背景：以下是MSWord表单的xslt修改的xml提取。来自MSWord表单的一些文本以某种方式被分成多个元素，并且需要重新组合成单个元素。下面是倒数第二个XML输入的实际片段
<Section coord="2.13" posn="2" of="13">
<Segment coord="1.25" rowno="1" of="25">
<Entry coord="1.1" colno="1" of="1" desgn="Table">QUAL</Entry>
<Entry coord="1.1" colno="1" of="1" desgn="Table">I</Entry>
<Entry coord="1.1" colno="1" of="1" desgn="Table">FICATIONS</Entry>
</Segment>
<Segment coord="2.25" rowno="2" of="25">
<Entry coord="1.1" colno="1" of="1" desgn="Table">ACADEMIC QUALIFICATIONS</Entry>
<Entry coord="1.1" colno="1" of="1" desgn="Table"> (Most recent first)</Entry>
</Segment>
<Segment coord="3.25" rowno="3" of="25">
<Entry coord="1.4" colno="1" of="4" desgn="Column">Degree/Diploma/Certificate</Entry>
<Entry coord="2.4" colno="2" of="4" desgn="Column">Institution</Entry>
<Entry coord="3.4" colno="3" of="4" desgn="Column">Date Conferred</Entry>
<Entry coord="3.4" colno="3" of="4" desgn="Column">(mm/yyyy)</Entry>
<Entry coord="4.4" colno="4" of="4" desgn="Column">SAQA Evaluated?</Entry>
<Entry coord="4.4" colno="4" of="4" desgn="Column">(If not SA qualification)</Entry>
</Segment>
<Segment coord="4.25" rowno="4" of="25"/>
<!-- remaining 21 Segments from Section deleted ... -->
</Section>
注意：@coord属性是由兄弟“position（）。last（）”构建的。
所需的合并输出： 例如，在带有@coord 1.25的段中，需要将三个条目合并为一个条目：
<Entry coord="1.1" colno="1" of="1" desgn="Table">QUALIFICATIONS</Entry>
将他们的文本连成一个。
同样，段1.26有两个条目应该合并为：
<Entry coord="1.1" colno="1" of="1" desgn="Table">ACADEMIC QUALIFICATIONS (Most recent first)</Entry>
这同样适用于第3.25节中的最后两个，具有不同的合并条目：
<Entry coord="3.4" colno="3" of="4" desgn="Column">Date Conferred(mm/yyyy)</Entry>
和
<Entry coord="4.4" colno="4" of="4" desgn="Column">SAQA Evaluated?(If not SA qualification)</Entry>
我能够（按文档顺序）测试@coord的重复，例如： test =“@ coord = following-sibling :: Entry / @ coord”开始连接 要么 test =“@ coord！= preceding-sibling :: Entry / @ coord”停止连接 但我的困难是在连接文本时推迟xsl：copy。它在文档顺序上变得混乱（我的失败和未完成的尝试只进行一次连接而不是根据需要进行连接）：
<xsl:template match="Segment">
<xsl:for-each select="Entry" >
<xsl:choose>
<xsl:when test="position()= 1 and (@coord = following-sibling::Entry/@coord)" >
<xsl:copy>
<xsl:value-of select="@*"/><xsl:value-of select="text()" /> <xsl:value-of select="following-sibling::Entry/text()" />
</xsl:copy>
</xsl:when>
<xsl:when test="@coord != preceding-sibling::Entry/@coord" >
<xsl:copy>
<xsl:value-of select="@*"/><xsl:value-of select="text()" />
</xsl:copy>
</xsl:when>
<xsl:otherwise>
<xsl:for-each select=".">
<xsl:if test="@coord = following-sibling::Entry/@coord" >
<xsl:value-of select="following-sibling::Entry/text()" />
</xsl:if>
</xsl:for-each>
</xsl:otherwise>
</xsl:choose>
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:for-each>
</xsl:template>
似乎在反向文档顺序中可能更自然地连接，但它只是在思考它时仍然是混乱的。攻击这个问题的最佳方法是什么？
根据我对答案2的评论，如何根据建议（）扩展其他父处理的答案。具有父属性的修改输入（ frags =子::文本片段的数量，大小 =连接文本片段的总字符串长度）需要填充并显示在下面的xml输入为空属性。
<Section coord="2.13" posn="2" of="13">
<Segment coord="1.25" rowno="1" of="25" frags="" size="">
<Entry coord="1.1" colno="1" of="1" desgn="Table" size="4">QUAL</Entry>
<Entry coord="1.1" colno="1" of="1" desgn="Table" size="1">I</Entry>
<Entry coord="1.1" colno="1" of="1" desgn="Table" size="9">FICATIONS</Entry>
</Segment>
<Segment coord="2.25" rowno="2" of="25" frags="" size="">
<Entry coord="1.1" colno="1" of="1" desgn="Table" size="23">ACADEMIC QUALIFICATIONS</Entry>
<Entry coord="1.1" colno="1" of="1" desgn="Table" size="20"> (Most recent first)</Entry>
</Segment>
<Segment coord="3.25" rowno="3" of="25" frags="" size="">
<Entry coord="1.4" colno="1" of="4" desgn="Column" size="26">Degree/Diploma/Certificate</Entry>
<Entry coord="2.4" colno="2" of="4" desgn="Column" size="11">Institution</Entry>
<Entry coord="3.4" colno="3" of="4" desgn="Column" size="14">Date Conferred</Entry>
<Entry coord="3.4" colno="3" of="4" desgn="Column" size="9">(mm/yyyy)</Entry>
<Entry coord="4.4" colno="4" of="4" desgn="Column" size="15">SAQA Evaluated?</Entry>
<Entry coord="4.4" colno="4" of="4" desgn="Column" size="25">(If not SA qualification)</Entry>
</Segment>
<!-- delete -->
</Section>
父级（细分）元素的额外处理的预期输出：
<!-- deleted prior input xml -->
<Segment coord="1.25" rowno="1" of="25" frags="3" size="14">
<!-- deleted collapsed Entries as transformed -->
</Segment>
<Segment coord="2.25" rowno="2" of="25" frags="2" size="43">
<!-- deleted collapsed Entries as transformed -->
</Segment>
<Segment coord="3.25" rowno="3" of="25" frags="6" size="100">
<!-- deleted collapsed Entries as transformed -->
</Segment>
<!-- deleted rest of input xml -->
答案 0 :(得分：1)
试试这个XSLT1.0样式表（这个XSLT2.0版本会简单得多）：
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:strip-space elements="*"/>
<!-- identity transform template -->
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Segment">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:for-each select="Entry[not(@coord = preceding-sibling::Entry/@coord)]">
<xsl:variable name="Coord" select="@coord"/>
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:for-each select="../Entry[@coord = $Coord]">
<xsl:value-of select="."/>
</xsl:for-each>
</xsl:copy>
</xsl:for-each>
</xsl:copy>
</xsl:template>
</xsl:transform>
答案 1 :(得分：1)
适用于xslt1和2，但仅当以下情况属实时：如果父级和坐标相同，则应折叠两个条目。
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="utf-8" indent="yes"/>
<xsl:key name="entry-by-coord" match="Entry" use="concat(generate-id(parent::*), '||', @coord)"/>
<xsl:template match="/">
<xsl:apply-templates select="*"/>
</xsl:template>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Entry">
<xsl:if test="generate-id()=generate-id(key('entry-by-coord', concat(generate-id(parent::*), '||', @coord))[1])">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:copy-of select="key('entry-by-coord', concat(generate-id(parent::*), '||', @coord))/text()"/>
</xsl:copy>
</xsl:if>
</xsl:template>
</xsl:stylesheet>