我有以下xml文件
<ScheduleProvider id="257" scheduleDate="2008-03-20T15:34:18Z" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="my.xsd">
<Content action="insert" duration="7200" id="2" title="movie-2">
<EpgDescription>
<EpgElement key="Year">2002</EpgElement>
<EpgElement key="Actors">Actor2 Actor22</EpgElement>
<EpgElement key="Directors">Director2</EpgElement>
<EpgElement key="Rating">2</EpgElement>
</EpgDescription>
<EpgDescription locale="en_US">
<EpgElement key="Title">Blockbuster-2</EpgElement>
</EpgDescription>
<Media comment="" fileName="Asset_2" format="AV_ClearTS" frameDuration="180000" id="LYS008168695" title="Asset_2">
<TechnicalMetadata key="ReadyForBroadcast">4</TechnicalMetadata>
<TechnicalMetadata key="Subtitle_Languages"/>
</Media>
</Content>
<Content action="insert" duration="7200" id="1" title="movie-1">
<EpgDescription>
<EpgElement key="Year">2001</EpgElement>
<EpgElement key="Actors">Actor1 Actor11</EpgElement>
<EpgElement key="Directors">Director1</EpgElement>
<EpgElement key="Rating">1</EpgElement>
</EpgDescription>
<EpgDescription locale="en_US">
<EpgElement key="Title">Blockbuster-1</EpgElement>
</EpgDescription>
<Media comment="" fileName="Asset_1" format="AV_ClearTS" frameDuration="180000" id="LYS008168695" title="Asset_1">
<TechnicalMetadata key="ReadyForBroadcast">4</TechnicalMetadata>
<TechnicalMetadata key="Subtitle_Languages"/>
</Media>
</Content>
</ScheduleProvider>
是否可以按 id 属性对其内容节点进行排序?我期待以下结果
<ScheduleProvider id="257" scheduleDate="2008-03-20T15:34:18Z" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="my.xsd">
<Content action="insert" duration="7200" id="1" title="movie-1">
<EpgDescription>
<EpgElement key="Year">2001</EpgElement>
<EpgElement key="Actors">Actor1 Actor11</EpgElement>
<EpgElement key="Directors">Director1</EpgElement>
<EpgElement key="Rating">1</EpgElement>
</EpgDescription>
<EpgDescription locale="en_US">
<EpgElement key="Title">Blockbuster-1</EpgElement>
</EpgDescription>
<Media comment="" fileName="Asset_1" format="AV_ClearTS" frameDuration="180000" id="LYS008168695" title="Asset_1">
<TechnicalMetadata key="ReadyForBroadcast">4</TechnicalMetadata>
<TechnicalMetadata key="Subtitle_Languages"/>
</Media>
</Content>
<Content action="insert" duration="7200" id="2" title="movie-2">
<EpgDescription>
<EpgElement key="Year">2002</EpgElement>
<EpgElement key="Actors">Actor2 Actor22</EpgElement>
<EpgElement key="Directors">Director2</EpgElement>
<EpgElement key="Rating">2</EpgElement>
</EpgDescription>
<EpgDescription locale="en_US">
<EpgElement key="Title">Blockbuster-2</EpgElement>
</EpgDescription>
<Media comment="" fileName="Asset_2" format="AV_ClearTS" frameDuration="180000" id="LYS008168695" title="Asset_2">
<TechnicalMetadata key="ReadyForBroadcast">4</TechnicalMetadata>
<TechnicalMetadata key="Subtitle_Languages"/>
</Media>
</Content>
</ScheduleProvider>
我听说过 xsl:sort ,我玩了一下。但我没有使用xslt的经验,我无法弄清楚如何使用 xsl:sort 。
答案 0 :(得分:3)
你可以试试这个:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="ScheduleProvider">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:apply-templates select="Content">
<xsl:sort data-type="number" select="@id"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="Content">
<xsl:copy-of select="."/>
</xsl:template>
</xsl:stylesheet>
答案 1 :(得分:0)
您是否尝试过<xsl:sort select="id" />
?