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在格式化节点内排序

xslt-2.0 xslt-1.0 xslt

orosco03 · 技术社区 · 7 年前

我需要显示XML,并对一天的总距离进行排序,以便首先显示公里数最多的一天我只能算出距离的总和

输入

<output>
    <cars>
        <car>
          <id>1</id>
          <license>B-01-TST</license>
        </car>
        <car>
          <id>2</id>
          <license>IF-02-TST</license>
        </car>
    </cars>
    <distances>
        <distance>
          <id_car>1</id_car>
          <date>20110901</date>
          <distance>111</distance>
        </distance>     
        <distance>
          <id_car>1</id_car>
          <date>20110903</date>
          <distance>56</distance>
        </distance>
        <distance>
          <id_car>2</id_car>
          <date>20110901</date>
          <distance>92</distance>
        </distance>
        <distance>
          <id_car>2</id_car>
          <date>20110902</date>
          <distance>97</distance>
        </distance>
    </distances>
</output>

预期输出

<output>
<cars>
    <car>
        <id>1</id>
        <license>B-01-TST</license>
        <distance totalKm="Day 01: 111"/>
        <distance totalKm="Day 03: 56"/>
    </car>
    <car>
        <id>2</id>
        <license>IF-02-TST</license>
        <distance totalKm="Day 01: 92"/>
        <distance totalKm="Day 02: 97"/>
    </car>
</cars>

我无法复制所有XSL,因为对于这个问题来说,它的代码太多了, 主要信息是

到目前为止,XSL如下所示:

<xsl:key name="byCarIdAndDate" match="distance" use="concat(id_car, '|', date)"/>
<xsl:key name="byCarId" match="distance" use="id_car"/>
<xsl:template match="car">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
            <xsl:apply-templates select="key('byCarId', id)"/>
        </xsl:copy>
    </xsl:template>
<xsl:template match="distance[generate-id()= generate-id(key('byCarIdAndDate', concat(id_car, '|', date))[1])]">
    <xsl:variable name="thisDate" select="key('byCarIdAndDate', concat(id_car, '|', date))"/>
    <xsl:variable name="sum" select="sum($thisDate/distance)"/>
    <!--make some sort by higher value -->
    <distance totalKm="Day {substring(date, 7, 2)}: {$sum}"/>
</xsl:template>

模板匹配=“距离|距离”

1 回复 | 直到 7 年前

Martin Honnen 7 年前

我认为最好求助于 xsl:for-each-group 然后按日期分组,以便我们轻松地对组进行求和:

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">

    <xsl:output indent="yes"/>
    <xsl:strip-space elements="*"/>

    <xsl:key name="byCarId" match="distances/distance" use="id_car"/>

    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="distances"/>

    <xsl:template match="car">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
            <xsl:for-each-group select="key('byCarId', id)" group-by="date">
                <xsl:sort select="sum(current-group()/distance)" order="descending"/>
                <xsl:apply-templates select="."/>
            </xsl:for-each-group>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="distance">
        <distance totalKm="Day {substring(date, 7, 2)}: {sum(current-group()/distance)}"/>
    </xsl:template>

</xsl:transform>

http://xsltransform.net/3MvmrAX