XSLT1.0在Quiz XML中选择元素
XSLT1.0在Quiz XML中选择元素
提问于 2015-02-12 15:53:46
下面是一个示例XML。
<Root>
<Story>
<body>
<quiz>1</quiz>How to select that rightanswer from another Story ?
</body>
<answer>answer1-1</answer>
<answer>answer1-2</answer>
<answer>answer1-3</answer>
<body>
<quiz>2</quiz>And how to push these <qitem/>s into <qitems/> ?
</body>
<answer>answer2-1</answer>
<answer>answer2-2</answer>
<answer>answer2-3</answer>
<answer>answer2-4</answer>
</Story>
<Story>
<h3>RIGHT_ANSWERS</h3>
<rightanswer>
RightAnswer1 IDK
</rightanswer>
<rightanswer>
RightAnswer2 IDK
</rightanswer>
</Story>
</Root>预期产出：
<Root>
<Story>
<quiz>
<qitems>
<qitem>
<question>
<qtitle>1</qtitle>
<qtext>How to select that rightanswer from another Story ?</qtext>
</question>
<answers>
<answer>answer1-1</answer>
<answer>answer1-2</answer>
<answer>answer1-3</answer>
</answers>
<rightanswer>RightAnswer1 IDK</rightanswer>
</qitem>
<qitem>
<question>
<qtitle>2</qtitle>
<qtext>And how to push these <qitem/>s into <qitems/> ?</qtext>
</question>
<answers>
<answer>answer2-1</answer>
<answer>answer2-2</answer>
<answer>answer2-3</answer>
<answer>answer2-4</answer>
</answers>
<rightanswer>RightAnswer2 IDK</rightanswer>
</qitem>
</qitems>
</quiz>
</Story>
</Root>我所做的就是将每个body[quiz]及其下面的兄弟姐妹answer放入<qitem>元素，并将这些<qitem>s推入父<qitems>并选择<rightanswer>。到目前为止，我有了这个xsl：
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" />
<xsl:strip-space elements="*" />
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="body[quiz]" >
<qitem>
<question>
<qtitle>
<xsl:value-of select="quiz" />
</qtitle>
<qtext>
<xsl:apply-templates select="node()" />
</qtext>
</question>
<answers>
<xsl:apply-templates select="following-sibling::*[1][self::answer]" mode="following" />
</answers>
<rightanswer></rightanswer>
</qitem>
</xsl:template>
<xsl:template match="answer" mode="following">
<answer>
<xsl:apply-templates select="node()" />
</answer>
<xsl:apply-templates select="following-sibling::*[1][self::answer]" mode="following" />
</xsl:template>
<xsl:template match="body/quiz" />
<xsl:template match="answer" />
<xsl:template match="h3"/>
</xsl:stylesheet>看看这个：http://xsltransform.net/nc4NzQr
回答 1
Stack Overflow用户
回答已采纳
发布于 2015-02-12 18:49:42
如果可以假设Root元素始终包含确切的2个Story元素，其中一个元素具有Q&As，另一个元素具有正确的答案，那么您可以简单地这样做：
XSLT1.0
<?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:strip-space elements="*"/>
<!-- identity transform -->
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Story[1]">
<qitems>
<xsl:apply-templates select="body"/>
</qitems>
</xsl:template>
<xsl:template match="body">
<xsl:variable name="i" select="position()" />
<qitem>
<question>
<qtitle>
<xsl:value-of select="quiz" />
</qtitle>
<qtext>
<xsl:apply-templates/>
</qtext>
</question>
<answers>
<xsl:apply-templates select="following-sibling::answer[1]" mode="following"/>
</answers>
<xsl:apply-templates select="following::Story/rightanswer[$i]"/>
</qitem>
</xsl:template>
<!-- sibling recursion -->
<xsl:template match="answer" mode="following">
<answer>
<xsl:apply-templates/>
</answer>
<xsl:apply-templates select="following-sibling::*[1][self::answer]" mode="following" />
</xsl:template>
<xsl:template match="Story[2]" />
<xsl:template match="quiz" />
</xsl:stylesheet>被困在将这些
<qitem>推入父级<qitems>中
我不知道为什么需要<qitems>包装器，它代表什么；在我看来，您请求的输出中已经有太多的包装器元素。在上面的示例中，我让<qitems>表示父Story，并删除Story包装器本身。
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原文链接：
https://stackoverflow.com/questions/28481894
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