我怎么知道xslt中根节点和当前节点之间的特定节点总数? [英] how do i know total number of specific nodes between root node and current node in xslt?

问题描述

Hi All,

I want to count the total number of specific nodes between root node and current node. for Example:

<pre lang="text"><w:body xmlns:w="#unknown"> <!-- root Node -->
     <w:p />
             <w:rect>
             <w:p>some text1</w:p>  
            </w:rect>
     <w:p />
             <w:rect>
              <w:p>some text2</w:p>  
            </w:rect>

     <w:p />
     <w:p />
     <w:p />
            <w:rect>
             <w:p>some text5</w:p>   <!-- This is current node now -->
            </w:rect>
     <w:p />
     <w:p />
     <w:p />
     <w:p />
</w:body>

这是我的情况.在这里，假设一些文本是我的当前节点.所以，现在我想获取当前节点的parent :: w:p(假设此节点名称为"targetNode").之后，我想计算从根节点到targetNode的w:p级别相同的总数.

我想要的输出是:5

请引导我解决此问题...

This is my scenario.Here, Assume some text is my current node.So, Now i want to get the parent::w:p of the current node (Assume this node name as "targetNode"). After that i want to calculate total number of the same level of w:p from root node to targetNode.

My wanted output is :5

Please Guide me to get out of this issue...

推荐答案

count(祖先:: w:p [1]/preceding-sibling :: w:p + 1)(第一个祖先w:p本身).

count(ancestor::w:p[1]/preceding-sibling::w:p + 1)(the first ancestor w:p itself).

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