通过在XSLT中对节点值进行硬编码对节点进行分组 [英] Grouping Nodes by hardcoding node values in XSLT

问题描述

<root>
   <Entry>
      <ID>1</ID>
      <Details>
         <Code>A1</Code>
         <Value>1000</Value>
      </Details>
   </Entry>

   <Entry>
      <ID>2</ID>
      <Details>
         <Code>A2</Code>
         <Value>2000</Value>
      </Details>
   </Entry>

   <Entry>
      <ID>3</ID>
      <Details>
         <Code>B1</Code>
         <Value>3000</Value>
      </Details>
   </Entry>

   <Entry>
      <ID>4</ID>
      <Details>
         <Code>B2</Code>
         <Value>4000</Value>
      </Details>
   </Entry>
</root>

我有此输入XML，希望通过XSLT进行分组，其中分组是通过硬编码节点值进行的.让我详细解释一下:

I have this input XML which I am looking to group via XSLT wherein the grouping happens by hardcoding node values. Let me explain that in detail:

需要根据出现在节点<Code>中的Code参数进行分组，如下所示:

The grouping needs to happen based on the Code parameter appearing in the node <Code> as follows:

1. 代码"A1"和"A2"需要分组在一起
2. 代码"B1"和"B2"需要分组在一起

我最终将这些组中<Value>个节点的值相加.因此输出如下:

I am eventually summing the values coming out of <Value> nodes in these groups. So the Output would be as follows:

<Output>
   <Code-group> A </Code-group>
   <Sum> 3000 </Sum>

   <Code-group> B </Code-group>
   <Sum> 7000 </Sum>
</Output>

对于此要求，需要对分组值进行硬编码(将A1，A2分组为A，将B1，B2分组为B).我之所以使用硬编码"一词，是因为代码(A1，A2，B1，B2)可以以任何顺序排列，所以我想对值进行硬编码以寻找分组，而不是寻找节点索引.

For this requirement there needs to be hardcoding of the grouping values (to group A1,A2 as A and B1, B2 as B). I am using the word 'hardcoded' because the Codes (A1,A2,B1,B2) can come in any order so I want to rather hardcode the values to look for grouping than looking for Node indices.

我研究了for-each-group方法以及Muenchian分组方法，但是无法实现上述组映射.任何帮助表示赞赏！

I looked at for-each-group method as well as the Muenchian Grouping method but wasn't able to achieve the above group mapping. Any help appreciated!

预先感谢

映射A1，A2-> & B1，B2-> B是一个通用示例，实际节点值与此完全不同，因此子字符串解决方案将不起作用.这就是为什么我一直专注于硬编码以实现该映射的原因.

The mapping A1,A2 --> A & B1,B2 --> B is a generic example, the actual node values are entirely different than this, so a substring solution wouldn't work. That is why I was focusing on hardcoding to achieve that mapping.

推荐答案

硬编码要求很难理解.也许您想做类似的事情:

The hardcoding requirement is difficult to understand. Perhaps you want to do something like:

XSLT 2.0

<xsl:stylesheet version="2.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<xsl:key name="entry" match="Entry" use="Details/Code"/>

<xsl:template match="/root">
    <Output>
        <Code-group> A </Code-group>
        <Sum>
            <xsl:value-of select="sum(key('entry', ('A1', 'A2'))/Details/Value)" />
        </Sum>
        <Code-group> B </Code-group>
        <Sum>
            <xsl:value-of select="sum(key('entry', ('B1', 'B2'))/Details/Value)" />
        </Sum>
    </Output>
</xsl:template>

</xsl:stylesheet>

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