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在使用 xslt 1.0 进行转换时,我发现实现以下逻辑有点困难。这是我的要求:我需要根据员工的工作时间定义以下逻辑。1. 任何超过 8 的东西都会在一天内 OT else Regular。2. 如果连续工作 3 天,那么第 3 天的时间将是双倍时间。

这是一个例子:

输入:

<Input>
          <Record>
            <EmpId>1</EmpId>
            <Date></Date>
            <Hours></Hours>
          </Record>
          <Record>
            <EmpId>1</EmpId>
            <Date>12/1/2012</Date>
            <Hours>6</Hours>
          </Record>
          <Record>
            <EmpId>1</EmpId>
            <Date>12/1/2012</Date>
            <Hours>4</Hours>
          </Record>
          <Record>
            <EmpId>1</EmpId>
            <Date>12/2/2012</Date>
            <Hours>4</Hours>
          </Record>
          <Record>
            <EmpId>1</EmpId>
            <Date>12/2/2012</Date>
            <Hours></Hours>
          </Record>
          <Record>
            <EmpId>1</EmpId>
            <Date>12/3/2012</Date>
            <Hours>3</Hours>
          </Record>
          <Record>
            <EmpId>2</EmpId>
            <Date>12/1/2012</Date>
            <Hours>4</Hours>
          </Record>
          <Record>
            <EmpId>2</EmpId>
            <Date>12/1/2012</Date>
            <Hours></Hours>
          </Record>
      </Input>

输出:

 <Output>
          <Record>
            <EmployeeId>1</EmployeeId>
            <Detail>
              <Date>12/1/2012</Date>
              <RegHours>8</Reghours>
               <OTHours>2</OTHours>
            </Detail>
            <Detail>
              <Date>12/2/2012</Date>
              <RegHours>4</Reghours>
             </Detail>
            <Detail>
              <Date>12/3/2012</Date>
              <DTHours>3</DThours>
            </Detail>
          </Record>
          <Record>
            <EmployeeId>2</EmployeeId>
            <Detail>
              <Date>12/1/2012</Date>
              <RegHours>4</Reghours>
             </Detail>
            <Detail>
          </Record>
    </Output>

将不胜感激任何帮助。

谢谢..阿比

4

1 回答 1

0

这种转变

<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="kRecByEmpId" match="Record[Date/node() and Hours/node()]"
          use="EmpId"/>
 <xsl:key name="kRecByEmpIdDate" match="Record[Date/node() and Hours/node()]"
          use="concat(EmpId,'+',Date)"/>

 <xsl:template match=
  "Record[generate-id()=generate-id(key('kRecByEmpId',EmpId)[1])]">
   <Record>
     <EmployeeId>
        <xsl:value-of select="EmpId"/>
     </EmployeeId>
     <xsl:apply-templates mode="datetime" select=
      "key('kRecByEmpId', EmpId)
               [generate-id()
               =
                generate-id(key('kRecByEmpIdDate',
                                 concat(EmpId,'+',Date)
                                 )[1]
                        )
            ]
      "/>
   </Record>
  </xsl:template>

  <xsl:template match="Record" mode="datetime">
    <Detail>
      <xsl:copy-of select="Date"/>
      <xsl:variable name="vTotal" select=
       "sum(key('kRecByEmpIdDate', concat(EmpId,'+',Date))/Hours)"/>
      <RegHours>
        <xsl:value-of select="8*($vTotal >= 8) + $vTotal*(not($vTotal >= 8))"/>
      </RegHours>
      <xsl:if test="$vTotal > 8">
          <OTHours>
            <xsl:value-of select="$vTotal -8"/>
          </OTHours>
      </xsl:if>
    </Detail>
  </xsl:template>
  <xsl:template match="text()"/>
</xsl:stylesheet>

应用于提供的 XML 文档时:

<Input>
    <Record>
        <EmpId>1</EmpId>
        <Date></Date>
        <Hours></Hours>
    </Record>
    <Record>
        <EmpId>1</EmpId>
        <Date>12/1/2012</Date>
        <Hours>6</Hours>
    </Record>
    <Record>
        <EmpId>1</EmpId>
        <Date>12/1/2012</Date>
        <Hours>4</Hours>
    </Record>
    <Record>
        <EmpId>1</EmpId>
        <Date>12/2/2012</Date>
        <Hours>4</Hours>
    </Record>
    <Record>
        <EmpId>1</EmpId>
        <Date>12/2/2012</Date>
        <Hours></Hours>
    </Record>
    <Record>
        <EmpId>1</EmpId>
        <Date>12/3/2012</Date>
        <Hours>3</Hours>
    </Record>
    <Record>
        <EmpId>2</EmpId>
        <Date>12/1/2012</Date>
        <Hours>4</Hours>
    </Record>
    <Record>
        <EmpId>2</EmpId>
        <Date>12/1/2012</Date>
        <Hours></Hours>
    </Record>
</Input>

产生想要的正确结果

<Record>
   <EmployeeId>1</EmployeeId>
   <Detail>
      <Date>12/1/2012</Date>
      <RegHours>8</RegHours>
      <OTHours>2</OTHours>
   </Detail>
   <Detail>
      <Date>12/2/2012</Date>
      <RegHours>4</RegHours>
   </Detail>
   <Detail>
      <Date>12/3/2012</Date>
      <RegHours>3</RegHours>
   </Detail>
</Record>
<Record>
   <EmployeeId>2</EmployeeId>
   <Detail>
      <Date>12/1/2012</Date>
      <RegHours>4</RegHours>
   </Detail>
</Record>
于 2012-12-23T15:26:34.887 回答