Question

我正在尝试使用XSLT 2.0对xml元素分类列表进行排序。每个元素都有唯一的ID，并且在包含这些元素和更多元素的另一个列表中定义了分类。这是一个起始XML文档的示例。我要排序的部分是/ Atlas / VisitedCities。应该根据世界范围和访问日期来排序：

<?xml version="1.0" encoding="UTF-8"?>
<Atlas>
    <Cities>
        <City id="1" worldPart="Africa">
            <Name>Luxor</Name>
            <Founded>-3200</Founded>
            <Location>Egypt</Location>
        </City>
        <City id="2" worldPart="Africa">
            <Name>Tripoli</Name>
            <Founded>-700</Founded>
            <Location>Libya</Location>
        </City>
        <City id="3" worldPart="Americas">
            <Name>Cholula</Name>
            <Founded>-200</Founded>
            <Location>Mexico</Location>
        </City>
        <City id="4" worldPart="Americas">
            <Name>Flores</Name>
            <Founded>-1000</Founded>
            <Location>Guatemala</Location>
        </City>
        <City id="5" worldPart="Europe">
            <Name>Argos</Name>
            <Founded>-5000</Founded>
            <Location>Greece</Location>
        </City>
        <City id="6" worldPart="Europe">
            <Name>Athens</Name>
            <Founded>-4000</Founded>
            <Location>Greece</Location>
        </City>
    </Cities>
    <VisitedCities lastUpdate="2018-09-10">
        <VisitedCity cityID="6">
            <Date>1883-08-26</Date>
            <Visitor>Dora</Visitor>
        </VisitedCity>
        <VisitedCity cityID="3">
            <Date>1907-01-02</Date>
            <Visitor>Nemo</Visitor>
        </VisitedCity>
        <VisitedCity cityID="4">
            <Date>1940-02-08</Date>
            <Visitor>Jimenez</Visitor>
        </VisitedCity>
        <VisitedCity cityID="2">
            <Date>1886-06-10</Date>
            <Visitor>James T. Kirk</Visitor>
        </VisitedCity>
    </VisitedCities>
</Atlas>

所需的输出是这样

<?xml version="1.0" encoding="UTF-8"?>
<Atlas>
    <Cities>
        <City id="1" worldPart="Africa">
            <Name>Luxor</Name>
            <Founded>-3200</Founded>
            <Location>Egypt</Location>
        </City>
        <City id="2" worldPart="Africa">
            <Name>Tripoli</Name>
            <Founded>-700</Founded>
            <Location>Libya</Location>
        </City>
        <City id="3" worldPart="Americas">
            <Name>Cholula</Name>
            <Founded>-200</Founded>
            <Location>Mexico</Location>
        </City>
        <City id="4" worldPart="Americas">
            <Name>Flores</Name>
            <Founded>-1000</Founded>
            <Location>Guatemala</Location>
        </City>
        <City id="5" worldPart="Europe">
            <Name>Argos</Name>
            <Founded>-5000</Founded>
            <Location>Greece</Location>
        </City>
        <City id="6" worldPart="Europe">
            <Name>Athens</Name>
            <Founded>-4000</Founded>
            <Location>Greece</Location>
        </City>
    </Cities>
    <VisitedCities lastUpdate="2018-09-10">
        <VisitedCity cityID="2">
            <Date>1886-06-10</Date>
            <Visitor>James T. Kirk</Visitor>
        </VisitedCity>
        <VisitedCity cityID="6">
            <Date>1883-08-26</Date>
            <Visitor>Dora</Visitor>
        </VisitedCity>
        <VisitedCity cityID="3">
            <Date>1907-01-02</Date>
            <Visitor>Nemo</Visitor>
        </VisitedCity>
        <VisitedCity cityID="4">
            <Date>1940-02-08</Date>
            <Visitor>Jimenez</Visitor>
        </VisitedCity>
    </VisitedCities>
</Atlas>

我正在努力的样式表（XSLT 2.0）看起来像这样：

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

    <!-- Format output -->
    <xsl:output method="xml" indent="yes"/>
    <xsl:strip-space elements="*" />

    <!-- Copy everything that does not match later templates. -->
    <xsl:template match="node()|@*" priority="-1">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
    </xsl:template>

    <xsl:variable name="city.list" select="/Atlas/Cities"/>

    <xsl:variable name="sort-order" as="element()*">
        <wPart>Africa</wPart>
        <wPart>Europe</wPart>
        <wPart>Americas</wPart>
    </xsl:variable>

    <xsl:template match="/Atlas/VisitedCities">
        <xsl:variable name="city-list" select="."/>
        <xsl:copy>
            <xsl:apply-templates select="@*"/>
            <xsl:for-each select="$sort-order">
                <xsl:variable name="this-wpart" select="./text()"/>
                <!-- How to select VisitedCity based on info in other list??? -->
                <xsl:apply-templates select="$city-list/VisitedCity[$city.list/City[@cityID=$city-list/VisitedCity/@cityID]/@worldPart=$this-wpart]">
                    <xsl:sort select="./Date"/>
                </xsl:apply-templates>
            </xsl:for-each>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

我想我理解为什么此样式表不起作用（根本无法排序），因为我不知道如何在（最后一个）应用模板中进行选择。我看不到如何从表达式的内部引用最外部的元素。

Answer 1

通过min属性设置一个键来引用np.min(L + X[...,None]*(q1+q2) + Y[...,None]*q2,axis=2)元素，然后在City id表达式中引用{{1} }属性。另外，您可以使用

的xsl:sort调用替换大陆订单上的select

worldPart

https://xsltfiddle.liberty-development.net/eiZQaFJ

完整的示例是XSLT 3，但是要将其与XSLT 2一起使用，只需将其中的for-each声明替换为模板，并在模板中添加注释index-of()，即带有标识转换模板。

从参考列表收集的属性排序

1 个答案: