如何用Perl構建一個家族樹？

Question

我在Perl編程作業，要求我做到以下幾點：

1. 創建在MySQL數據庫中的表，並插入這些記錄到其中：

2. 負載從表中的數據到Son類的實例數組中。

3. 使用該數組創建表示父子樹的HTML代碼，並將該html代碼打印到STDOUT。沒有必要讓樹看起來不錯。這樣的事情就可以了：

tree http://i25.tinypic.com/314t177.png

我跑出來的想法，請幫助。我的代碼如下：

#!/usr/bin/perl 

use strict; 
use Son; 
use CGI; 
use Data::Dumper; 
use DBI; 
my $q = new CGI; 

#DB connect vars 
my $user = "##"; 
my $pass = "##"; 
my $db = "##"; 
my $host = "localhost"; 

my $dsn = "DBI:mysql:database=$db;host=$host"; 

my $dbh = DBI->connect($dsn,$user,$pass); 
eval { $dbh->do("DROP TABLE sons") }; 
print "Drop failed: $@\n" if $@; 

$dbh->do("CREATE TABLE sons (son VARCHAR(30) PRIMARY KEY, father VARCHAR(30))"); 

my @rows = (["bill", "sam"], 
     ["bob", ""], 
     ["jack", "sam"], 
     ["jone", "mike"], 
     ["mike", "bob"], 
     ["sam", "bob"] 
); 

for my $i (0 .. $#rows) { 
    $dbh->do("INSERT INTO sons (son, father) VALUES (?,?)", {}, $rows[$i][0], $rows[$i][1]); 
} 

our @sons_array; 
my $sth = $dbh->prepare("SELECT * FROM sons"); 
$sth->execute(); 
while (my $ref = $sth->fetchrow_hashref()) { 
    $sons_array[++$#sons_array] = Son->new($ref->{'son'}, $ref->{'father'}); 
} 
$sth->finish(); 
$dbh->disconnect(); 


print $q->header("text/html"),$q->start_html("Perl CGI"); 
print "\n\n"; 
constructFamilyTree(@sons_array, ''); 
print $q->end_html; 

sub constructFamilyTree { 
    my @sons_array = @_[0..$#_ -1]; 
    my $print_father; 
    my $print_son; 
    my $print_relation; 
    my $current_parent = @_[$#_]; 
    my @new_sons_array; 
    my @new_siblings; 

    #print $current_parent."\n"; 
    foreach my $item (@sons_array){ 
     if(!$item->{'son'} || $item->{'son'} eq $item->{'father'}) { # == ($item->{'son'} eq '') 
      print "\n List contains bad data\n"; 
      return 0; 
     } 

     if($item->{'father'} eq $current_parent) { 
      my $temp_print_relation; 
      foreach my $child (@sons_array) { 
       if($child->{'father'} eq $item->{'son'}) { 
        if(!$temp_print_relation) { 
         $temp_print_relation .= ' |'; 
        } 
        else { 
         $temp_print_relation .= '-----|'; 
        } 
       } 
      } 
      $print_relation .= $temp_print_relation." "; 
      $print_son .= '('.$item->{'son'}.') '; 
      @new_siblings[++$#new_siblings] = $item; 
      $print_father = $item->{'father'}; 
     } 
     else { 
      $new_sons_array[++$#new_sons_array] = $item; 
     } 
    } 

    print $print_son. "\n". $print_relation."\n"; 
    #print $print_father."\n"; 
    #print $print_relation . "\n". $print_son; 
    foreach my $item (@new_siblings) { 
     constructFamilyTree(@new_sons_array, $item->{'son'}); 
    } 
} 


perl module: 
#File Son.pm, module for class Son 

package Son; 

sub new { 
    my($class, $son, $father) = @_; 
    my $self = {'son' => $son, 
       'father' => $father}; 

    bless $self, $class; 
    return $self; 
} 

1; 

Answer 1

在等待澄清的問題是什麼，我想看到你在某種學習機構的越來越給出的Perl有關的工作的時候，我的理由有沒有更好的時間來介紹你到穆斯和CPAN，你真的應該在現實世界中使用的東西。

它及其各種擴展將使您的生活更輕鬆，並使面向對象的設計更加直接和可維護。

#!/usr/bin/perl 
use strict; 
use warnings; 
use Data::Dumper; 
use Moose::Autobox; 
use 5.010; 

sub Moose::Autobox::SCALAR::sprintf { 
    my $self = shift; 
    sprintf($self, @_); 
} 

{ 

    package Son; 
    use Moose; 
    use MooseX::Types::Moose qw(:all); 
    use MooseX::ClassAttribute; 
    use MooseX::Has::Sugar 0.0300; 
    use Moose::Autobox; 

    class_has 'Ancestry' => (isa => HashRef, rw, default => sub { {} }); 
    class_has 'People' => (isa => HashRef, rw, default => sub { {} }); 
    has 'name'   => (isa => Str,  rw, required); 
    has 'father'   => (isa => Str,  rw, required); 

    sub BUILD { 
    my $self = shift; 
    $self->Ancestry->{ $self->name } //= {}; 
    $self->Ancestry->{ $self->father } //= {}; 
    $self->People->{ $self->name }  //= $self; 
    $self->Ancestry->{ $self->father }->{ $self->name } = $self->Ancestry->{ $self->name }; 
    } 

    sub children { 
    my $self = shift; 
    $self->subtree->keys; 
    } 

    sub subtree { 
    my $self = shift; 
    $self->Ancestry->{ $self->name }; 
    } 

    sub find_person { 
    my ($self, $name) = @_; 
    return $self->People->{$name}; 
    } 

    sub visualise { 
    my $self = shift; 
    '<ul><li class="person">%s</li></ul>'->sprintf($self->visualise_t); 
    } 

    sub visualise_t { 
    my $self = shift; 
    '%s <ul>%s</ul>'->sprintf(
     $self->name, 
     $self->children->map(
     sub { 
      '<li class="person">%s</li>'->sprintf($self->find_person($_)->visualise_t); 
     } 
     )->join('') 
    ); 
    } 
    __PACKAGE__->meta->make_immutable; 
} 

my @rows = ([ "bill", "sam" ], [ "bob", "" ], [ "jack", "sam" ], [ "jone", "mike" ], [ "mike", "bob" ], [ "sam", "bob" ],); 

for (@rows) { 
    Son->new(
    father => $_->at(1), 
    name => $_->at(0), 
); 
} 

<<'EOX'->sprintf(Son->find_person('bob')->visualise)->say; 
<html> 
    <head> 
    <style> 
     li.person { 
border: 1px solid #000; 
padding: 4px; 
margin: 3px; 
background-color: rgba(0,0,0,0.05); 
     } 
    </style> 
    </head> 
    <body> 
    %s 
    </body> 
</html> 
EOX 

Answer 2

雖然我很喜歡從Kent Fredric's answer（看，我幾乎沒有寫任何超出使用穆斯簡單的練習），我想你可能會通過查看一個較爲傳統的解決方案，以顯示問題了解更多學習數據結構。 它不直接解決你的問題（我假設你的問題是基於家庭作業）。如果代碼證明是有用的，我相信你的老師會很感激你，如果你引用任何你收到的外部幫助。

#!/usr/bin/perl 

use strict; 
use warnings; 

my @rows = (
    [ bill => 'sam' ], 
    [ bob => ''  ], 
    [ jack => 'sam' ], 
    [ jone => 'mike' ], 
    [ mike => 'bob' ], 
    [ sam => 'bob' ], 
    [ jim => ''  ], 
    [ ali => 'jim' ], 
); 

my %father_son; 

for my $pair (@rows) { 
    push @{ $father_son{ $pair->[1] } }, $pair->[0]; 
} 

for my $root (@{ $father_son{''} }) { 
    print_branch($root, 0); 
} 

sub print_branch { 
    my ($branch, $level) = @_; 
    print "\t" x $level, $branch, "\n"; 
    if (exists $father_son{$branch}) { 
     for my $next_branch (@{ $father_son{$branch} }) { 
      print_branch($next_branch, $level + 1); 
     } 
    } 
    return; 
} 

__END__ 

輸出：

C:\Temp> tkl 
bob 
     mike 
       jone 
     sam 
       bill 
       jack 
jim 
     ali 

Answer 3

使用GraphViz。這比自己製作圖片要容易得多。