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Re: Deleting a matching string in an array

by thezip (Vicar)
on Oct 30, 2007 at 17:35 UTC ( #648069=note: print w/replies, xml ) Need Help??

in reply to Deleting a matching string in an array

If I may, I have a couple of other constructive points to mention that don't really pertain to the question at hand, but rather coding style.

There are escaped escaped backslashes in the declarations section, and this is not necessary.


can be rewritten as:


Also, it is preferable to use singles quotes instead of double-quotes, but only when you don't need to interpolate any variables within the string.

# This: # my $x="ex"; # Becomes: my $x= 'ex'; # ... since there are no variables to interpolate # This is wrong: # my $foo = '/foo/$bar/baz'; # $bar is a variable and does not get int +erpolated # instead, it's literally the string '$ba +r' # It must be written with the double-quotes to interpolate: # my $foo = "/foo/$bar/baz";

Where do you want *them* to go today?

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Re^2: Deleting a matching string in an array
by Shamaeso (Initiate) on Oct 30, 2007 at 19:51 UTC

    Thanks, I did the double slashes since I am using a unc path and did it the way I learned like 9 years ago in a PERL class.

    I got the code to remove the unwanted lines but it is not elegant, but I wanted to post it for more comments and advice on how to improve.

    I did try a switch statement but could not get it to work so just built the ugly if statement

    Thanks again for the help and direction

    #!perl #use strict; use Text::CSV_XS; use warnings; use FileHandle; # declarations my($log_file_path)="\\\\dt00mx84\\LogArchive\\\\dt00m +h77\\"; my($robot_file)="\\\\dt00mx84\\LogArchive\\Webrobots.txt"; my($second, $minute, $hour, $dayOfMonth, $month, $yearOffset, $day +OfWeek, $dayOfYear, $daylightSavings) = localtime(); my$x="ex"; my($extension)=".log"; my($year)=1900+$yearOffset; my($month_new)=1+$month; my $day=$dayOfMonth-1; my @logmessages; # build filename format # this is how log files are named using the yy/mm/dd formatex040101.lo +g if (length($month_new)< 2) { $month_new="0".$month_new }; if (length($dayOfMonth)< 2) { $dayOfMonth="0".$dayOfMonth }; # build input file name and file path to read from my($filename)=$x.substr($year,2).$month_new.$day; my($log_file)=$log_file_path.$filename.$extension; # build output file name and path to write file my($file_name)=substr($year,2).$month_new.$day; my($out_file)=$log_file_path.$file_name.$extension; # Declare the FileHandles and open the input and output files my $fh= new FileHandle; open(LOG, "<$log_file") or die "Could not open file"; @logmessages=<LOG>; close(LOG); my $outfile= new FileHandle; open(OUTFILE, ">$out_file") or die "Could not open file"; foreach $LogLine (@logmessages){ if(($LogLine!~/Slurp/) && ($LogLine!~/Jeeves/)&&($LogLine!~/Go +oglebot/)&&($LogLine!~/FunWebProducts/)&&($LogLine!~/msnbot.htm/)&&($ +LogLine!~/PeoplePal/)&&($LogLine!~/ventura5/)&&($LogLine!~/Speedy/)&& +($LogLine!~/GovDelivery/)&&($LogLine !~ /gif/)&&($LogLine !~ /jpg/)&& +($LogLine !~ /ico/)&&($LogLine !~ /css/)&&($LogLine !~ /js/)&&($LogLi +ne !~ /archive/)&&($LogLine !~ /CazoodleBot/)&&($LogLine !~ /WebTrend +s/)&&($LogLine !~ /ShopWiki/)&&($LogLine !~ /Ultraseek/)&&($LogLine ! +~ /msrbot/)&&($LogLine !~ /Moskow/)&&($LogLine !~ /Gigabot/)) { print OUTFILE $LogLine; } }
      Why not use grep as Moritz pointed out? If you loaded the robot file into the @robots array, you would need to remove the newlines like chomp(@robots) before you used it in the grep. Your date string could be stated like:
      my ($day, $month, $year) = (localtime)[3..5]; # $date is YYMMDD format - you may want $day - 1? my $date = sprintf "%02d%02d%02d", $year % 100, $month + 1, $day; my $log_file_path = '/dt00mx84/LogArchive/'; my $log_file = $log_file_path . 'ex' . $date. '.log'; my $out_file = $log_file_path . $date. '.log';


      Update: Not_a_Number nailed it. Didn't think about day-1 being yesterday.

        I kept getting an error: Nested quantifiers in regex; marked by <-- HERE in m/#Software: Microsoft Internet Information Services 6.0

        I figured I would get it to remove one of the filters and not get the error then move ahead from there.

        Being new to using PERL for this stuff I figured to get the code working and improve on it from there.

        I am going to use these suggestions and get it working but the ugly fix I came up with satisfies my supervisor.

      I wanted to post it for more comments and advice on how to improve ;)

      Well, I can see several areas for improvement, notably the huge if block at the end.

      (BTW, why do you declare a variable $robot_file that you never subsequently use? And why have you commented out use strict; at the beginning of your program?)

      Immediately, however, one thing that springs to mind is this line:

      my $day=$dayOfMonth-1;

      It seems to me, in the context, that you're trying here to use this to get the date of the previous day. If I'm wrong, please excuse me, but if not, what do you think will happen, for example on November 1st (not to mention Jan 1st, or March 1st, when the previous day might be either Feb 28 or Feb 29)?

      In fact, 'How do I find yesterday's date?' is a faq (How do I find yesterday's date?). With Perl (not PERL, by the way), there are modules out there that deal with heaps of similar problems, and save you from reinventing the proverbial wheel :).

      Here's one way, based on the answers to the faq, to create the sort of string that you seem to require, using the DateTime module (Don't hesitate, BTW, to use whitespace in your code, to make it more human-legible):

      use DateTime; my $yesterday = DateTime->today->subtract( days => 1 )->ymd( '' ); my ( $prefix, $extension ) = ( 'ex', '.log' ); my $log_file = $prefix . substr( $yesterday, 2 ) . $extension;

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