perl help to parse log file based on time input - perl

Perl newbie here. I have a log file that I need to parse out "Backup succeeded" and any "Error:" entries. I tried parsing the log file by using unix cat and piping it to grep. I got the information that I want, but I would like try this in perl and to also have the option to pass a date parameter and give me the lines based on the date I need.
Sample of log file output: (Backup succeeded)
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: flush-logs-time=00:00:00
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: backup-time=06:14:23
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: backup-status=Backup succeeded
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: Backup succeeded
Sample of log file output: (Error:)
Wed Jun 09 05:00:03 2010: rip1.mil.mad:backup:ERROR: mysql-zrm appears to be already running for this backupset
Wed Jun 09 05:00:03 2010: rip1.mil.mad:backup:ERROR: If you are sure mysql-zrm is not running, please remove the file /etc/mysql-zrm/rip1.mail.mad/.mysql-zrm.pid and restart mysql-zrm
**I would like a text and/or email with this information. Like so, but with the option to pass in the date I need.
Wed Jun 09 05:00:03 2010: rip1.mil.mad:backup:ERROR: mysql-zrm appears to be already running for this backupset
Wed Jun 09 05:00:03 2010: rip1.mil.mad:backup:ERROR: If you are sure mysql-zrm is not running, please remove the file /etc/mysql-zrm/rip1.mail.mad/.mysql-zrm.pid and restart mysql-zrm
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: backup-status=Backup succeeded
If you would please provide me with some perl code and/or ideas to get started. I would appreciate the help. Thank you.

#!/usr/bin/perl
# usage example: <this script> Jun 09 2010 <logfile>
use strict;
use warnings;
my ($mon,$day,$year) = ($ARGV[0],$ARGV[1],$ARGV[2]);
open(FH,"< $ARGV[3]") or die "can't open log file $ARGV[3]: $!\n";
while (my $line = <FH>) {
if ($line =~ /.* $mon $day \d{2}:\d{2}:\d{2} $year:.*(ERROR:|Backup succeeded)/) {
print $line;
}
}

Here's a simple script. The file name to scan and the target date are hard-coded. Matches are printed to STDOUT.
BTW, this code is totally untested. I typed it into the text box in my browser.
use strict;
use warnings;
my $logpath = './bar/log';
my $target = 'Jun 09 2010';
open my $fh, '<', $logpath or die "Error opening $logpath $!\n";
while (my $line = <$fh> ) {
next unless date_match( $target, $line );
next unless my $result = got_error($line) // got_backup($line);
print $result;
}
sub got_backup {
my $line = shift;
return unless $line =~ /backup-status=Backup succeeded/;
return $line;
}
sub got_error {
my $line = shift;
return unless $line =~ /:ERROR:/;
return $line;
}
# Take a line and a target date. Compare the date derived from the line to
# the target, and returns true if they match.
# Also always returns true if target is not defined
sub date_match {
my $target = shift;
my $line = shift;
return 1 unless defined $target; # Always true if target is undefined.
# Where did that god-awful date format come from? Yech.
my $date = extract_date($line);
return $date eq $target;
}
# Simple extract of date using split and join with extra variables
# to make it newbie friendly.
# IMO, it would be a good idea to switch to using DateTime objects and
# DateTime::Format::Strptime
sub extract_date {
my $line = shift;
my #parts = split /:/, $line;
my $date = join ':' #parts[0..2];
#parts = split /\s+/, $date;
$date = #parts[1,2,4];
return $date;
}
You can use Getopt::Long to get a filename and target date.
It would be a good idea to use a more robust date/time parsing and comparison scheme. DateTime and friends are very good, powerful modules for date manipulation. Check them out.
If you are processing tons of data and need to be more efficient, you can avoid copying $line everywhere in a number of ways.
For future reference, if you post a little code, you'll get better responses

Related

Perl Get File Last Modified Date Time No Module

I'm creating a script where I need to get the Last Modified Date of the files
I checked this thread How do I get a file's last modified time in Perl?
So I used the script below to get the last modified, at first it was working but when I try to run it again, the timestamp returns 00:00 January 1, 1970.
Why is that happening and how can I get the correct last modified date and time?
my $dir = '/tmp';
opendir(DIR, $dir) or die $!;
#content=readdir(DIR);
foreach(#content)
{
next unless ($_ =~ m/\bfile.txt|file2.csv\b/);
my $epoch_timestamp = (stat($_))[9];
my $timestamp = localtime($epoch_timestamp);
$f_detail = $_ .' '.$timestamp;
print "$f_detail\n";
}
closedir(DIR);
exit 0;
When I tried to run the perl, I will get this result
file.txt Thu Jan 1 00:00:00 1970
file2.csv Thu Jan 1 00:00:00 1970
Ok, last update, it is working now, I try to run all of the scripts you've given to me, standalone script. I found what's causing the default time, see the script below, I remove that in my program and it works, didn't notice this at first, sorry. But still, it feels weird because I was sure that it is working when I first run it, but now it is working so yeah thank you guys!
if (($month = ((localtime)[4] + 1)) < 10)
{
$month = '0' . $month;
}
if (($day = ((localtime)[3])) < 10)
{
$day = '0' . $day;
}
if (($year = ((localtime)[5]+1900)) >= 2000)
{
if (($year = $year - 2000) < 10)
{
$year = '0' . $year;
}
}
else
{
$year = $year - 1900;
}
$date = $month . $day . $year;
readdir returns file names without the full path. You need to prepend the path manually:
for (#content) {
next unless /^(?:file\.txt|file2\.csv)\z/;
my $epoch_timestamp = (stat("$dir/$_"))[9];
# ~~~~~~~~~
Also note how I changed the regex to match the file names.
If you have a directory name, and you want to see if some files whose names you already know exist in that directory, there's really no need for opendir/readdir - that's more helpful if you don't know the filenames ahead of time. When you do, you can just build a path using both parts and use file test operators/stat/etc. on it.
#!/usr/bin/env perl
use strict;
use warnings;
use feature qw/say/;
my $dir = '/tmp';
my #files = qw/file.txt file2.csv/;
for my $file (#files) {
# Better to use File::Spec->catfile($dir, $file), but your question
# title said no modules...
my $name = "$dir/$file";
if (-e $name) { # Does the file exist?
# _ to re-use the results of the above file test operator's stat call
my $epoch_timestamp = (stat _)[9];
my $timestamp = localtime $epoch_timestamp;
say "$file $timestamp";
}
}
Example execution:
$ perl demo.pl
file.txt Tue Feb 8 07:26:07 2022
file2.csv Tue Feb 8 07:26:10 2022
Following demo code utilizes glob to obtain modification time for specified files in a directory.
use strict;
use warnings;
use feature 'say';
my $dir = '/tmp';
my #files = qw(file.txt file2.csv);
my $mask = join ' ', map { "$dir/$_" } #files;
say "$_\t" . localtime((stat($_))[9]) for glob($mask);

Issue with Perl File::stat output, need to show date

I am having an issue with getting File::stat to output the last modified date of the file. This is my code so far:
#!/usr/bin/perl
use Time::localtime;
use File::stat;
use warnings;
use File::Find;
my $dirloc = 'E:\tmp\testdir';
sub find_txt {
my $F = $File::Find::name;
if ( ! -d $F && $F =~ /.tar|.exe|.zip/ ) {
my #result = $F;
foreach my $result (#result){
my $timestamp;
$timestamp = (stat("$result"))->[9] or die "No $_: $!";
print "$result : $timestamp\n";
}
}
}
find({wanted => \&find_txt}, $dirloc);
It is outputing something like this:
C:/tmp/testdir/foo/bar/test.tar : 1415305933
I need it to output instead (date format doesn't have to be what is listed, i just want to see the date):
C:/tmp/testdir/foo/bar/test.tar : 11/07/2014
I know that the output it is giving me is the time since epoch but I thought stat was supposed to give the date. Am I doing something wrong? Thanks!
edit: I have tried localtime, and i get: Time::tm=ARRAY(0x245b220), not sure what is happening there
You can use the localtime (Note: not Time::localtime) function to convert the timestamp into something useful
my $date = localtime $timestamp
which will make it a human readable string like Fri Nov 7 15:33:00 2014
Or you can use it in a list context to spit it into individual fields:
my($sec, $min, $hour, $day, $month, $year, $weekday, $yearOfDay, $isDST) = localtime $timestamp

endless loop in perl

I want to extract the content except http header from tcp flow files
the content is like the following
the http header ends when two ^M are met
HTTP/1.1 200 OK^M
Last-Modified: Sat, 20 Mar 2010 09:43:12 GMT^M
Content-Type: video/x-flv^M
Date: Wed, 24 Oct 2012 14:34:13 GMT^M
Expires: Wed, 24 Oct 2012 14:34:13 GMT^M
Cache-Control: private, max-age=22124^M
Accept-Ranges: bytes^M
Content-Length: 29833281^M
Connection: close^M
X-Content-Type-Options: nosniff^M
Server: gvs 1.0^M
^M
FLV^A^E^#^#^# ^#^#^#^#^R^#^CK^#^#^#^#^#^#^#^B^#
onMetaData^H^#^#^#^O^#^Hduration^##i<97>
=p£×^# starttime^#^#^#^#^#^#^#^#^#^#^Mtotalduration^##i<97>
my code for extraction is as follows,
and I run : extract.pl < tcp.flow
but it seems the loop is endless,
what is wrong with the codes? thanks!
#!/usr/bin/perl
$start=0;
$data="";
while(<STDIN>)
{
if ( $start eq 0 && $_ =~ /^\r\n/) { $start = 1; }
elsif ( $start eq 1 ) { $data = $data . $_; }
}
open(FH, ">sample.flv");
print FH $data;
close(FH);
This is a one-liner. I see no reason for any endless loop, however.
perl -00 -lne '$i++ and print' file > sample.flv
Which deparsed looks like this:
>perl -MO=Deparse -00 -lne '$i++ and print' input.txt
BEGIN { $/ = ""; $\ = "\n\n"; } # from -l and -00
LINE: while (defined($_ = <ARGV>)) { # from -n
chomp $_; # from -l, removes "\n\n" now
print $_ if $i++; # skips the first line
}
-e syntax OK
If you need to clean your file up first, just do
perl -pi -le 's/[\r\n]+$//' input.txt
Call binmode() on STDIN before reading the data, it's possible that the contents of the file are interfering with the file reading. You'll want to use it on FH as well before writing the data. HTH

How to read a file and save the contents till we encounter the first blank line in perl script

I am trying to read a file and save the lines which starts with $path till it encounters first balnk line in an array. I have the below code, bt it only prints the path name and not the lines. Could some-one have a look.
Below are the contents of the $file:
\sbd\archieve\date\form
-rwxrwxrwx 1 etd maadm 4354270 Aug 16 21:56 COMAHCUT.dat.20120816.ftpd.201208162156*
-rw-r--r-- 1 etd maadm 0 Aug 16 21:56 COMAHCUT.DONE.20120816.ftpd.201208162156
\sbd\single\archieve\date\form
-rwxr-xr-x 1 etd maadm 1362780 Aug 15 22:02 COMAINS.dat.ftpd.201208152203*
-rwxr-xr-x 1 etd maadm 0 Aug 15 22:02 COMAINS.DONE.ftpd.201208152203*
Below is the code i tried:
#!/usr/bin/perl
my $file = "/home/pauler/practice/DataIt/line.txt";
open (INFO, $file) or die "Cannot open the file $file :$! \n";
my $path = "\sbd\archieve\date\form";
foreach $line (<INFO>) {
if ($line =~ m/$path/) {
push (#array1, $line);
last if ($line =~ m/^$/);
print #array1;
}
}
You can take advantage of the fact, that filehandles remember their position in the file.
use strict;
use warnings;
my #array;
my $path = '\sbd\archieve\date\form';
while ( my $line = <DATA> ) {
next unless $line =~ /\Q$path\E/;
push #array, $line;
while ( my $line = <DATA> ) {
last if $line =~ /^\s*$/;
push #array, $line;
}
}
print #array;
__DATA__
\sbd\archieve\date\form
-rwxrwxrwx 1 etd maadm 4354270 Aug 16 21:56 COMAHCUT.dat.20120816.ftpd.201208162156*
-rw-r--r-- 1 etd maadm 0 Aug 16 21:56 COMAHCUT.DONE.20120816.ftpd.201208162156
\sbd\single\archieve\date\form
-rwxr-xr-x 1 etd maadm 1362780 Aug 15 22:02 COMAINS.dat.ftpd.201208152203*
-rwxr-xr-x 1 etd maadm 0 Aug 15 22:02 COMAINS.DONE.ftpd.201208152203*
The flip-flop operator .. saves life ... our you code. It stays false until the expression on the left returns true, and remains true until the expression on the right turns true ... then it is false again until the left expressions evaluates to true again.
# read lines into $_ for cleaner code
while (<INFO>) {
if (/$path/ .. /^$/) {
push #array1, $_;
}
}
print #array1;
Oh, and a note on paths ... I know no single Operating System that really needs backslashes, not even Windows … Using normal slashes / will save you from weird escape sequences and other magic that lurks in the dark

How do I convert epoch time to normal time in Perl?

I am attempting to write a Perl script that parses a log where on each line the second value is the date. The script takes in three arguments: the input log file, the start time, and the end time. The start and end time are used to parse out a certain value on each line that that falls between those two times. But to properly run this I am converting the start and end time to epoch time. The problem I am having is that to convert the loops 'i' value back to normal time to compare against the log file. After running localtime($i) I print the value and only see a reference printed not the actual value.
Here is the script I have so far (it is a work in progress):
#!/usr/bin/perl
use strict;
use warnings;
use Time::Local;
use Time::localtime;
use File::stat;
my $sec = 0;
my $min = 0;
my $hour = 0;
my $mday = 0;
my $mon = 0;
my $year = 0;
my $wday = 0;
my $yday = 0;
my $isdst = 0;
##########################
# Get the engine log date
##########################
my $date = `grep -m 1 'Metric' "$ARGV[0]" | awk '{print \$2}'`;
($year,$mon,$mday) = split('-', $date);
$mon--;
#########################################
# Calculate the start and end epoch time
#########################################
($hour,$min,$sec) = split(':', $ARGV[1]);
my $startTime = timelocal($sec,$min,$hour,$mday,$mon,$year);
($hour,$min,$sec) = split(':', $ARGV[2]);
my $endTime = timelocal($sec,$min,$hour,$mday,$mon,$year);
my $theTime = 0;
for (my $i = $startTime; $i <= $endTime + 29; $i++) {
#print "$startTime $i \n";
$theTime = localtime($i);
#my $DBInstance0 = `grep "$hour:$min:$sec" "$ARGV[0]"`;# | grep 'DBInstance-0' | awk '{print \$9}'`;
#print "$DBInstance0\n";
print "$theTime\n";
}
print "$startTime $endTime \n";
The output looks like:
Time::tm=ARRAY(0x8cbbd40)
Time::tm=ARRAY(0x8cbc1a0)
Time::tm=ARRAY(0x8cbbe80)
Time::tm=ARRAY(0x8cbc190)
Time::tm=ARRAY(0x8bbb170)
Time::tm=ARRAY(0x8cbc180)
Time::tm=ARRAY(0x8cbbf30)
Time::tm=ARRAY(0x8cbc170)
Time::tm=ARRAY(0x8cbc210)
Time::tm=ARRAY(0x8cbc160)
1275760356 1275760773
I only have access to the core Perl modules and am unable to install any others.
You can use ctime, depending on your definition of "Normal time":
Example code:
use Time::Local;
use Time::localtime;
my $time=timelocal(1,2,3,24,6,2010);
print "$time\n";
$theTime = ctime($time);
print "$theTime\n";
Result:
1279954921
Sat Jul 24 03:02:01 2010
Also, you don't need to use Time::Localtime (which is why you get Time::tm instead of a standard array/string from Perl's internal localtime):
use Time::Local;
my $time=timelocal(1,2,3,24,6,2010);
print "$time\n";
$theTime = localtime($time);
print "$theTime\n";
1279954921
Sat Jul 24 03:02:01 2010
Don't forget to subtract 1900 from the year!
Remember that in scalar context, localtime and gmtime returns a ctime-formatted string, so you could use it as in the following. If that's unsuitable, you might want to use strftime from the POSIX module.
#! /usr/bin/perl
use warnings;
use strict;
use Time::Local;
my $start = "01:02:03";
my $end = "01:02:05";
my $date = "2010-02-10";
my($year,$mon,$mday) = split /-/, $date;
$mon--;
$year -= 1900;
my($startTime,$endTime) =
map { my($hour,$min,$sec) = split /:/;
timelocal $sec,$min,$hour,$mday,$mon,$year }
$start, $end;
for (my $i = $startTime; $i <= $endTime + 29; $i++) {
print scalar localtime($i), "\n";
}
print "$startTime $endTime \n";
Tail of the output:
Wed Feb 10 01:02:26 2010
Wed Feb 10 01:02:27 2010
Wed Feb 10 01:02:28 2010
Wed Feb 10 01:02:29 2010
Wed Feb 10 01:02:30 2010
Wed Feb 10 01:02:31 2010
Wed Feb 10 01:02:32 2010
Wed Feb 10 01:02:33 2010
Wed Feb 10 01:02:34 2010
1265785323 1265785325