xslt-1.0 - 如何在xslt中选择特定的节点集

Question

我目前使用的 xslt 在根目录上生成所有标签。
我需要获取 <row>套和<config>放。

源 XML:

<root>
    <postdate>2011-03-30</postdate>
    <location>84</location>
    <meal>07:36</meal>
    <config>
        <postdate>2011-03-30</postdate>
        <location>84</location>
        <meal>07:36</meal>
        <checknumber>91339082011-03-30T07:36:12</checknumber>
    </config>
    <items>
        <row>
            <descriptor>7297364</descriptor>
            <qty>1</qty>
            <price>33</price>
            <value>33</value>
            <recordtype>1</recordtype>
            <postdate>2011-03-30</postdate>
            <location>84</location>
        </row>
        <row>
            <descriptor>7794473</descriptor>
            <qty>1</qty>
            <price>60</price>
            <value>60</value>
            <recordtype>1</recordtype>
            <postdate>2011-03-30</postdate>
            <location>84</location>
        </row>
    </items>
    <tenders>
        <row>
            <id>13</id>
            <value>117.99</value>
            <recordtype>2</recordtype>
            <postdate>2011-03-30</postdate>
            <location>84</location>
        </row>
    </tenders>
    <taxes>
        <row>
            <id>2</id>
            <value>8.25</value>
            <recordtype>3</recordtype>
            <postdate>2011-03-30</postdate>
            <location>84</location>
        </row>
    </taxes>
</root>

尝试 Xslt:

<xsl:stylesheet version="1.0" exclude-result-prefixes="msxsl" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt">
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>
    <xsl:template match="row/*">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

期望输出:

<root>
    <config>
        <postdate>2011-03-30</postdate>
        <location>84</location>
        <meal>07:36</meal>
        <checknumber>91339082011-03-30T07:36:12</checknumber>
    </config>
    <row>
        <descriptor>7297364</descriptor>
        <qty>1</qty>
        <price>33</price>
        <value>33</value>
        <recordtype>1</recordtype>
        <postdate>2011-03-30</postdate>
        <location>84</location>
    </row>
    <row>
        <descriptor>7794473</descriptor>
        <qty>1</qty>
        <price>60</price>
        <value>60</value>
        <recordtype>1</recordtype>
        <postdate>2011-03-30</postdate>
        <location>84</location>
    </row>
    <row>
        <id>13</id>
        <value>117.99</value>
        <recordtype>2</recordtype>
        <postdate>2011-03-30</postdate>
        <location>84</location>
    </row>
    <row>
        <id>2</id>
        <value>8.25</value>
        <recordtype>3</recordtype>
        <postdate>2011-03-30</postdate>
        <location>84</location>
    </row>
</root>

Answer 1

这个简短而简单的转换 :

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

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

 <xsl:template match=
  "node()[not(self::root or ancestor-or-self::config or ancestor-or-self::row)]">
  <xsl:apply-templates/>
 </xsl:template>
</xsl:stylesheet>

当应用于提供的 XML 文档时:

<root>
    <postdate>2011-03-30</postdate>
    <location>84</location>
    <meal>07:36</meal>
    <config>
        <postdate>2011-03-30</postdate>
        <location>84</location>
        <meal>07:36</meal>
        <checknumber>91339082011-03-30T07:36:12</checknumber>
    </config>
    <items>
        <row>
            <descriptor>7297364</descriptor>
            <qty>1</qty>
            <price>33</price>
            <value>33</value>
            <recordtype>1</recordtype>
            <postdate>2011-03-30</postdate>
            <location>84</location>
        </row>
        <row>
            <descriptor>7794473</descriptor>
            <qty>1</qty>
            <price>60</price>
            <value>60</value>
            <recordtype>1</recordtype>
            <postdate>2011-03-30</postdate>
            <location>84</location>
        </row>
    </items>
    <tenders>
        <row>
            <id>13</id>
            <value>117.99</value>
            <recordtype>2</recordtype>
            <postdate>2011-03-30</postdate>
            <location>84</location>
        </row>
    </tenders>
    <taxes>
        <row>
            <id>2</id>
            <value>8.25</value>
            <recordtype>3</recordtype>
            <postdate>2011-03-30</postdate>
            <location>84</location>
        </row>
    </taxes>
</root>

产生想要的、正确的结果:

<root>
   <config>
      <postdate>2011-03-30</postdate>
      <location>84</location>
      <meal>07:36</meal>
      <checknumber>91339082011-03-30T07:36:12</checknumber>
   </config>
   <row>
      <descriptor>7297364</descriptor>
      <qty>1</qty>
      <price>33</price>
      <value>33</value>
      <recordtype>1</recordtype>
      <postdate>2011-03-30</postdate>
      <location>84</location>
   </row>
   <row>
      <descriptor>7794473</descriptor>
      <qty>1</qty>
      <price>60</price>
      <value>60</value>
      <recordtype>1</recordtype>
      <postdate>2011-03-30</postdate>
      <location>84</location>
   </row>
   <row>
      <id>13</id>
      <value>117.99</value>
      <recordtype>2</recordtype>
      <postdate>2011-03-30</postdate>
      <location>84</location>
   </row>
   <row>
      <id>2</id>
      <value>8.25</value>
      <recordtype>3</recordtype>
      <postdate>2011-03-30</postdate>
      <location>84</location>
   </row>
</root>

说明 :

使用和覆盖 identity rule .

正确使用ancestor-or-self::轴。