使用XSLT

Question

我有以下XML结构：

 <?xml version="1.0" encoding="UTF-8"?>
    <globalRequesteeDepartment>
        <listItems>
            <description type="varchar">HstnOperations</description>
            <indexValue type="varchar">JOB COST COORDNTR</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnOperations</description>
            <indexValue type="varchar">ENGINEERING COORD</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnOperations</description>
            <indexValue type="varchar">OPERATIONS COORDINAT</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnSafety</description>
            <indexValue type="varchar">WC COORDINATOR</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnSafety</description>
            <indexValue type="varchar">SAFETY DIRECTOR</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnSafety</description>
            <indexValue type="varchar">SAFETY SPECIALIST IV</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnEquipment</description>
            <indexValue type="varchar">EQUIP MANAGER</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnAccounting</description>
            <indexValue type="varchar">CONTROLLER</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnAccounting</description>
            <indexValue type="varchar">SR ACCOUNTANT</indexValue>
        </listItems>

        <listItems>
            <description type="varchar">HstnAccounting</description>
            <indexValue type="varchar">FINANCIAL ANALYST</indexValue>
        </listItems>

  </globalRequesteeDepartment>

有可能以及如何使用XSLT获取以下所需的XML输出：

<globalRequesteeDepartment>
    <HstnOperations>
        <indexValue>JOB COST COORDNTR</indexValue>
        <indexValue>ENGINEERING COORD</indexValue>
        <indexValue>OPERATIONS COORDINAT</indexValue>
    </HstnOperations>
    <HstnSafety>
        <indexValue>WC COORDINATOR</indexValue>
        <indexValue>SAFETY DIRECTOR</indexValue>
        <indexValue>SAFETY SPECIALIST IV</indexValue>
    </HstnSafety>
    <HstnEquipment>
        <indexValue>EQUIP MANAGER</indexValue>
    </HstnEquipment>
    <HstnAccounting>
        <indexValue>CONTROLLER</indexValue>
        <indexValue>SR ACCOUNTANT</indexValue>
        <indexValue>FINANCIAL ANALYST</indexValue>
    </HstnAccounting>
</globalRequesteeDepartment>

总结：每个“listItems”都有一个“indexValue”与“description”的联系。我希望能够将每个“描述”值作为自己的元素，但没有重复。然后，每个“listItems”的“indexValue”应该是匹配“description”元素的子元素。

Answer 1

对于此任务，请使用Muenchian grouping 所以你的XSLT看起来像这样：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <!-- create a key to get all elements grouped -->
    <xsl:key name="distinct" match="listItems" use="description" />

    <xsl:template match="/globalRequesteeDepartment">
      <xsl:copy>
        <!-- apply to the first of each group with muenchian grouping-->
        <xsl:apply-templates select="listItems[generate-id() = generate-id(key('distinct',description)[1])]" />
      </xsl:copy>
    </xsl:template>

    <xsl:template match="listItems">
      <xsl:element name="{description}">
        <!-- for-each over one whole group -->
        <xsl:for-each select="key('distinct',description)">
          <indexValue><xsl:value-of select="indexValue"/></indexValue>
        </xsl:for-each>
      </xsl:element>
    </xsl:template>

</xsl:stylesheet>