排除所选元素的某些部分的 XPath? [英] XPath that excludes some part of selected element?

问题描述

XML:

    <root>
      <rows>
        <row hash="156458">
          <column name="Id">1</column>
          <column name="Nome">Evandro</column>
          <column name="CPF">98765432100</column>
        </row>
        <row hash="52458">
          <column name="Id">2</column>
          <column name="Nome">Everton</column>
          <column name="CPF">12345678900</column>
        </row>
      </rows>
    </root>

XPath 查询:

<代码>./root/rows/row/column[@name='Nome'] |./root/rows/row/column[@name='CPF']

XPath 返回:

    <root>
      <column name="Nome">Evandro</column>
      <column name="CPF">98765432100</column>
      <column name="Nome">Everton</column>
      <column name="CPF">12345678900</column>
    </root>

我希望 XPath 返回什么:

What I wish the XPath return:

    <root>
      <rows>
        <row hash="156458">
          <column name="Nome">Evandro</column>
          <column name="CPF">98765432100</column>
        </row>
        <row hash="52458">
          <column name="Nome">Everton</column>
          <column name="CPF">12345678900</column>
        </row>
      </rows>
    </root>

我想让 xpath 在我进行查询时保留文档结构.

I want to make the xpath preserve the document structure when I make the query.

推荐答案

XPath 非常适合选择，但不适用于结构化.加强对两者的完整 XSLT.您只需要一个简单的基于身份的转换...

XPath's great for selecting but not for structuring. Step up to full XSLT for both. A simple identity-based transformation is all you need...

鉴于此 XML 输入:

<?xml version="1.0" encoding="utf-8" ?>
<root>
  <rows>
    <row hash="156458">
      <column name="Id">1</column>
      <column name="Nome">Evandro</column>
      <column name="CPF">98765432100</column>
    </row>
    <row hash="52458">
      <column name="Id">2</column>
      <column name="Nome">Everton</column>
      <column name="CPF">12345678900</column>
    </row>
  </rows>
</root>

这个 XSLT 转换:

<xsl:stylesheet version="1.0"
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="@*|node()">
    <xsl:copy>
      <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="column[@name='Id']"/>

</xsl:stylesheet>

将产生所需的 XML 输出:

<?xml version="1.0" encoding="UTF-8"?>
<root>
   <rows>
      <row hash="156458">
         <column name="Nome">Evandro</column>
         <column name="CPF">98765432100</column>
      </row>
      <row hash="52458">
         <column name="Nome">Everton</column>
         <column name="CPF">12345678900</column>
      </row>
   </rows>
</root>

注意事项:

• 第一个模板是标识模板；它将从中复制节点输入到输出，除非有更具体的模板覆盖它.
• 第二个模板是省略Id columns 的覆盖.

• The first template is the identify template; it will copy nodes from input to output unless a more specific template overrides it.
• The second template is an override to omit Id columns.

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