
> sourceip = line.scan(/\b(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})\b/)
> And as you might expect, it is pulling both IP addresses. Is there a
> way I can adjust the expression to grab the second IP testing for the
> ")" or is there another method I can use? Short of dissecting the
> entire string backwards and testing whether I have a number or a char,
> decimal and at most 3 chars from it, etc?

What you want is an IP address, possibly followed by ')' and
definitely coming at the end of the string (give or take a newline
character after it). That can be expressed like this:

/((\d{1,3}\.){3}\d{1,3})(?=\)?\Z)/

I've got 3 occurences of (\d{1,3}\.), followed by the same thing
without a dot. I've stipulated that this submatch be "looking at"
(i.e., positioned just before) an optional ')' followed by the end of
the string. (\Z gives you end of string, ignoring a possible terminal
newline.)

With your line, it gives you:

irb(main):039:0> line[re] # re.match(line)[0], or whatever
=> "222.222.222.22"


David

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David A. Black wrote:
> Hi --
>
> On Sat, 29 Mar 2008, Tony De wrote:
>
>> count on it being enclosed in paren's. Although I can expect the right
>> paren to always be there.
>>
>> So here's my regex exp:
>> sourceip = line.scan(/\b(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})\b/)
>> And as you might expect, it is pulling both IP addresses. Is there a
>> way I can adjust the expression to grab the second IP testing for the
>> ")" or is there another method I can use? Short of dissecting the
>> entire string backwards and testing whether I have a number or a char,
>> decimal and at most 3 chars from it, etc?
>
> What you want is an IP address, possibly followed by ')' and
> definitely coming at the end of the string (give or take a newline
> character after it). That can be expressed like this:
>
> /((\d{1,3}\.){3}\d{1,3})(?=\)?\Z)/
>
> I've got 3 occurences of (\d{1,3}\.), followed by the same thing
> without a dot. I've stipulated that this submatch be "looking at"
> (i.e., positioned just before) an optional ')' followed by the end of
> the string. (\Z gives you end of string, ignoring a possible terminal
> newline.)
>
> With your line, it gives you:
>
> irb(main):039:0> line[re] # re.match(line)[0], or whatever
> => "222.222.222.22"
>
>
> David

David, you rock. I'll give it a try. Those expressions make my head
hurt. But I've been taking in
http://www.regular-expressions.info/tutorial.html. It seems to cover a
lot of foundation and application. Thanks again!

tonyd
--
Posted via http://www.ruby-forum.com/.

On Sat, Mar 29, 2008 at 8:16 AM, Tony De <tonydema@gmail.com> wrote:
> David A. Black wrote:
> > Hi --
> >
> > On Sat, 29 Mar 2008, Tony De wrote:
> >
>
> >> count on it being enclosed in paren's. Although I can expect the right
> >> paren to always be there.
> >>
> >> So here's my regex exp:
> >> sourceip = line.scan(/\b(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})\b/)
> >> And as you might expect, it is pulling both IP addresses. Is there a
> >> way I can adjust the expression to grab the second IP testing for the
> >> ")" or is there another method I can use? Short of dissecting the
> >> entire string backwards and testing whether I have a number or a char,
> >> decimal and at most 3 chars from it, etc?
> >
> > What you want is an IP address, possibly followed by ')' and
> > definitely coming at the end of the string (give or take a newline
> > character after it). That can be expressed like this:
> >
> > /((\d{1,3}\.){3}\d{1,3})(?=\)?\Z)/
> >
> > I've got 3 occurences of (\d{1,3}\.), followed by the same thing
> > without a dot. I've stipulated that this submatch be "looking at"
> > (i.e., positioned just before) an optional ')' followed by the end of
> > the string. (\Z gives you end of string, ignoring a possible terminal
> > newline.)
> >
> > With your line, it gives you:
> >
> > irb(main):039:0> line[re] # re.match(line)[0], or whatever
> > => "222.222.222.22"
> >
> >
> > David
>
> David, you rock. I'll give it a try. Those expressions make my head
> hurt. But I've been taking in
> http://www.regular-expressions.info/tutorial.html. It seems to cover a
> lot of foundation and application. Thanks again!

Another possibility (if I understood correctly): check for the
mandatory ')' in the HELO part, followed by any character (could be
changed by the specific spaces), followed by the numbers and dots for
the IP:

a = "Received: from mmds-111-19-22-30.twm.ca.internet.net (HELO
?192.168.1.2?) (222.222.222.22)"
a.match(/\).*?(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})/)[1]

gives: "222.222.222.22"

Jesus.

Jesús Gabriel y Galán wrote:
> On Sat, Mar 29, 2008 at 8:16 AM, Tony De <tonydema@gmail.com> wrote:
>> >> sourceip = line.scan(/\b(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})\b/)
>> > /((\d{1,3}\.){3}\d{1,3})(?=\)?\Z)/
>> > => "222.222.222.22"
>> >
>> >
>> > David
>>
>> David, you rock. I'll give it a try. Those expressions make my head
>> hurt. But I've been taking in
>> http://www.regular-expressions.info/tutorial.html. It seems to cover a
>> lot of foundation and application. Thanks again!
>
> Another possibility (if I understood correctly): check for the
> mandatory ')' in the HELO part, followed by any character (could be
> changed by the specific spaces), followed by the numbers and dots for
> the IP:
>
> a = "Received: from mmds-111-19-22-30.twm.ca.internet.net (HELO
> ?192.168.1.2?) (222.222.222.22)"
> a.match(/\).*?(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})/)[1]
>
> gives: "222.222.222.22"
>
> Jesus.


Thanks Jesus,

I appraciate your imput as well. You guys have been a great deal of
.

tonyd
--
Posted via http://www.ruby-forum.com/.

On 29 Mar 2008, at 07:16, Tony De wrote:
> Those expressions make my head
> hurt. But I've been taking in
> http://www.regular-expressions.info/tutorial.html. It seems to
> cover a
> lot of foundation and application. Thanks again!

You may find Rubular ful for concocting your regular expressions
and decoding other people's.

http://www.rubular.com/

Regards,
Andy Stewart

-------
http://airbladesoftware.com