英文:
How to transform an html ul tree to 1D xml using xslt 1.0?
问题
你想要将HTML的嵌套ul树转换为一维XML，使用XSLT 1.0。以下是期望的XML输出（注意：pid是父级id）：
<?xml version="1.0" encoding="UTF-8"?>
<items>
<item id="1" pid="0" name="Name 1">Hi!</item>
<item id="2" pid="1" name="Name 2">Hello</item>
<item id="3" pid="2" name="Name 3">Ho ho</item>
<item id="4" pid="2" name="Name 4">How do you do?</item>
<item id="5" pid="4" name="Name 5">I'm fine</item>
<item id="6" pid="5" name="Name 4">Ok</item>
<item id="7" pid="1" name="Name 3">Hi. How do you do?</item>
<item id="8" pid="7" name="Name 1">Fine</item>
</items>
你提供的XSLT看起来基本正确，但它可能无法处理完整的HTML。你需要确保XSLT可以从给定的HTML中提取所需的数据。如果你遇到问题，可能需要进一步调整XSLT以适应你的HTML结构。
英文:
How to transform an html ul tree to 1D xml using xslt 1.0?
I want to transform the chat tree into a flat list. This is my first time working with xml and I have already done 2 other transformations. But I couldn't make xml from html. How is this done?
An input html structure with a nested chat tree with messages and names.
<!DOCTYPE html>
<html>
<head>
<title>Chat</title>
</head>
<body>
<ul>
<li>
<b>Name 1</b> say: Hi!<ul>
<li>
<b>Name 2</b> say: Hello<ul>
<li>
<b>Name 3</b> say: Ho ho
</li>
<li>
<b>Name 4</b> say: How do you do?<ul>
<li>
<b>Name 5</b> say: I'm fine<ul>
<li>
<b>Name 4</b> say: Ok
</li>
</ul>
</li>
</ul>
</li>
</ul>
</li>
<li>
<b>Name 3</b> say: Hi. How do you do?<ul>
<li>
<b>Name 1</b> say: Fine
</li>
</ul>
</li>
</ul>
</li>
</ul>
</body>
</html>
Expected xml output (clarification: pid is the parent id):
<?xml version="1.0" encoding="UTF-8"?>
<items>
<item id="1" pid="0" name="Name 1">Hi!</item>
<item id="2" pid="1" name="Name 2">Hello</item>
<item id="3" pid="2" name="Name 3">Ho ho</item>
<item id="4" pid="2" name="Name 4">How do you do?</item>
<item id="5" pid="4" name="Name 5">I'm fine</item>
<item id="6" pid="5" name="Name 4">Ok</item>
<item id="7" pid="1" name="Name 3">Hi. How do you do?</item>
<item id="8" pid="7" name="Name 1">Fine</item>
</items>
Maybe like this? But it doesn't work with full html.
<xsl:stylesheet version="1.0" encoding="utf-8"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="b">
<item>
<xsl:attribute name="name">
<xsl:value-of select="@*|node()"/>
</xsl:attribute>
</item>
</xsl:template>
</xsl:stylesheet>
答案1
得分: 2
请看这个作为你的起点：
XSLT 1.0
<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="*"/>
<xsl:template match="/">
<items>
<xsl:for-each select="//li">
<item id="{generate-id()}" pid="{generate-id(ancestor::li[1])}" name="{b[1]}">
<xsl:value-of select="normalize-space(substring-after(text()[1], 'say: '))" />
</item>
</xsl:for-each>
</items>
</xsl:template>
</xsl:stylesheet>
请注意，id 和 pid 值的格式取决于处理器。
英文:
Try this as your starting point:
XSLT 1.0
<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="*"/>
<xsl:template match="/">
<items>
<xsl:for-each select="//li">
<item id="{generate-id()}" pid="{generate-id(ancestor::li[1])}" name="{b[1]}">
<xsl:value-of select="normalize-space(substring-after(text()[1], 'say: '))" />
</item>
</xsl:for-each>
</items>
</xsl:template>
</xsl:stylesheet>
Note that the format of the id and pid values is processor-dependent.
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