K, I tried this and it didn't work as expected:
$code =~ / HTTP\/\d\.\d\" (\d+)/;
if (!$code) {
print "NEXT\n";
next;
}
print "$code\n";
The loop just printed NEXT 300 or so times. I was thinking that $code would
equal whatever was in the parentheses. Am I still not getting this?
-----Original Message-----
From: George P. [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 9:48 AM
To: Romeyn, Derek
Cc: '[EMAIL PROTECTED]'
Subject: Re: Regular Expressions http error code
On Wed, 12 Mar 2003, Romeyn, Derek wrote:
> I have a DB that records HTTP log data like this. I'm horrible at regular
> expressions and could use some real help in pulling out the HTTP error
code
> from lines that look like this.
>
> 65.248.129.126 - xm1721 [11/Mar/2003:05:41:35 -0800] "POST
> /applogic/brpulseappletinfonew HTTP/1.1" 500 16 7 "-" "
> 12.235.196.99 - - [11/Mar/2003:05:55:35 -0800] "GET /.com HTTP/1.1" 500
184
> "-" "Mozilla/4.0 (compatible; MSIE 6.0; Win
> 12.229.35.165 - - [12/Mar/2003:04:34:11 -0800] "GET
/e/t/applogic/spacer.gif
> HTTP/1.1" 404 0 "https://www.here.com/t
> 24.98.188.254 - - [12/Mar/2003:04:43:19 -0800] "GET
> /e/t/kc/images/liquidation_rights.gif HTTP/1.1" 404 0 "https://www.h
> 207.5.195.98 - - [12/Mar/2003:04:56:32 -0800] "GET
> /e/t/invest/img/spacer.gif HTTP/1.1" 404 0 "https://www.here.com/t
>
> I'm guessing I need the three digits that are located on each line between
> 'HTTP/\d.\d" ' and another ' ' but not sure how to do that.
You've almost done it yourself.
I've never really worked with HTTP log data, but from the lines that
you've given, I would say:
my ($error_code);
if ($ln =~ / HTTP\/\d\.\d\" (\d+)/)
{
($error_code) = $1;
}
bye
George P.
>
> Thanks much,
> Derek
>
>
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