Question

类似这个问题(http://stackoverflow.com/q/1333558/9404) 我想利用XPath,在类似结构中计算出一笔高于产品的款项:

<items>
    <item>
        <value>1.0</value>
        <quantity>3</quantity>
    </item>
    <item>
        <value>2.5</value>
        <quantity>2</quantity>
    </item>
    <!-- ... -->
</items>

是否有计算每个项目产品总和的XPath表述?

Update: The solution has to work with PHPs XSLTProcessor class meaning that it probably has to be XSLT 1.0 compliant. This is why I did not yet accept the two probably correct answers using XSLT 2.0. I could neither test them in my PHP implementation nor in my browser nor in the Tryit Editor [1] from w3schools. Sorry!

http://w3schools.com/xsl/tryxslt.asp?xmlfile=cdcatalog&xsltfile=cdcatalog

Answer 1

www.un.org/spanish/ecosoc 采用2.0代语:

sum(/items/item/(value * quantity))

www.un.org/spanish/ecosoc 这里的XSLT 2.0转化成核查:

<xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="text"/>

    <xsl:template match="/">
        <xsl:sequence select="sum(/items/item/(value * quantity))"/>
    </xsl:template>
</xsl:stylesheet>

www.un.org/spanish/ecosoc 当这一转变适用于所提供的XML文件:。

<items>
    <item>
        <value>1.0</value>
        <quantity>3</quantity>
    </item>
    <item>
        <value>2.5</value>
        <quantity>2</quantity>
    </item>
    <!-- ... -->
</items>

XPath expression is evaluation and the results:

<>Explanation:

在XPath 2.0中,地点步骤是合法的

http://www.un.org/Depts/DGACM/index_spanish.htm

or even

。

II.XSLT 1.0 Solutions:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="/*">
  <xsl:call-template name="sumProducts">
    <xsl:with-param name="pNodes" select="item"/>
  </xsl:call-template>
 </xsl:template>

 <xsl:template name="sumProducts">
   <xsl:param name="pNodes" select="/.."/>
   <xsl:param name="pAccum" select="0"/>

   <xsl:choose>
    <xsl:when test="not($pNodes)">
     <xsl:value-of select="$pAccum"/>
    </xsl:when>
    <xsl:otherwise>
     <xsl:call-template name="sumProducts">
      <xsl:with-param name="pNodes" select="$pNodes[position() >1]"/>
      <xsl:with-param name="pAccum" select=
      "$pAccum + $pNodes[1]/value * $pNodes[1]/quantity"/>
     </xsl:call-template>
    </xsl:otherwise>
   </xsl:choose>
 </xsl:template>
</xsl:stylesheet>

当这一转变适用于所提供的XML文件时(上文),也是想要的,即产生正确的结果:

transform-and-sum。

Answer 2

引证:

<xsl:stylesheet version="2.0"xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="text"/>
    <xsl:template match="/">
        <xsl:sequence select="sum(/items/item/(value * quantity))"/>
    </xsl:template>
</xsl:stylesheet>

Answer 3

根据这项建议,。我收到了符合XSLT 1.0和XPath 1.0的解决办法,这些解决办法由PHP用于实施<代码>XSLTProcessor的校准/校准实施。 (签名至@Dimitre Novatchev, 待确认)。

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

    <xsl:template match="/items">
        <xsl:call-template name="recursivesum">
            <xsl:with-param name="items" select="*" />
        </xsl:call-template>
    </xsl:template>

    <xsl:template name="recursivesum">
        <xsl:param name="items" />
        <xsl:param name="sum" select="0" />
        <xsl:variable name="head" select="$items[1]" />
        <xsl:variable name="tail" select="$items[position()>1]" />
        <xsl:variable name="thissum" select="$head/value * $head/quantity" />
        <xsl:choose>
            <xsl:when test="not($tail)">
                <xsl:value-of select="$sum+$thissum" />
            </xsl:when>
            <xsl:otherwise>
                <xsl:call-template name="recursivesum">
                    <xsl:with-param name="sum" select="$sum+$thissum" />
                    <xsl:with-param name="items" select="$tail" />
                </xsl:call-template>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>

</xsl:transform>

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