如何计算xsl中相同的紧邻兄弟的数量 [英] How to count the number of same immediate preceding siblings in xsl

问题描述

我使用的是 xslt 版本 2，我正在尝试将 xml 转换为 fo 输出，但我遇到了一个特定的问题.

I'm using xslt version 2, I'm trying to transform an xml to a fo output, and I'm stuck on a specific problematic.

这是我的输入:

    <a1/>
    <a1/>
    <b/>
    <c/>
    <d/>
    <a2/>
    <b/>
    <c/>
    <a1/>
    <a1/>
    <a1/>
    <a1/>
    <b/>
    <c/>
    <d/>

从功能上讲，此数据包含由 a1|a2,b?,c?,d? 定义的集合"列表.

This data, functionally speaking, contains a list of 'sets' defined by a1|a2,b?,c?,d?.

我的问题是我不知道如何计算特定集合"的 a1 标签数量.

My problem is that I don't see how I can count the number of a1 tags for a specific 'set'.

确实，我已经编写了我的 xsl 并且得到了这样的输出:

Indeed, I have written my xsl and I get an output like that:

<fo:table>
    <fo:row>
        <fo:cell>b: </fo:cell>
        <fo:cell>b value</fo:cell>
    </fo:row>
    <fo:row>
        <fo:cell>a1: </fo:cell>
        <fo:cell>number of a1 ???</fo:cell> <-- what I am trying to retrieve
    </fo:row>
    <fo:row>
        ...
    </fo:row>
    ...
</fo:table>

我在 a1+|a2 标签上做了一个应用模板，如果 a1 标签有一个等于 a1 的后续兄弟，我什么都不做.我认为必须有一种方法来计算具有前面兄弟的标签(但是如何确保只计算相应的标签?)

I have done an apply-template on a1+|a2 tags, and I do nothing if a1 tag has a following sibling that equals to a1. I think there must be a way to count the tags with preceding sibling (but then how to insure to count only the corresponding one?)

任何提示将不胜感激！

在上面的输入示例中，第一个计数应该是 2:

On the above example of input, the first count should be 2:

    <a1/>
    <a1/>
    <b/>
    <c/>
    <d/>

那么它应该是 4，而不是 6:

then it should be 4, and not 6:

    <a1/>
    <a1/>
    <a1/>
    <a1/>
    <b/>
    <c/>
    <d/>

推荐答案

你的问题不是很清楚.
相应的"应该是什么?在当前之前计算所有 a1 将是:

Your question is not really clear.
What should "the corresponding one" be? Counting all a1 before the current one would be:

 count(preceding-sibling::a1) 

如果需要，您可以添加以下谓词:

if needed you may add predicates like:

 count(preceding-sibling::a1[/corresponding one/]) 

要仅计算 a1 节点序列中的主控兄弟 a1，请尝试以下操作:找到第一个不是 a1 的节点.

To count only the presiding sibling a1 which are in a sequence of a1 nodes try this: Find the first node which is not an a1.

<xsl:variable name="firstnota1" select="preceding-sibling::*[not (self::a1)][1]" />

得出的结果是，计算当前 a1 之前的所有节点数减去第一个之前的节点数，而不是 a1 + 这个节点本身.

The wonted result than, is count all nodes before the current a1 minus the count of nodes before the first not a1 + this node it self.

<xsl:value-of select="count(preceding-sibling::*) 
       -  count($firstnota1/preceding-sibling::* | $firstnota1)"/>

或不带变量:

<xsl:value-of 
      select="count(preceding-sibling::*)
             -  count( preceding-sibling::*[not (self::a1)][1]
                      /preceding-sibling::*
                      | preceding-sibling::*[not (self::a1)][1] )"/>

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