CODEWITHSUNDEEP
perlsearchfilegrep
user36457
user36457

Reputation:

How can I scan multiple log files to find which ones have a particular IP address in them?

Recently there have been a few attackers trying malicious things on my server so I've decided to somewhat "track" them even though I know they won't get very far.

Now, I have an entire directory containing the server logs and I need a way to search through every file in the directory, and return a filename if a string is found. So I thought to myself, what better of a language to use for text & file operations than Perl? So my friend is helping me with a script to scan all files for a certain IP, and return the filenames that contain the IP so I don't have to search for the attacker through every log manually. (I have hundreds) 

#!/usr/bin/perl

$dir = ".";

opendir(DIR, "$dir");
@files = grep(/\.*$/,readdir(DIR));
closedir(DIR);

foreach $file(@files) {
    open FILE, "$file" or die "Unable to open files";

    while(<FILE>) {
        print if /12.211.23.200/;
    }

}

although it is giving me directory read errors. Any assistance is greatly appreciated.

EDIT: Code edited, still saying permission denied cannot open directory on line 10. I am just going to run the script from within the logs directory if you are questioning the directory change to "."

Mike.

Upvotes: 1

Views: 5423

Answers (14)

impurethinker
impurethinker

Reputation: 1

Use perl from the command line, like a better grep

perl -wnl -e '/12.211.23.200/ and print;' *.log > output.txt

the benefit here is that you can chain logic far easier

perl -wnl -e '(/12.211.23.20[1-11]/ or /denied/i ) and print;' *.log

if you are feeling wacky you can also use more advanced command line options to feed perl one liner result into other perl one liners.

You really need to read "Minimal Perl: For UNIX and Linux People", awesome book on this very sort of thing.

Upvotes: 0

mattsmith321
mattsmith321

Reputation: 6851

I know I am way late to this discussion (ran across it while searching for grep related posts) but I am going to answer anyway:

It isn't specified clearly if these are web server logs (Apache, IIS, W3SVC, etc.) but the best tool for mining those for data is the LogParser tool from Microsoft. See logparser.com for more info.

LogParser will allow you to write SQL-like statements against the log files. It is very flexible and very fast.

Upvotes: 0

jrockway
jrockway

Reputation: 42704

BTW, I thought I would throw in a mention for File::Next. To iterate over all files in a directory (recursively):

use Path::Class; # always useful.
use File::Next;

my $files = File::Next::files( dir(qw/path to files/) ); # look in path/to/files
while( defined ( my $file = $files->() ) ){
    $file = file( $file );
    say "Examining $file";
    say "found foo" if $file->slurp =~ /foo/;
}

File::Next is taint-safe.

Upvotes: 1

Osama Al-Maadeed
Osama Al-Maadeed

Reputation: 5695

To get all the lines with the IP, I would directly use grep, no need to show a list of files, it's a simple command:

grep 12\.211\.23\.200 *

I like to pipe it to another file and then open that file in an editor...

If you insist on wanting the filenames, it's also easy

grep -l 12\.211\.23\.200 *

grep is available on all Unix//Linux with the GNU tools, or on windows using one of the many implementations (unxutils, cygwin, ...etc.)

Upvotes: 5

j_random_hacker
j_random_hacker

Reputation: 51326

First, use grep.

But if you don't want to, here are two small improvements you can make that I haven't seen mentioned yet:

1) Change:

@files = grep(/\.*$/,readdir(DIR));

to

@files = grep({ !-d "$dir/$_" } readdir(DIR));

This way you will exclude not just "." and ".." but also any other subdirectories that may exist in the server log directory (which the open downstream would otherwise choke on).

2) Change:

print if /12.211.23.200/;

to

print if /12\.211\.23\.200/;

"." is a regex wildcard meaning "any character". Changing it to "\." will reduce the number of false positives (unlikely to change your results in practice but it's more correct anyway).

Upvotes: -1

Adam Bellaire
Adam Bellaire

Reputation: 110539

You have to concatenate $dirname with $filname when using files found through readdir, remember you haven't chdir'ed into the directory where those files resides.

open FH, "<", "$dirname/$filname" or die "Cannot open $filname:$!";

Incidentally, why not just use grep -r to recursively search all subdirectories under your log dir for your string?

EDIT: I see your edits, and two things. First, this line:

@files = grep(/\.*$/,readdir(DIR));

Is not effective, because you are searching for zero or more . characters at the end of the string. Since it's zero or more, it'll match everything in the directory. If you're trying to exclude files ending in ., try this:

@files = grep(!/\.$/,readdir(DIR));

Note the ! sign for negation if you're trying to exclude those files. Otherwise (if you only want those files and I'm misunderstanding your intent), leave the ! out.

In any case, if you're getting your die message on line 10, most likely you're hitting a file that has permissions such that you can't read it. Try putting the filename in the die output so you can see which file it's failing on:

open FILE, "$file" or die "Unable to open file: $file";

But as with other answers, and to reiterate: Why not use grep? The unix command, not the Perl function.

Upvotes: 3

PolyThinker
PolyThinker

Reputation: 5218

Am I reading this right? Your line 10 that gives you the error is

open FILE, "$file" or die "Unable to open files";

And the $file you are trying to read, according to line 6,

@files = grep(/\.*$/,readdir(DIR));

is a file that ends with zero or more dot. Is this what you really wanted? This basically matches every file in the directory, including "." and "..". Maybe you don't have enough permission to open the parent directory for reading?

EDIT: if you only want to read all files (including hidden ones), you might want to use something like the following:

opendir(DIR, ".");
@files = readdir(DIR);
closedir(DIR);

foreach $file (@files) {
  if ($file ne "." and $file ne "..") {
    open FILE, "$file" or die "cannot open $file\n";
    # do stuff with FILE
  }
}

Note that this doesn't take care of sub directories.

Upvotes: 0

Jonathan Leffler
Jonathan Leffler

Reputation: 755094

Have you looked on CPAN for log parsers? I searched with 'log parse' and it yielded over 200 hits. Some (probably many) won't be relevant - some may be. It depends, in part, on which web server you are using.

Upvotes: 0

Alnitak
Alnitak

Reputation: 340055

Also, if you must use readdir(), make sure you guard the expression thus:

while (defined(my $filename = readdir(DH))) {
    ...
}

If you don't do the defined() test, the loop will terminate if it finds a file called '0'.

Upvotes: 0

Barry Brown
Barry Brown

Reputation: 20634

My first suggest would be to use grep instead. The right tool for the job, they say...

But to answer your question:

readdir just returns the filenames from the directory. You'll need to concatenate the directory name and filename together.

$path = "$dirname/$filname";
open FH, $path or die ...

Then you should ignore files that are actually directories, such as "." and "..". After getting the $path, check to see if it's a file.

if (-f $path) {
    open FH, $path or die ...
    while (<FH>)

Upvotes: 1

Kent Fredric
Kent Fredric

Reputation: 57414

~ doesn't auto-expand in Perl.

opendir my $fh,  '~/' or die("Doin It Wrong");  # Doing It Wrong. 

opendir my $fh, glob('~/') and die( "Thats right!" );

Upvotes: 0

A. Rex
A. Rex

Reputation: 32001

Can you use grep instead?

Upvotes: 14

benPearce
benPearce

Reputation: 38383

Firstly get a list of files within your source directory:

opendir(DIR, "$dir");
@files = grep(/\.log$/,readdir(DIR));
closedir(DIR);

And then loop through those files

foreach $file(@files)
{
  // file processing code
}

Upvotes: 1

ng.
ng.

Reputation: 7189

This will get the file names you are looking for in perl, and probably do it much faster than running and doing a perl regex.

@files = `find ~/ServerLogs -name "*.log" | xargs grep -l "<ip address>"`'

Although, this will require a *nix compliant system, or Cygwin on Windows.

Upvotes: 2

Related Questions