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Re^4: WWW::Mechanize problem

by Anonymous Monk
on Oct 20, 2005 at 16:04 UTC ( #501718=note: print w/replies, xml ) Need Help??


in reply to Re^3: WWW::Mechanize problem
in thread WWW::Mechanize problem

HI,

Thank you for your help.
Can you explain how this method can be usedd to walk up and down the links?
I dont think i understand how the link depth comment you made works.
Thanks again

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Re^5: WWW::Mechanize problem
by Limbic~Region (Chancellor) on Oct 20, 2005 at 16:24 UTC
    Anonymous Monk,
    Sometimes, it helps to stop thinking about the problem at hand. You get too close to the problem and you miss the forest through the trees. The process I am about to explain can be applied to any similar problem and is not unique to your situation. Once you understand that - you have the tools you need to solve it yourself in the future.

    Problem Description:

    You start out with some number of jobs to perform. In working on those jobs, you discover that you have new jobs to work on. The total number of jobs to perform can't be known ahead of time. Additionally, it is possible for one job to lead to another job which leads back to the original job. If we think of the number of known jobs at any given time as a stack or a queue, then we know we can stop work when it is empty.
    my @work = fetch_jobs($starting_condition); while ( @work ) { # ... }

    When we evaluate an array in this context, it will be false when the array is empty. The while loop will terminate because all work has been completed. Inside the loop, we add to our known work queue/stack by checking to see if the job leads to more work.

    my @work = fetch_jobs($starting_condition); while ( @work ) { # Remove 1 item from our stack/queue my $job = shift @work; # Possibly add new jobs to our stack/queue if ( more_jobs($job) ) { push @work, fetch_jobs($job); } # process job }
    We now need to consider that one job may lead back to itself and break the infinite loop.
    my @work = fetch_jobs($starting_condition); my %seen; while ( @work ) { # Remove 1 item from our stack/queue my $job = shift @work; # Skip this job if we have already done it next if $seen{$job}++; # Possibly add new jobs to our stack/queue if ( more_jobs($job) ) { push @work, fetch_jobs($job); } # process job }
    We can additionally decide to abandon our work if we discover that our queue/stack has grown larger than we anticipated. We rely on the fact that when an array is evaluated in scalar context it returns the number of elements present.
    my @work = fetch_jobs($starting_condition); my %seen; while ( @work && @work < 1000 ) { # ... }
    Now it may be important to process the work in a specific order. A depth first approach is when one job leads to another job which leads to another job and they need to be processed it that order. A breadth first approach is when secondary and tertiary jobs are only executed after all primary jobs are complete. The way to control this is by adjusting what end of the stack/queue you take work off and put on. See push, pop, shift, unshift for more details.

    Cheers - L~R

    Update: Minor oversights corrected per the astute herveus.

      Howdy!

      A couple of emendations that don't change the essential meaning but do affect whether it works as presented...

      my @work = fetch_jobs($starting_condition); my %seen; while ( @work ) { # Remove 1 item from our stack/queue my $job = shift @work; # Skip this job if we have already done it next if $seen{$_}++; ## $_ should probably be $job, no? # Possibly add new jobs to our stack/queue if ( more_jobs($_) ) { ## $_ should probably be $job, no? push @work, fetch_jobs($_); ## $_ should probably be $job, no? } # process job }
      ...because $_ isn't set or otherwise given meaning in this snippet.

      while ( @work || @work < 1000 ) { # should ought to be && not || }
      ...otherwise the loop will never exit, as an empty @work will always be less than 1000...

      yours,
      Michael

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