Question

我正在尝试编写一个脚本，该脚本遍历三个现有XML文档并编译第四个XML文档，其中包含现有三个文档中的所有语素（说语言的部分单词）。我试图确保这个新的语素数据库不包含任何重复项，并且我很难使它不添加重复项。我将在下面立即发布相关代码段，并在底部发布整个相关代码块。

对重复项的检查如下：((string)$source == (string)$storySource)，其中$ source和$ storySource都是simpleXMLElement，如下所示：<m>text</m>。谁能告诉我我哪里出问题了？

最好，吉米

这是遍历其中一个XML文件的整个循环。

$storycorpus = new SimpleXMLElement($file,null,true);
$storyEntries = $storycorpus->xpath("//morpheme");
foreach($storyEntries as $entry){
    // check to see if in morpheme database. we will match the Pomo and the English, hence, if either is not a match,
    // we will add a new morpheme
    $storySource = $entry->m;
    $storyGloss = $entry->g;
    // set a variable equal to false
    $foundInDB = false; 

    //we will loop through the database looking for a match.    
    foreach($morphemeEntries as $existingMorpheme){
        $source = $existingMorpheme->source;
        $gloss = $existingMorpheme->gloss;

        // if we find a match, we will set our variable to be true and break out of the morpheme DB loop
        if(((string)$source == (string)$storySource) && ((string)$gloss == (string)$storyGloss)){
            $foundInDB = true; // problem: this line isn't firing
            break;
        }
    }
    // after the morphemeDB loop, we will check to see if the var is true. 
    if($foundInDB == true){
        // if it is true, we don't need to enter anything and can 
        // go to the next entry
        continue;
    } else{
        // if we didn't find a match, create a new morpheme
        $newMorphemeEntry = $morphemeDB->addChild("morpheme");
        $newMorphemeEntry->addChild("source", $storySource);
        $newMorphemeEntry->addChild("gloss", $storyGloss);
        $newMorphemeEntry->addChild("root", $storySource);
        $newMorphemeEntry->addChild("hypernym", $storySource);
        $newMorphemeEntry->addChild("link", "S");
        if(substr($storySource, 0, 1) == "-"){
            $newMorphemeEntry->addChild("affix", "suffix");
        } elseif(substr($storySource, -1, 1) == "-"){
            $newMorphemeEntry->addChild("affix", "prefix");
        } else{
            $newMorphemeEntry->addChild("affix", "root");
        }
    }
}

好的，所以我重写了代码块并使用了DOMDocument而不是SimpleXML，而且在防止重复方面我还是没有运气。这是新代码

    // check to see if in morpheme database. we will match the Pomo and the English, hence, if either is not a match,
    // we will add a new morpheme
    $phraseSource = $entry->nodeValue;
    $phraseGlossId = $entry->getAttribute("id");
    $phraseGloss = $xpath2->query("//g[@id =\"$phraseGlossId\"]")->item(0)->nodeValue;
    // set a variable equal to false
    $foundInDB = false; 

    //we will loop through the database looking for a match.    
    foreach($morphemeEntries as $existingMorpheme){
        $source = $existingMorpheme->getElementsByTagName("source")->item(0)->nodeValue;
        $gloss = $existingMorpheme->getElementsByTagName("gloss")->item(0)->nodeValue;
        // if we find a match, we will set our variable to be true and break out of the morpheme DB loop
        if(($source == $phraseSource) && ($gloss == $phraseGloss)){
            $foundInDB = true; // problem: this line isn't firing
            break;
        }
    }
    // after the morphemeDB loop, we will check to see if the var is true. 
    if($foundInDB == true){
        // if it is true, we don't need to enter anything and can 
        // go to the next entry
        continue;
    } else{
        // if we didn't find a match, create a new morpheme
        $newMorphemeEntry = $morphemeXmlDoc->createElement("morpheme");

        $newMorphemeSource = $morphemeXmlDoc->createElement("source");
        $newMorphemeSource->nodeValue = $phraseSource;
        $newMorphemeEntry->appendChild($newMorphemeSource);

        $newMorphemeGloss = $morphemeXmlDoc->createElement("gloss");
        $newMorphemeGloss->nodeValue = $phraseGloss;
        $newMorphemeEntry->appendChild($newMorphemeGloss);

        $newMorphemeRoot = $morphemeXmlDoc->createElement("root");
        $newMorphemeRoot->nodeValue = $phraseSource;
        $newMorphemeEntry->appendChild($newMorphemeRoot);

        $newMorphemeHypernym = $morphemeXmlDoc->createElement("hypernym");
        $newMorphemeHypernym->nodeValue = $phraseSource;
        $newMorphemeEntry->appendChild($newMorphemeHypernym);

        $newMorphemeLink = $morphemeXmlDoc->createElement("link");
        $newMorphemeLink->nodeValue = "P";
        $newMorphemeEntry->appendChild($newMorphemeLink);

        $newMorphemeAffix = $morphemeXmlDoc->createElement("affix");
        $newMorphemeAffix->nodeValue = $phraseGloss;

        if(substr($phraseSource, 0, 1) == "-"){
            $newMorphemeAffix->nodeValue = "suffix";
        } elseif(substr($phraseSource, -1, 1) == "-"){
            $newMorphemeAffix->nodeValue = "prefix";
        } else{
            $newMorphemeAffix->nodeValue = "root";
        }
        $newMorphemeEntry->appendChild($newMorphemeAffix);

        $morphemeRootNode->appendChild($newMorphemeEntry);
    }
}

以下是脚本正在搜索的内容，以创建新的XML表：

<phrasicon>
<phrase id="4">
    <ref1>ES</ref1>
    <source>t̪o: xa jo: k'ala:</source>
    <morpheme>
      <m id="4.1">t̪o:</m>
      <m id="4.2">xa</m>
      <m id="4.3">jo:</m>
      <m id="4.4">k'ala:</m>
    </morpheme>
    <gloss lang="en">
      <g id="4.1">me</g>
      <g id="4.2">water</g>
      <g id="4.3">for</g>
      <g id="4.4">die</g>
    </gloss>
    <translation lang="en">I'm dying for water.</translation>
    <media1 mimeType="audio/wav" url="im_dying_for_water.wav"/>
    <ref2/>
    <media2 mimeType="" url=""/>
    <ref3/>
    <media3 mimeType="" url=""/>
  </phrase>
</phrasicon>

这是新的词素XML工作表应该是什么样子

<?xml version="1.0" encoding="UTF-8"?>
<morphemedatabase>
<morpheme>
  <source>t̪o:</source>
  <gloss>me</gloss>
  <root>t̪o:</root>
  <hypernym>t̪o:</hypernym>
  <link>P</link>
  <affix>root</affix>
</morpheme>
</morphemedatabase>

Answer 1

我想象$morphemeEntries是SimpleXMLElement对象的固定列表，并且不会使用添加的节点进行更新。我建议使用$morphemeDB对象进行检查。另外，您可以用Xpath表达式替换循环。

$storySource = $entry->m;
$storyGloss = $entry->g;

$foundInDB = count(
  $morphemeDB->xpath(
    sprintf('.//morpheme[source="%s" and gloss="%s"]', $storySource, $storyGloss)
  )
) > 0;

在DOM中，DOMXpath::evaluate()也是可能的：

$phraseSource = $xpathSource->evaluate('string(m)', $entry);
$phraseGloss = $xpathSource->evaluate('string(g)', $entry);

$foundInDB = $xpathTarget->evaluate(
  sprintf(
    'count(//morpheme[source="%s" and gloss="%s"]) > 0', 
    $storySource, 
    $storyGloss
  )
);

在DOM实现中，您可以将createElement()嵌套到appendChild()中，但是您应该将内容添加为文本节点（以进行适当的转义）：

$newMorphemeEntry = $morphemeRootNode->appendChild(
  $morphemeXmlDoc->createElement("morpheme")
);
$newMorphemeEntry
  ->appendChild($morphemeXmlDoc->createElement("source"))
  ->appendChild($morphemeXmlDoc->createTextNode($phraseSource));
$newMorphemeEntry
  ->appendChild($morphemeXmlDoc->createElement("gloss"))
  ->appendChild($morphemeXmlDoc->createTextNode($phraseGloss));

Answer 2

在比较之前不要尝试强制转换为（字符串）。而是在每个元素上调用->asXML()方法。替换为：

if(((string)$source == (string)$storySource) && ((string)$gloss == (string)$storyGloss))

与此：

if(($source->asXML() == $storySource->asXML()) && ($gloss->asXML() == $storyGloss->asXML()))

或比较包含的字符串（不包括标签）

if(($source->__toString() == $storySource->__toString()) && ($gloss->__toString() == $storyGloss->__toString()))

问题是SimpleXMLElement不是“经典” PHP对象。 SimpleXML是通过“实时” API构建的，该API链接到XML文档的内部表示。

Comparing Objects上的手册页指出：“两个对象实例具有相同的属性和值，并且属于同一类，则它们是相等的。”

在SimpleXMLElement的print_r（）或var_dump（）中，

显示为代表子节点和属性的属性。但是，实际的实现只包含一个指向在解析XML时创建的内存结构的指针，即使您两次解析相同的字符串也将有所不同。因此，仅将两个SimpleXMLElement对象与==进行比较就永远不会返回true。

PHP和XML：如何比较两个XML元素的文本内容？

2 个答案: