非常有趣的Python bounty question我知道可以用XSLT 1.0解决。请注意，这不是一个重复的问题，因为之前的帖子以Python方法为中心，而这是尝试针对同一问题的XSLT解决方案。以下是我的尝试，但是受限于预设数量的父/子组合，这里有四个级别，并且有条件地遍历每个级别。
有没有办法在任何组合级别推广我的解决方案？我知道这可能需要使用-->分隔符标记值。预期输出是电流输出，但需要动态解决方案。我包含Python脚本以显示最终的最终结果。为了明确利益冲突，我不会在上面的帖子中使用任何答案，但请您这样做！
XML 输入
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<nodes>
<node name="Car" child="Engine"/>
<node name="Car" child="Wheel"/>
<node name="Engine" child="Piston"/>
<node name="Engine" child="Carb"/>
<node name="Carb" child="Bolt"/>
<node name="Spare Wheel"/>
<node name="Bolt" child="Thread"/>
<node name="Carb" child="Foat"/>
<node name="Truck" child="Engine"/>
<node name="Engine" child="Bolt"/>
<node name="Wheel" child="Hubcap"/>
</nodes>
XSLT
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output version="1.0" encoding="UTF-8" indent="yes" />
<xsl:strip-space elements="*"/>
<xsl:template match="nodes">
<data>
<xsl:apply-templates select="node[not(@name=ancestor::nodes/node/@child)]"/>
</data>
</xsl:template>
<xsl:template match="node">
<xsl:variable select="@name" name="currname"/>
<xsl:variable select="@child" name="currchild"/>
<xsl:variable select="/nodes/node" name="nodeset1"/>
<xsl:variable select="/nodes/node[@name=$currchild]" name="nodeset2"/>
<xsl:variable select="/nodes/node[@name=$nodeset2/@child]" name="nodeset3"/>
<xsl:variable select="/nodes/node[@name=$nodeset3/@child]" name="nodeset4"/>
<xsl:for-each select="$nodeset2">
<xsl:variable select="@child" name="nodeset2child"/>
<xsl:for-each select="$nodeset3">
<xsl:variable select="@child" name="nodeset3child"/>
<xsl:if test="@name=$nodeset2child">
<xsl:for-each select="$nodeset4">
<xsl:if test="@name=$nodeset3child">
<xsl:value-of select="$currname"/> --> <xsl:value-of select="$currchild"/> --> <xsl:value-of select="$nodeset2child"/> --> <xsl:value-of select="$nodeset3child"/> --> <xsl:value-of select="@child"/><xsl:text>
</xsl:text>
</xsl:if>
</xsl:for-each>
<xsl:if test="$nodeset2child!=$nodeset3/@child and $nodeset3child != $nodeset4/@name">
<xsl:value-of select="$currname"/> --> <xsl:value-of select="$currchild"/> --> <xsl:value-of select="$nodeset2child"/> --> <xsl:value-of select="$nodeset3child"/><xsl:text>
</xsl:text>
</xsl:if>
</xsl:if>
</xsl:for-each>
<xsl:if test="not($nodeset2child=$nodeset3/@child or ancestor::nodes/node[@name=$nodeset2child]/@child)">
<xsl:value-of select="$currname"/> --> <xsl:value-of select="$currchild"/> --> <xsl:value-of select="$nodeset2child"/><xsl:text>
</xsl:text>
</xsl:if>
</xsl:for-each>
<xsl:value-of select="@name[not(ancestor::node/@child=$nodeset2/@name)]"/><xsl:text>
</xsl:text>
</xsl:template>
</xsl:transform>
XML 转换后的输出
<?xml version='1.0' encoding='UTF-8'?>
<data>Car --> Engine --> Piston
Car --> Engine --> Carb --> Bolt --> Thread
Car --> Engine --> Carb --> Foat
Car --> Engine --> Bolt --> Thread
Car --> Wheel --> Hubcap
Spare Wheel
Truck --> Engine --> Piston
Truck --> Engine --> Carb --> Bolt --> Thread
Truck --> Engine --> Carb --> Foat
Truck --> Engine --> Bolt --> Thread
</data>
Python 脚本（在转换后的输出根节点上运行xpath）
import lxml.etree as ET
# LOAD XML AND XSL DOCS
dom = ET.parse('Input.xml')
xslt = ET.parse('XSLTScript.xsl')
# TRANSFORM XML
transform = ET.XSLT(xslt)
newdom = transform(dom)
# XPATH NEW DOM ROOT NODE (<data>)
print(newdom.xpath('/data')[0].text.replace("\n\n", "\n"))
# Car --> Engine --> Piston
# Car --> Engine --> Carb --> Bolt --> Thread
# Car --> Engine --> Carb --> Foat
# Car --> Engine --> Bolt --> Thread
# Car --> Wheel --> Hubcap
# Spare Wheel
# Truck --> Engine --> Piston
# Truck --> Engine --> Carb --> Bolt --> Thread
# Truck --> Engine --> Carb --> Foat
# Truck --> Engine --> Bolt --> Thread
答案 0 :(得分：3)
这是一个更短（23行）且更有效的解决方案
这在计算上也是最简单的 - 比较嵌套级别1到嵌套级别3 - 4 ...
此解决方案尾递归意味着任何优秀的XSLT处理器都会通过迭代对其进行优化，从而避免堆栈溢出的可能性，因为最大调用堆栈深度保持不变（1）：< / p>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:key name="kNodeByChild" match="node" use="@child"/>
<xsl:key name="kNodeByName" match="node" use="@name"/>
<xsl:template match="/*">
<xsl:apply-templates select="node[not(key('kNodeByChild', @name))]"/>
</xsl:template>
<xsl:template match="node[not(key('kNodeByName', @child))]">
<xsl:param name="pParentPath"/>
<xsl:value-of select="concat($pParentPath, @name, ' ---> ', @child, '
')"/>
</xsl:template>
<xsl:template match="node">
<xsl:param name="pParentPath"/>
<xsl:apply-templates select="key('kNodeByName', @child)">
<xsl:with-param name="pParentPath" select="concat($pParentPath, @name, ' ---> ')"/>
</xsl:apply-templates>
</xsl:template>
</xsl:stylesheet>
在提供的XML文档上应用此转换时：
<nodes>
<node name="Car" child="Engine"/>
<node name="Car" child="Wheel"/>
<node name="Engine" child="Piston"/>
<node name="Engine" child="Carb"/>
<node name="Carb" child="Bolt"/>
<node name="Spare Wheel"/>
<node name="Bolt" child="Thread"/>
<node name="Carb" child="Foat"/>
<node name="Truck" child="Engine"/>
<node name="Engine" child="Bolt"/>
<node name="Wheel" child="Hubcap"/>
</nodes>
产生了想要的正确结果：
Car ---> Engine ---> Piston
Car ---> Engine ---> Carb ---> Bolt ---> Thread
Car ---> Engine ---> Carb ---> Foat
Car ---> Engine ---> Bolt ---> Thread
Car ---> Wheel ---> Hubcap
Spare Wheel --->
Truck ---> Engine ---> Piston
Truck ---> Engine ---> Carb ---> Bolt ---> Thread
Truck ---> Engine ---> Carb ---> Foat
Truck ---> Engine ---> Bolt ---> Thread
答案 1 :(得分：2)
如果你只想要叶子节点
<xsl:transform version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output method="text"/>
<xsl:strip-space elements="*"/>
<xsl:template match="node">
<xsl:param name="already" select="''"/>
<xsl:choose>
<xsl:when test="$already = '' and not(@child)">
<!-- no child (sparewheel) -->
<xsl:value-of select="concat(@name,'
')"/>
</xsl:when>
<xsl:when test="not(../node[@child = current()/@name])">
<!-- else will already have prefix -->
<xsl:choose>
<xsl:when test="../node[@name = current()/@child]">
<xsl:apply-templates select="../node[@name = current()/@child]">
<xsl:with-param name="already">
<xsl:value-of select="concat(@name, ' --> ', @child)"/>
</xsl:with-param>
</xsl:apply-templates>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="concat(@name, ' --> ', @child,'
')"/>
</xsl:otherwise>
</xsl:choose>
</xsl:when>
<xsl:when test="$already != ''">
<!-- called with prefix -->
<xsl:choose>
<xsl:when test="../node[@name = current()/@child]">
<xsl:apply-templates select="../node[@name = current()/@child]">
<xsl:with-param name="already">
<xsl:value-of select="concat($already, ' --> ', @child)"/>
</xsl:with-param>
</xsl:apply-templates>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="concat($already, ' --> ', @child,'
')"/>
</xsl:otherwise>
</xsl:choose>
</xsl:when>
</xsl:choose>
</xsl:template>
</xsl:transform>
答案 2 :(得分：0)
这应该执行任何递归级别，但也会输出所有中间步骤：
<xsl:transform version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output method="text"/>
<xsl:strip-space elements="*"/>
<xsl:template match="node">
<xsl:param name="already" select="''"/>
<xsl:choose>
<xsl:when test="$already = '' and not(@child)">
<!-- no child (sparewheel) -->
<xsl:value-of select="concat(@name,'
')"/>
</xsl:when>
<xsl:when test="not(../node[@child = current()/@name])">
<!-- else will already have prefix -->
<xsl:value-of select="concat(@name, ' --> ', @child,'
')"/>
<xsl:apply-templates select="../node[@name = current()/@child]">
<xsl:with-param name="already">
<xsl:value-of select="concat(@name, ' --> ', @child)"/>
</xsl:with-param>
</xsl:apply-templates>
</xsl:when>
<xsl:when test="$already != ''">
<!-- called with prefix -->
<xsl:value-of select="concat($already, ' --> ', @child,'
')"/>
<xsl:apply-templates select="../node[@name = current()/@child]">
<xsl:with-param name="already">
<xsl:value-of select="concat($already, ' --> ', @child)"/>
</xsl:with-param>
</xsl:apply-templates>
</xsl:when>
</xsl:choose>
</xsl:template>
</xsl:transform>
Ups，必须在第三种情况下调整输出，@ name已经不再需要@name了......