2017-03-15 35 views
1

在下面的示例中,我想确定具有XPath(2.0)表达式的节点的“嵌套级别”。这个“嵌套级别”将是连续后代的数量,例如,如果“跨度/ SPAN /跨度”的存在,这将是3,预计嵌套水平给出意见:如何确定XPath中的嵌套级别?

<?xml version="1.0" encoding="UTF-8"?> 
<text> 
    <div>Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aenean commodo ligula eget 
     dolor. Aenean massa. 
     <span><!--nesting level:2-->Cum sociis natoque penatibus et magnis dis parturient montes, 
      nascetur ridiculus mus. 
      <span><!--nesting levels:1-->Donec quam felis, ultricies nec, pellentesque eu, pretium quis, sem. 
       <span><!--nesting levels:0-->Nulla consequat massa quis enim.</span> 
      </span> 
      <span><!--nesting levels:0-->Donec pede justo, fringilla vel, aliquet nec, vulputate eget, arcu.</span> 
      In enim justo, rhoncus ut, imperdiet a, venenatis vitae, justo. 
     </span> 
     <span><!--nesting levels:0-->Nullam dictum felis eu pede mollis pretium. Integer tincidunt. Cras dapibus. Vivamus 
      elementum semper nisi. 
     </span> 
     <span><!--nesting levels:0-->Aenean vulputate eleifend tellus. Aenean leo ligula, 
      porttitor eu, consequat vitae, eleifend ac, enim. Aliquam lorem ante, dapibus in, viverra 
      quis, feugiat a, tellus. 
     </span> 
    </div> 
    <div>Phasellus viverra nulla ut metus varius laoreet. 
     <span><!--nesting levels:0-->Quisque rutrum. Aenean imperdiet. Etiam ultricies nisi vel augue. 
     </span> 
     <span><!--nesting levels:2-->Curabitur ullamcorper ultricies nisi. 
      <span><!--nesting levels:0-->Nam eget dui.</span> 
      Etiam rhoncus. 
      <span><!--nesting levels:1-->Maecenas tempus, tellus eget condimentum rhoncus, sem quam semper libero, sit amet 
       adipiscing sem neque sed ipsum. 
       <span><!--nesting levels:0-->Nam quam nunc, blandit vel, luctus pulvinar, hendrerit id, lorem.</span> 
       <span><!--nesting levels:0-->Maecenas nec odio et ante tincidunt tempus.</span> 
       Donec vitae sapien ut libero venenatis faucibus. 
       <span><!--nesting levels:0-->Nullam quis ante.</span> 
      </span> 
      Etiam sit amet orci eget eros faucibus tincidunt. Duis leo. Sed fringilla mauris sit amet 
      nibh. 
     </span> 
     Donec sodales sagittis magna. 
    </div> 
</text> 

现在,我试过count(descendant::span)),但很明显,这也将包括任何兄弟姐妹,并产生一个错误导致很多情况。我也尝试count(descendant::span[1]))count(descendant::span[position() = 1])),这也给了erraneous结果。我还无法弄清楚如何从总数中排除兄弟姐妹的数量。任何暗示是赞赏。

回答

2

在XSLT我得到正确的价值观与表达

  max(
       for $leaf in descendant-or-self::span[not(span)] 
       return count($leaf/ancestor-or-self::span except ancestor-or-self::span) 
      ) 

例如用样式表

<?xml version="1.0" encoding="UTF-8"?> 
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
    xmlns:xs="http://www.w3.org/2001/XMLSchema" 
    exclude-result-prefixes="xs" 
    version="2.0"> 

    <xsl:template match="@* | node()"> 
     <xsl:copy> 
      <xsl:apply-templates select="@* | node()"/> 
     </xsl:copy> 
    </xsl:template> 

    <xsl:template match="span"> 
     <xsl:copy> 
      <xsl:attribute name="nesting-level" 
       select=" 
       max(
        for $leaf in descendant-or-self::span[not(span)] 
        return count($leaf/ancestor-or-self::span except ancestor-or-self::span) 
       )"/> 
      <xsl:apply-templates select="@* | node()"/> 
     </xsl:copy> 
    </xsl:template> 

</xsl:stylesheet> 

我得到的输出

<?xml version="1.0" encoding="UTF-8"?><text> 
    <div>Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aenean commodo ligula eget 
     dolor. Aenean massa. 
     <span nesting-level="2"><!--nesting level:2-->Cum sociis natoque penatibus et magnis dis parturient montes, 
      nascetur ridiculus mus. 
      <span nesting-level="1"><!--nesting levels:1-->Donec quam felis, ultricies nec, pellentesque eu, pretium quis, sem. 
       <span nesting-level="0"><!--nesting levels:0-->Nulla consequat massa quis enim.</span> 
      </span> 
      <span nesting-level="0"><!--nesting levels:0-->Donec pede justo, fringilla vel, aliquet nec, vulputate eget, arcu.</span> 
      In enim justo, rhoncus ut, imperdiet a, venenatis vitae, justo. 
     </span> 
     <span nesting-level="0"><!--nesting levels:0-->Nullam dictum felis eu pede mollis pretium. Integer tincidunt. Cras dapibus. Vivamus 
      elementum semper nisi. 
     </span> 
     <span nesting-level="0"><!--nesting levels:0-->Aenean vulputate eleifend tellus. Aenean leo ligula, 
      porttitor eu, consequat vitae, eleifend ac, enim. Aliquam lorem ante, dapibus in, viverra 
      quis, feugiat a, tellus. 
     </span> 
    </div> 
    <div>Phasellus viverra nulla ut metus varius laoreet. 
     <span nesting-level="0"><!--nesting levels:0-->Quisque rutrum. Aenean imperdiet. Etiam ultricies nisi vel augue. 
     </span> 
     <span nesting-level="2"><!--nesting levels:2-->Curabitur ullamcorper ultricies nisi. 
      <span nesting-level="0"><!--nesting levels:0-->Nam eget dui.</span> 
      Etiam rhoncus. 
      <span nesting-level="1"><!--nesting levels:1-->Maecenas tempus, tellus eget condimentum rhoncus, sem quam semper libero, sit amet 
       adipiscing sem neque sed ipsum. 
       <span nesting-level="0"><!--nesting levels:0-->Nam quam nunc, blandit vel, luctus pulvinar, hendrerit id, lorem.</span> 
       <span nesting-level="0"><!--nesting levels:0-->Maecenas nec odio et ante tincidunt tempus.</span> 
       Donec vitae sapien ut libero venenatis faucibus. 
       <span nesting-level="0"><!--nesting levels:0-->Nullam quis ante.</span> 
      </span> 
      Etiam sit amet orci eget eros faucibus tincidunt. Duis leo. Sed fringilla mauris sit amet 
      nibh. 
     </span> 
     Donec sodales sagittis magna. 
    </div> 
</text> 
+0

非常感谢!这令人印象深刻。我没有想到解决方案会变得如此复杂。我将不得不在稍后检查 - 对于延迟抱歉! –

+2

这可能给出给出的例子的正确答案,但它并不试图满足路径中的span元素必须连续的要求:它会报告“span/div/div/span/span'' –

+2

为了解决这个问题,我认为你必须用'[祖先:: *中的每个$ a除了$ origin/ancestor :: *满足$ a/self :: span]的谓词来限定'descendant :: span'。 '$原点被绑定到最外层的上下文节点。 –