在查询所有子节点时，如何确定遍历的XPath？

2 回答

• 0

  The following XPath 2.0 expression ：

  //file/concat(string-join(ancestor::*[parent::*]
                                     /concat(name(.), '/'),
                              ''),
                   @name, '
'
                   )
  

  when evaluated against the provided XML document ：

  <xml>
      <home>
          <mysite>
              <www>
                  <images>
                      <file name="logo.gif"/>
                  </images>
                  <file name="index.html"/>
                  <file name="about_us.html"/>
              </www>
          </mysite>
      </home>
  </xml>
  

  produces the wanted, correct result ：

  home/mysite/www/images/logo.gif
   home/mysite/www/index.html
   home/mysite/www/about_us.html
  

  In case you cannot use XPath 2.0, it isn't possible to produce the wanted result only with an XPath 1.0 expression .

  然后必须使用托管XPath的编程语言（例如XSLT，C＃，php，...）来生成结果 .

  Here is an XSLT 1.0 solution ：

  <xsl:stylesheet version="1.0"
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
   <xsl:output method="text"/>
  
   <xsl:template match="file">
     <xsl:for-each select="ancestor::*[parent::*]">
       <xsl:value-of select="concat(name(),'/')"/>
     </xsl:for-each>
     <xsl:value-of select="concat(@name, '&#xA;')"/>
   </xsl:template>
  
   <xsl:template match="text()"/>
  </xsl:stylesheet>
  

  when this transformation is applied on the same XML document, the same correct result is produced ：

  home/mysite/www/images/logo.gif
  home/mysite/www/index.html
  home/mysite/www/about_us.html
  

  回复于 2024-04-29T13:43:59+08:00
• 2

  你也可以试试这个

  <?php 
  
  $dom = new DOMDocument();
  $dom->loadXML($xml);
  
  $xpath = new DOMXPath($dom);
  $arrNodes = $xpath->query('//file');
  foreach($arrNodes as $node) {
  
  $tmpNode = $node->parentNode;
  $arrPath = array();
  while ($tmpNode->parentNode) {
      $arrPath[] = $tmpNode->tagName;     
      $tmpNode = $tmpNode->parentNode;
  }
  unset($arrPath[count($arrPath)-1]); 
  printf('%s/%s<BR>',implode('/',array_reverse($arrPath)),$node->getAttribute('name'));
  
  }
  

  回复于 2024-04-29T13:43:59+08:00