Question about conditional regex capture

TASdvlper has asked for the wisdom of the Perl Monks concerning the following question:

Hello all,

I have another regexp question (seems to be my favorite topic).

I'm trying to parse the output of a cli command, and I'm looking for string, in each output line, that consists of digits seperated by periods. i.e.

0.1.4.1 
0.1.5.2 
0.1.6.3 
0.1.7.4
[download]

but, my problem is, I don't want to capture the output if the 3 field, from the left, is a 0,1,2, or 3. But, it can start with it. I have this snippet:


if ( /.*(\d+\.\d+\.[4-9]\.\d+).*/ )  {
         print "$1 <br> \n";
      }
[download]

But this will not work if the 3rd field is a double digit.

So, to elaborate more, the following I want to ignore:

0.1.1.1
0.1.2.2
0.1.3.3
[download]

but, these I do want to capture:

0.1.10.1
0.1.20.2
0.1.30.3
[download]

Basically, if the 3rd field is > 3, I want to capture it.

Any thoughts on to do this ???? Thx.

20040519 Edit by Corion: Changed title from 'Another regular expression'

Comment on Question about conditional regex capture Select or Download Code

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Re: Question about conditional regex capture by Fletch (Bishop) on May 18, 2004 at 20:02 UTC
So just use capturing parens in your regexp and compare the number to 3. You don't have to do everything in a regexp. `if( /(\d+\.\d+\.(\d+)\.\d+)/ and $2 > 3 ) { print $1, "<br />\n"; }` [download]	[reply] [d/l]
Re: Question about conditional regex capture by kvale (Monsignor) on May 18, 2004 at 20:03 UTC
You are on th right track. Gropuing lets you have alternatives `if ( /.(\d+\.\d+\.(\d\d+\|[4-9])\.\d+)./ ) { print "$1 <br> \n"; }` [download] This matches either a multidigit number or a single digit in 4..9. Note it could still be fooled by 01, etc, so check your input for that problem. -Mark	[reply] [d/l]
Re: Question about conditional regex capture by cLive ;-) (Prior) on May 18, 2004 at 20:16 UTC
`if ( /(\d+\.\d+\.(?:\d{2,}\|[4-9])\.\d+)/ ) { print "$1 <br> \n"; }` [download] cLive ;-)	[reply] [d/l]
Re: Question about conditional regex capture by saskaqueer (Friar) on May 18, 2004 at 20:16 UTC
update: I missed the fact that the string of digits with periods needs to be extracted from a line of additional text. The solution I posted below would only work once you have already extracted the digit/period string. `# why use a regex? splitting is enough while ( <DATA> ) { print $_ if ( (split( /\./ ))[2] > 3 ); } __DATA__ 0.1.1.1 0.1.2.2 0.1.3.3 0.1.10.1 0.1.20.2 0.1.30.3` [download]	[reply] [d/l]
Re: Question about conditional regex capture by Abigail-II (Bishop) on May 19, 2004 at 08:57 UTC
For starters, anchor your regex, for two reasons: efficiency, and you prevent a match to happen by the regex engine starting to match at the second group. `/^(\d+.\d+\.(\d{2,}\|[4-9])\.\d+)/` [download] Abigail	[reply] [d/l]
Re: Question about conditional regex capture by hv (Prior) on May 19, 2004 at 13:03 UTC
You can also express this slightly more naturally, at the cost of a slightly less efficient regexp: `if (/^(\d+\.\d+\.(?![0-3]\.)\d+\.\d+)/) { print ... }` [download] that is: match two numbers then (as long as the next number isn't in the range 0 .. 3) match two more numbers. Hugo	[reply] [d/l]


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