XPath 获取没有父节点的所有子节点(元素、注释和文本) [英] XPath to get all child nodes (elements, comments, and text) without parent

问题描述

我需要一个 XPath 来获取所有没有父元素的子节点(包括文本元素、注释元素和子元素).任何帮助

I need an XPath to fetch all ChildNodes ( including Text Element, Comment Element & Child Elements ) without Parent Element. Any help

示例:

<DOC>
<PRESENTEDIN>
    <X>
        First Text Node #1 
        <y> Y can Have Child Nodes # 
            <child> deep to it </child> 
         </y>
         Second Text Node #2 <z/> 
    </X>
    <EVTS>
        <evt/>
        <evt>
            <mtg_descr> SAE 2006 World Congress &amp; Exhibition </mtg_descr>
            <sess_descr> Advanced Hybrid Vehicle Powertrains (Part 1 of 5) </sess_descr>
            <loc> Detroit,MI,United States </loc>
            <sess_prod_grp_cd> TSESS </sess_prod_grp_cd>
            <sess_evt_name> P13 </sess_evt_name>
            <sess_gen_num> 138352 </sess_gen_num>
            <mtg_start_dt> 04/03/2006 </mtg_start_dt>
            <mtg_end_dt> 04/06/2006 </mtg_end_dt>
            <desig> CONGRESS-2006 </desig>
        </evt>
    </EVTS>
    <EVTTYPE>PAPER</EVTTYPE>
    <SUPERTECH>
        <![CDATA[C8585]]>
    </SUPERTECH>
</PRESENTEDIN>

XPATH 尝试

   1. $doc/PRESENTEDIN/X
   2. $doc/PRESENTEDIN/X/descendant::*
   2. $doc/PRESENTEDIN/X/self::*

预期输出

    First Text Node #1 
    <y> Y can Have Child Nodes # 
        <child> deep to it </child> 
     </y>
     Second Text Node #2 <z/> 

我不想

<X>
  First Text Node #1 
        <y> Y can Have Child Nodes # 
            <child> deep to it </child> 
         </y>
         Second Text Node #2 <z/> 
</X>

推荐答案

来自 XPath 的文档 ( http://www.w3.org/TR/xpath/#location-paths ):

From the documentation of XPath ( http://www.w3.org/TR/xpath/#location-paths ):

child::* 选择所有元素上下文节点的子节点

child::* selects all element children of the context node

child::text() 选择所有文本节点上下文节点的子节点

child::text() selects all text node children of the context node

child::node() 选择所有上下文节点的子节点，无论他们的节点类型

child::node() selects all the children of the context node, whatever their node type

所以我猜你的答案是:

$doc/PRESENTEDIN/X/child::node()

如果你想要一个所有嵌套节点的扁平数组:

And if you want a flatten array of all nested nodes:

$doc/PRESENTEDIN/X/descendant::node()

这篇关于XPath 获取没有父节点的所有子节点(元素、注释和文本)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！