XSLT将两个属性的乘积相加

Question

我有以下XML源结构:

<turnovers>
    <turnover repid="1" amount="500" rate="0.1"/>
    <turnover repid="5" amount="600" rate="0.5"/>
    <turnover repid="4" amount="400" rate="0.2"/>
    <turnover repid="1" amount="700" rate="0.05"/>
    <turnover repid="2" amount="100" rate="0.15"/>
    <turnover repid="1" amount="900" rate="0.25"/>
    <turnover repid="2" amount="1000" rate="0.18"/>
    <turnover repid="5" amount="200" rate="0.55"/>
    <turnover repid="9" amount="700" rate="0.40"/>
</turnovers>

我需要一个XSL:value-of select语句,它将返回给定代表ID的rate属性和amount属性的乘积之和.所以对于代表5,我需要((600 x 0.5)+(200 x 0.55)).

Answer 1

<xsl:stylesheet 
  version="1.0" 
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
>
  <xsl:template match="/turnovers">
    <val>
      <!-- call the sum function (with the relevant nodes) -->
      <xsl:call-template name="sum">
        <xsl:with-param name="nodes" select="turnover[@repid='5']" />
      </xsl:call-template>
    </val>
  </xsl:template>

  <xsl:template name="sum">  
    <xsl:param name="nodes" />
    <xsl:param name="sum" select="0" />

    <xsl:variable name="curr" select="$nodes[1]" />

    <!-- if we have a node, calculate & recurse -->
    <xsl:if test="$curr">
      <xsl:variable name="runningsum" select="
        $sum + $curr/@amount * $curr/@rate
      " />
      <xsl:call-template name="sum">
        <xsl:with-param name="nodes" select="$nodes[position() &gt; 1]" />
        <xsl:with-param name="sum"   select="$runningsum" />
      </xsl:call-template>
    </xsl:if>

    <!-- if we don't have a node (last recursive step), return sum -->
    <xsl:if test="not($curr)">
      <xsl:value-of select="$sum" />
    </xsl:if>

  </xsl:template>
</xsl:stylesheet>

得到:

<val>410</val>

这两个<xsl:if>可以用一个替换<xsl:choose>.这意味着在递归期间少检查一次,但这也意味着另外两行代码.