使用 XSLT 1.0 对相似标签进行分组的 XSLT 转换 [英] XSLT transformation to group similar tags with XSLT 1.0

问题描述

我有这个带有 players 和 citizens 部分的 XML.每个部分都有多个 person 标签.

I have this XML with players and citizens sections. Each section has multiple person tags.

<?xml version="1.0" encoding="UTF-8"?>
<test>
   <players>
      <person>
         <name>joe</name>
         <age>20</age>
      </person>
      <person>
         <name>sam</name>
         <age>23</age>
      </person>
   </players>
   <citizens>
      <person>
         <name>joe</name>
         <city>ny</city>
      </person>
      <person>
         <name>sam</name>
         <city>london</city>
      </person>
   </citizens>
</test>

现在我想对它进行转换，以便根据 name<将 person 标签、players 和 citizens 部分合并在一起/code> 标签.

Now I want to transform this so that person tags of, players and citizens sections are merged together based on name tag.

这是我需要的输出.

<?xml version="1.0" encoding="UTF-8"?>
<test>
   <players>
      <person>
         <name>joe</name>
         <age>20</age>
         <city>ny</city>
      </person>
      <person>
         <name>sam</name>
         <age>23</age>
         <city>london</city>
      </person>
   </players>
</test> 

我想为此做一个 XSLT 1.0 转换.

I want to do an XSLT 1.0 transformation for this.

我试过了.

<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:variable name="citizens" select="/test/citizens"/>
    <xsl:template match="/test/players">
        <players>
            <xsl:apply-templates select="person"/>
        </players>
    </xsl:template>

    <xsl:template match="person">
        <xsl:variable name="data1" select="."/>
        <xsl:variable name="data2" select="/test/citizens/person[name=current()/name]/."/>
        <person>
            <xsl:copy-of select="$data1/*"/>
            <xsl:for-each select="$data2/*">
                <xsl:variable name="element2" select="name(.)"/>
                <xsl:if test="count($data1/*[name()=$element2])=0">
                    <xsl:copy-of select="."/>
                </xsl:if>
            </xsl:for-each>
        </person>
    </xsl:template>
</xsl:stylesheet>

这几乎是正确的.我只想去掉最后 2 个 person 标签.请指导我.

It's almost correct. I just want to get rid of last 2 person tags. Please guide me.

   <players>
      <person>
         <name>joe</name>
         <age>20</age>
         <city>ny</city>
      </person>
      <person>
         <name>sam</name>
         <age>23</age>
         <city>london</city>
      </person>
   </players>
   <person>
      <name>joe</name>
      <city>ny</city>
   </person>
   <person>
      <name>sam</name>
      <city>london</city>
   </person>

注意:我收到了一个关于 XSLT 2.0 的答案，但现在我正在寻找 XSLT 1.0 的解决方案.

Note: I received an answer for XSLT 2.0, But now I'm looking for a solution with XSLT 1.0.

推荐答案

在 XSLT 1.0 中执行此操作的一种方法是:

One way to do this in XSLT 1.0 would be:

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

<xsl:key name="citizen-by-name" match="citizens/person" use="name" />

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

<xsl:template match="/test">
    <xsl:copy>
        <xsl:apply-templates select="players"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="person">
    <xsl:copy>
        <xsl:apply-templates/>
        <xsl:copy-of select="key('citizen-by-name', name)/city "/>
    </xsl:copy>
</xsl:template>

</xsl:stylesheet>

<小时>

另一种选择是使用慕尼黑方法来分组 来自两个分支的 person 元素，按它们的 name.

---

Another option would be to use the Muenchian method to group the person elements from both branches by their name.

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