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我希望有人可以帮助我，因为我一直在试图弄清楚如何将数据集转换为按多个级别分组的表结构。第一层是事件日期，第二层是公司名称，最后是用户。

以下是 XML。

<xmlData>
    <records>
      <record>
        <userid>1</userid>
        <usersname>Jane Doe</usersname>
        <companyname>Company A</companyname>
        <eventdate>01 FEB 2017</eventdate>
        <jeventdate>2457786</jeventdate>
      </record>
      <record>
        <userid>3</userid>
        <usersname>Jane Doe</usersname>
        <companyname>Company B</companyname>
        <eventdate>01 FEB 2017</eventdate>
        <jeventdate>2457786</jeventdate>
      </record>
      <record>
        <userid>2</userid>
        <usersname>Joe Smith</usersname>
        <companyname>Company B</companyname>
        <eventdate>01 DEC 2016</eventdate>
        <jeventdate>2457724</jeventdate>
      </record>
      <record>
        <userid>2</userid>
        <usersname>Joe Smith</usersname>
        <companyname>Company B</companyname>
        <eventdate>01 JAN 2017</eventdate>
        <jeventdate>2457755</jeventdate>
      </record>
      <record>
        <userid>2</userid>
        <usersname>Joe Smith</usersname>
        <companyname>Company B</companyname>
        <eventdate>01 FEB 2017</eventdate>
        <jeventdate>2457786</jeventdate>
      </record>
    </records>
</xmlData>

我试图得到的结果显示在以下简单的 HTML 输出中。

<h1>01 DEC 2016</h1>
  <h2>Company B</h2>
  <table>
    <tr><td>2</td><td>Joe Smith</td></tr>
  </table>
<h1>01 JAN 2017</h1>
  <h2>Company B</h2>
  <table>
    <tr><td>2</td><td>Joe Smith</td></tr>
  </table>
<h1>01 FEB 2017</h1>
  <h2>Company A</h2>
  <table>
    <tr><td>1</td><td>Jane Doe</td></tr>
  </table>
  <h2>Company B</h2>
  <table>
    <tr><td>3</td><td>Dave Dodd</td></tr>
    <tr><td>2</td><td>Joe Smith</td></tr>
  </table>

我遇到的问题实际上是两个部分。首先是获得三个层次的深度。第二个是获取所有记录，而不仅仅是前几个。

这是我一直在使用的 XSL。

<xsl:key name="monthof" match="record" use="eventdate"/>
<xsl:key name="companyof" match="record" use="concat(eventdate,'|',companyname)"/>
<xsl:key name="userof" match="record" use="concat(eventdate,'|',companyname,'|',usersname)"/>

<xsl:template match="xmlData/records">
    <xsl:for-each select="record[key('monthof',eventdate)]">
        <xsl:sort select="jeventdate"/>
        <xsl:variable name="lstEventDate" select="key('monthof',eventdate)" />
        <h2><xsl:value-of select="eventdate"/></h2>
        <xsl:for-each select="key('companyof',concat(eventdate,'|',companyname))">
            <xsl:sort select="companyname"/>
            <h3>
                <xsl:value-of select="companyname"/>
            </h3>
            <table>
                <xsl:for-each select="key('userof',concat(eventdate,'|',companyname,'|',usersname))">
                    <tr>
                        <td><xsl:value-of select="userid"/></td>
                        <td><xsl:value-of select="usersname"/></td>
                    </tr>                                   
                </xsl:for-each>
            </table>
        </xsl:for-each>
    </xsl:for-each>
</xsl:template>

这是我得到的输出。

01 DEC 2016

Company B

2 Jane Doe 

01 JAN 2017

Company B

2 Jane Doe 

01 FEB 2017

Company A

1 Joe Smith 

01 FEB 2017

Company B

1 Joe Smith 

Company B

3 Dave Dodd 

01 FEB 2017

Company B

1 Joe Smith 

Company B

2 Jane Doe
4

1 回答 1

0

您定义的键很好，但您需要使用它们，正如评论中已经指出的那样，使用Muenchian 分组方法：

<xsl:template match="xmlData/records">
    <xsl:for-each select="record[generate-id() = generate-id(key('monthof',eventdate)[1])]">
        <xsl:sort select="jeventdate"/>
        <h2><xsl:value-of select="eventdate"/></h2>
        <xsl:for-each select="key('monthof', eventdate)[generate-id() = generate-id(key('companyof',concat(eventdate,'|',companyname))[1])]">
            <xsl:sort select="companyname"/>
            <h3>
                <xsl:value-of select="companyname"/>
            </h3>
            <table>
                <xsl:for-each select="key('companyof',concat(eventdate,'|',companyname))[generate-id() = generate-id(key('userof',concat(eventdate,'|',companyname,'|',usersname))[1])]">
                    <tr>
                        <td><xsl:value-of select="userid"/></td>
                        <td><xsl:value-of select="usersname"/></td>
                    </tr>                                   
                </xsl:for-each>
            </table>
        </xsl:for-each>
    </xsl:for-each>
</xsl:template>

您可能想检查您是否不能迁移到 XSLT 2.0，它变得更容易做

<xsl:template match="xmlData/records">
    <xsl:for-each-group select="record" group-by="eventdate">
        <xsl:sort select="jeventdate"/>
        <h2><xsl:value-of select="eventdate"/></h2>
        <xsl:for-each-group select="current-group()" group-by="companyname">
            <xsl:sort select="companyname"/>
            <h3>
                <xsl:value-of select="companyname"/>
            </h3>
            <table>
                <xsl:for-each-group select="current-group()" group-by="usersname">
                    <tr>
                        <td><xsl:value-of select="userid"/></td>
                        <td><xsl:value-of select="usersname"/></td>
                    </tr>                                   
                </xsl:for-each-group>
            </table>
        </xsl:for-each-group>
    </xsl:for-each-group>
</xsl:template>
于 2017-02-27T11:25:47.557 回答