xslt - 如何在XSL中防止重复？

如何防止重复的条目进入列表，然后理想地对列表进行排序？我正在做的是，当某个级别的信息丢失时，将信息从其下一级获取，以建立上一级的缺失列表。当前，我有类似于以下内容的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>

但我得到的“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>

在此先感谢您的帮助!

最佳答案

尝试以下代码:

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

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

  <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">
            <xsl:variable name="foo" select="../c04/did/container[@type='Box']/text()"/>
            <!-- 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="distinct-values($foo)">
                <xsl:sort />
                <xsl:if test="position() &gt; 1">, </xsl:if>
                <xsl:value-of select="." />
              </xsl:for-each>
            </container>
            <xsl:apply-templates select="*" />
          </xsl:if>
        </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>

</xsl:stylesheet>

它看起来几乎与您想要的输出一样:

<?xml version="1.0" encoding="UTF-8"?>
<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>

诀窍是一起使用<xsl:sort>和distinct-values()。请参阅Michael Key撰写的(IMHO)很棒的书“XSLT 2.0和XPATH 2.0”

关于xslt - 如何在XSL中防止重复？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/2669813/