Perl將xml解碼為hash

Question

我需要解碼復雜的XML結構。 XML看起來像這樣：

<?xml version="1.0" encoding="ISO-8859-1"?>
    <MainNode comment="foo">
      <FirstMainBranch>
        <Struct>
          <String name="aStringValueUnderMainBranch" comment="Child node under first main branch"/>
          <String name="anotherStringValueUnderMainBranch" comment="Child node under first main branch"/>
          <Integer name="anIntegerValueUnderMainBranch" comment="Child node under first main branch"/>
          <List name="aList" comment="According to me this node should be an array, it could contain one or more child elements">
            <Struct comment="The node name means that, the child nodes are grouped, I think that the most appropriate structure here is hash. 
        The node itself doesn't have name attribute, which means that it only shows the type of the element">
          <String name="first" comment="
            Default Value: 0 
                        "/>
          <Long name="second" comment="
            Default Value: 0 

                          "/>
          <Long name="third" comment="
            Default Value: 0 

                        "/>
        </Struct>
      </List>
      <List name="secondList" comment="According to me this node should be array, it could contain one or more child elements">
        <Struct comment="The node name means that, the child nodes are grouped, I think that the most appropriate structure here is hash. 
        The node itself doesn't have name attribute, which means that it only shows the type of the element
                    ">
          <String name="first" comment="
            Default Value: 0 

                          "/>
          <Long name="second" comment="
            Default Value: 0 

                          "/>        
        </Struct>
      </List>
      <Struct name="namedStruct" comment="Here the struct element has a name, which means that it should be decoded
                    ">
        <List name="thirdList" comment="Again list, but now it is inside struct element, and it contains struct element
                ">
          <Struct comment="The node name means that, the child nodes are grouped, I think that the most appropriate structure here is hash.">
            <Integer name="first" comment="Child element of the struct"/>
          </Struct>
        </List>

      </Struct>

    </Struct>
  </FirstMainBranch>
  <SecondMainBranch>
    <Struct comment="">
      <Struct name="namedStructAgain" comment="
                ">
        <String name="First" comment="
                  "/>
        <String name="Second" comment=""/>

      </Struct>
    </Struct>
  </SecondMainBranch>
</MainNode>

我認為最合適的容器是哈希（如果您的意見不同，請告訴我）。 我發現難以解碼它，因為：

1. 主節點沒有“name”屬性，但它們應該存在於最終結構中

2. 只有存在“name”屬性時才應讀取子節點，但它們的數據類型（結構）取決於未解碼的父元素。

3. 其中一些父元素具有“name”屬性 - 在這種情況下，它們應該存在於最終結構中。

4. 我不關心整數，長整數，日期時間等數據類型，它們將被讀作字符串。 這里的主要問題是List和Struct類型

這是我愚蠢地嘗試應對任務：

use XML::LibXML;
use Data::Dumper;
use strict;
use warnings;
my $parser=XML::LibXML->new();
my $file="c:\\joro\\Data.xml";
my $xmldoc=$parser->parse_file($file);

sub buildHash{
my $mainParentNode=$_[0];
my $mainHash=\%{$_[1]};
my ($waitNextNode,$isArray,$arrayNode);
$waitNextNode=0;
$isArray=0;
sub xmlStructure{
my $parentNode=$_[0];
my $href=\%{$_[1]};
my ($name, %tmp);
my $parentType=$parentNode->nodeName();
$name=$parentNode->findnodes('@name');
foreach my $currentNode($parentNode->findnodes('child::*')){
my $type=$currentNode->nodeName();
if ($type&&$type eq 'List'){
$isArray=1;
}
elsif($type&&$type ne 'List'&&$parentType ne 'List'){
$isArray=0;
$arrayNode=undef;
}
if ($type&&!$currentNode->findnodes('@name')&&$type eq 'Struct'){
$waitNextNode=1;
}
else{
$waitNextNode=0;
}
if ($type&&$type ne 'List'&&$type ne 'Struct'&&!$currentNode->findnodes('@name')){
#$href->{$currentNode->nodeName()}={};
xmlStructure($currentNode,$href->{$currentNode->nodeName()});
}
# elsif ($type&&$type eq 'List'&&$currentNode->findnodes('@name')){
# print "2\n";
# $href->{$currentNode->findnodes('@name')}=[];
# xmlStructure($currentNode,$href->{$currentNode->findnodes('@name')});
# }
elsif ($type&&$type ne 'List'&&$currentNode->findnodes('@name')&&$parentType eq 'List'){
push(@{$href->{$currentNode->findnodes('@name')}},$currentNode->findnodes('@name'));
xmlStructure($currentNode,$href->{$currentNode->findnodes('@name')});

}
# elsif ($type&&$type ne 'List'&&!$currentNode->findnodes('@name')&&$parentType eq 'List'){
# print "4\n";
# push(@{$$href->{$currentNode->findnodes('@name')}},{});
##print Dumper %{$arrayNode};
# xmlStructure($currentNode,$href->{$currentNode->findnodes('@name')});
# }
else{
xmlStructure($currentNode,$href->{$currentNode->findnodes('@name')});
}
}

}
xmlStructure($mainParentNode,$mainHash);
}
my %href;
buildHash($xmldoc->findnodes('*'),\%href);
print "Printing the real HASH\n";
print Dumper %href;

但還有很長的路要走，因為：1。鑰匙和價值之間有一個寄生蟲，可能是未定義的元素。 2.我找不到在需要的地方將數據類型從哈希更改為子數組的方法。

這是輸出：

$VAR1 = 'FirstMainBranch';
$VAR2 = {
          '' => {
                  'aList' => {
                             '' => {
                                     'third' => {},
                                     'second' => {},
                                     'first' => {}
                                   }
                           },
                  'namedStruct' => {
                                   'thirdList' => {
                                                  '' => {
                                                          'first' => {}
                                                        }
                                                }
                                 },
                  'anotherStringValueUnderMainBranch' => {},
                  'secondList' => {
                                  '' => {
                                          'second' => {},
                                          'first' => {}
                                        }
                                },
                  'aStringValueUnderMainBranch' => {},
                  'anIntegerValueUnderMainBranch' => {}
                }
        };
$VAR3 = 'SecondMainBranch';
$VAR4 = {
          '' => {
                  'namedStructAgain' => {
                                        'First' => {},
                                        'Second' => {}
                                      }
                }
        };

任何幫助將不勝感激。 先感謝您。

編輯：關於Sobrique的評論 - XY問題：

這是我要解析的示例字符串：

(1,2,"N/A",-1,"foo","bar",NULL,3,2016-03-18 08:12:00.000,2016-03-18 08:12:00.559,2016-03-18 08:12:00.520,0,0,NULL,"foo","123456789",{NULL,NULL,NULL,NULL,NULL,NULL,2016-04-17 11:59:59.999,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,null,NULL,NULL,NULL,NULL,3,0,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,T,0,NULL,NULL,NULL,"9876543210",NULL,"foo","0","bar","foo","a1820000264d979c","0,0",NULL,"foo","192.168.1.82","SOAP",NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL},{INPUT="bar"},{aStringValueUnderMainBranch="ET", aList[{"first", "second", "third"}, {"first", "second", "third"}], secondList[{"first", "second"}, {"first", "second"}],namedStruct{thirdList[{first},{first}]}},{namedStructAgain{"first", "second"}},NULL,NULL,NULL,NULL,NULL)

不知何故，我應該將所有值分開，然后確定這一部分：

{aStringValueUnderMainBranch="ET", aList[{"first", "second", "third"}, {"first", "second", "third"}], secondList[{"first", "second"}, {"first", "second"}],namedStruct{thirdList[{first},{first}]}}

作為FirstMainBranch並解析XML中顯示的相應值。 在那之后，我應該確定：

{namedStructAgain{"first", "second"}}

作為SecondMainBranch並獲得各自的值。 主要數據分離還存在一個額外的問題，當它們在括號之間時，我不應該記住逗號。

Answer 1

我會用另一種方法。 我不是將XML轉換為哈希，而是使用XML :: Rabbit將其映射到對象。 我寫了一篇關於如何使用它的完整工作示例的小文章 。

XML :: Rabbit具有一系列優點：

• 使用簡單的Moose對象。
• 使用XPath以聲明方式定義要獲取的對象。
• 解析/定義你想要的東西。 無需從XML中獲取所有信息。

如果您的XML文件足夠小以便使用XPath和DOM，我發現這種方法非常簡潔且易於維護。