matze77 has asked for the wisdom of the Perl Monks concerning the following question:

Hello Monks!
I am a little stuck, maybe you could help ...
I have to parse a logfile on a windows machine which contains (at the last line)
$string as mentioned below, i want to match against this last line and $string should contain the actual date ... 

my $string='19.09.2011   ======== Ende Update';

my $date = `Date /T`;
my $regex ="$date  Ende  ";

[download]
To simplify i only want to check for $date and "Ende" on this line ...
My problem is there are whitespaces or other characters between $date and Ende
So how to change the regex to match? Any help?

Thanks
MH

Replies are listed 'Best First'.
Re: Parse Log File for Date and string?
by Caio (Acolyte) on Sep 19, 2011 at 10:27 UTC
    IF i did get this straight, you want to extract the date and "Ende" from the last line.. so I'd extract it with

    /(\d{2}\.\d{2}\.\d{4})\s*=*\s*(Ende)/

    which goes for 2 digits dot 2 digits dot 4 digits any number of spaces followed by any number of = followed by any number os spaces again followed by Ende

    In the end you can capture the date and "ende" separately with $1 and $2, while discarding the:
    " ============ " in between.

    Hope it helps ;)
[reply]
[d/l]
      Thanks for all the help now it works 

      use strict;
use warnings;
my $returncode;
$returncode = 0;

open (my $IN,"<E:/scripts/eva/allupdates.log") || die "can not open da
+tei: $!";
#regex für Datumsanalyse matcht auf Datum z.B. 30.11.2010
my $redate='((?:(?:[0-2]?\\d{1})|(?:[3][0,1]{1}))[-:\\/.](?:[0]?[1-9]|
+[1][012])[-:\\/.](?:(?:[1]{1}\\d{1}\\d{1}\\d{1})|(?:[2]{1}\\d{3})))(?
+![\\d])';    # DDMMYYYY 1


#Datumsbestimmung tatsächliches Datum:
(my $sec, my $min, my $hour, my $mday,my $mon, my $year, my $wday, my 
+$ydat, my $isdst)=localtime();
             my $jahr=$year;
             my $monat=$mon+1;
             my $tag=$mday;
             $jahr=$year;

$jahr=$year +1900;

if (length($monat) == 1)
{
    $monat="0$monat";
}
if(length($tag) == 1)
{
   $tag="0$tag";
}
if(length($hour) == 1)
{
   $hour="0$hour";
}
if(length($min) == 1)
{
   $min="0$min";
}
if(length($sec) == 1)
{
   $sec="0$sec";
}

my $Xdatum=$tag.".".$monat.".".$jahr;
my $Xzeit=$hour.":".$min.":".$sec;

#$Xdatum="$jahr$monat$tag";
#$Xzeit="$hour$min$sec";

#print $Xdatum."  ".$Xzeit;

#Datum im Logfile evawp.log
my $datelogfile;

#Vergleich Datum in Logdatei mit Systemdatum, wenn dies gleich ist, ok
+, $redate matcht auf Datumsstring, der 1. String ist das Datum ...
while(<$IN>){
  if( /($redate)(\s+).*?(Ende)/i){
          print " ,1. Variable $1\n";
         $datelogfile = $1;
         print " Datelogfile $datelogfile\n";
         print " Var Xdatum $Xdatum\n";
         if ( $datelogfile eq $Xdatum)

                { print " Variable datelogfile $datelogfile\n";
                    $returncode++; }

print;
         
  print "$returncode\n";
 }

 }
#wenn error nicht gefunden wurde dann auch die Datei nicht öffnen zum 
+schreiben ...
#für jeden Fund wird returncode um 1 erhöht ...


 if ( $returncode == 1 ){
  

print "Update TS ok!\n";
}


#Wenn Returncode 0 ist der String nicht gefunden, dann Fehler ...
if ( $returncode <= 0 ){
  

print "Update TS FEHLER!\n";

}
#print "$returncode\n";

close $IN;



#print "$datelogfile \n";
system("pause");

      [download]

      Thanks
      MH
[reply]
[d/l]

        sprintf is your friend:

        my ( $sec, $min, $hour, $mday, $mon, $year, $wday, $ydat, $isdst ) = l
+ocaltime();
my $jahr = $year + 1900;
my $monat = sprintf "%02d", $mon + 1;
my $tag = sprintf "%02d", $mday;
$hour = sprintf "%02d", $hour;
$min = sprintf "%02d", $min;
$sec = sprintf "%02d", $sec;

        [download]
[reply]
[d/l]
Re: Parse Log File for Date and string?
by RichardK (Parson) on Sep 19, 2011 at 10:36 UTC
    Perhaps reading the perlretut regex tutorial might help ?
[reply]
Re: Parse Log File for Date and string?
by norbert.csongradi (Beadle) on Sep 19, 2011 at 09:58 UTC

    To tell the truth, I do not clearly see it as a regexp...

    Maybe you are looking for something like this (it consumes any other characters in between, trying to match as little as possible, so not consuming your Ende):

    my $regex = "$date.*?Ende";

    [download]
[reply]
[d/l]

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