如何使用XPath查找所有紧邻的兄弟姐妹(How to find all immediately adjacent siblings with XPath)

我希望使用单个XPath表达式找到节点的所有紧邻的兄弟节点，如果可能的话。 鉴于输入

<a id="1"/>
<start/>
<a id="2"/>
<a id="3"/>
<b/>
<a id="4"/>
 
 和一个类似于//start/following-sibling::a的XPath表达式，我想选择[2]和[3] ，但不是[4] 。 此外，如果start和a [2]之间有任何介入元素，则不应选择任何内容。 
I want to find all the immediately adjacent siblings of a node using a single XPath expression, if at all possible. Given the input
 
<a id="1"/>
<start/>
<a id="2"/>
<a id="3"/>
<b/>
<a id="4"/>
 
and an XPath expression similar to //start/following-sibling::a, I want to select a[2], and a[3], but not a[4]. Also, if there are any intervening elements between start and a[2], nothing should be selected.
                
最满意答案
                
                    
                        
 我能找到的最简单的是： 
 
//start/following-sibling::a intersect //start/following-sibling::*[name()!='a'][1]/preceding-sibling::a
 
 这样做是： 
 
 
  在start ： //start/following-sibling::a 。 （结果：a2，a3，a4。）现在将其设置为一侧。  
  然后在start ： //start/following-sibling::*[name()!='a'][1] （结果：b。）之后取第一个非兄弟姐妹。  
  并找到它之前的所有节点： /preceding-sibling::a 。 （结果：a1，a2，a3）  
  取1和3的交点。（结果：a2，a3）  
 
 更新：另一种表达它的方法是//start/following-sibling::*[name()!='a'][1]/preceding-sibling::a[preceding-sibling::start] ，这大致翻译to：在start取第一个非兄弟姐妹，向后计数，但只选择仍然以start为先的元素。 
 
 更新2：如果你知道b将永远被称为b ，你当然可以用following-sibling::*[name()!='a'][1]代替替换难以阅读的following-sibling::*[name()!='a'][1]部分following-sibling::b[1] 。 
The simplest one I can find was this:
 
//start/following-sibling::a intersect //start/following-sibling::*[name()!='a'][1]/preceding-sibling::a
 
What this does is:
 
 
 Take all the a siblings following start: //start/following-sibling::a. (Result: a2, a3, a4.) Set this to one side for now. 
 Then take the first non-a sibling following start: //start/following-sibling::*[name()!='a'][1] (Result: b.) 
 And find all the a nodes that precede it: /preceding-sibling::a. (Result: a1, a2, a3) 
 Take the intersection of 1 and 3. (Result: a2, a3) 
 
Update: Another way to phrase it is //start/following-sibling::*[name()!='a'][1]/preceding-sibling::a[preceding-sibling::start], this roughly translates to: take the first non-a sibling following start, count backwards but only choose elements that are still preceded by start.
 
Update 2: If you know that b will always be called b, you can of course replace the rather hard to read following-sibling::*[name()!='a'][1] part with following-sibling::b[1].