xslt - 在 XSL 中如何防止重复？

Question

如何防止重复条目进入列表，然后理想情况下对该列表进行排序？我正在做的是，当一个级别的信息丢失时，从低于它的级别获取信息，在上面的级别中构建丢失的列表。目前，我有与此类似的 XML:

<c03 id="ref6488" level="file">
    <did>
        <unittitle>Clinic Building</unittitle>
        <unitdate era="ce" calendar="gregorian">1947</unitdate>
    </did>
    <c04 id="ref34582" level="file">
        <did>
            <container label="Box" type="Box">156</container>
            <container label="Folder" type="Folder">3</container>
        </did>
    </c04>
    <c04 id="ref6540" level="file">
        <did>
            <container label="Box" type="Box">156</container>
            <unittitle>Contact prints</unittitle>
        </did>
    </c04>
    <c04 id="ref6606" level="file">
        <did>
            <container label="Box" type="Box">154</container>
            <unittitle>Negatives</unittitle>
        </did>
    </c04>
</c03>

然后我应用以下 XSL:

<xsl:template match="c03/did">
    <xsl:choose>
        <xsl:when test="not(container)">
            <did>
                <!-- If no c03 container item is found, look in the c04 level for one -->
                <xsl:if test="../c04/did/container">

                    <!-- If a c04 container item is found, use the info to build a c03 version -->
                    <!-- Skip c03 container item, if still no c04 items found -->
                    <container label="Box" type="Box">

                        <!-- Build container list -->
                        <!-- Test for more than one item, and if so, list them, -->
                        <!-- separated by commas and a space -->
                        <xsl:for-each select="../c04/did">
                            <xsl:if test="position() &gt; 1">, </xsl:if>
                            <xsl:value-of select="container"/>
                        </xsl:for-each>
                    </container>                    
            </did>
        </xsl:when>

        <!-- If there is a c03 container item(s), list it normally -->
        <xsl:otherwise>
            <xsl:copy-of select="."/>
        </xsl:otherwise>
    </xsl:choose>
</xsl:template>

但是我得到了“容器”结果

<container label="Box" type="Box">156, 156, 154</container>

当我想要的是

<container label="Box" type="Box">154, 156</container>

以下是我想要获得的完整结果:

<c03 id="ref6488" level="file">
    <did>
        <container label="Box" type="Box">154, 156</container>
        <unittitle>Clinic Building</unittitle>
        <unitdate era="ce" calendar="gregorian">1947</unitdate>
    </did>
    <c04 id="ref34582" level="file">
        <did>
            <container label="Box" type="Box">156</container>
            <container label="Folder" type="Folder">3</container>
        </did>
    </c04>
    <c04 id="ref6540" level="file">
        <did>
            <container label="Box" type="Box">156</container>
            <unittitle>Contact prints</unittitle>
        </did>
    </c04>
    <c04 id="ref6606" level="file">
        <did>
            <container label="Box" type="Box">154</container>
            <unittitle>Negatives</unittitle>
        </did>
    </c04>
</c03>

预先感谢您的帮助!

Answer 1

这个问题不需要 XSLT 2.0 解决方案。

这是一个 XSLT 1.0 解决方案，它比当前选择的 XSLT 2.0 解决方案更紧凑(35 行与 43 行):

<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:key name="kBoxContainerByVal"
     match="container[@type='Box']" use="."/>

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

 <xsl:template match="c03/did[not(container)]">
   <xsl:copy>

   <xsl:variable name="vContDistinctValues" select=
    "/*/*/*/container[@type='Box']
            [generate-id()
            =
             generate-id(key('kBoxContainerByVal', .)[1])
            ]
            "/>

    <container label="Box" type="Box">
      <xsl:for-each select="$vContDistinctValues">
        <xsl:sort data-type="number"/>

        <xsl:value-of select=
        "concat(., substring(', ', 1 + 2*(position() = last())))"/>
      </xsl:for-each>
    </container>
    <xsl:apply-templates/>
   </xsl:copy>
 </xsl:template>
</xsl:stylesheet>

当此转换应用于最初提供的 XML 文档时，将生成正确的所需结果:

<c03 id="ref6488" level="file">
   <did>
      <container label="Box" type="Box">156, 154</container>
      <unittitle>Clinic Building</unittitle>
      <unitdate era="ce" calendar="gregorian">1947</unitdate>
   </did>
   <c04 id="ref34582" level="file">
      <did>
         <container label="Box" type="Box">156</container>
         <container label="Folder" type="Folder">3</container>
      </did>
   </c04>
   <c04 id="ref6540" level="file">
      <did>
         <container label="Box" type="Box">156</container>
         <unittitle>Contact prints</unittitle>
      </did>
   </c04>
   <c04 id="ref6606" level="file">
      <did>
         <container label="Box" type="Box">154</container>
         <unittitle>Negatives</unittitle>
      </did>
   </c04>
</c03>

更新:

我没有注意到容器编号必须按顺序排列的要求。现在的解决方案反射(reflect)了这一点。