使用 XSLT 复制 XML 中的所有节点，支持特殊情况
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Using XSLT to copy all nodes in XML, with support for special cases
提问by del.ave
Say I have a large XML file that the following structure:
假设我有一个大型 XML 文件，其结构如下：
<MyXml>
<Data1>
<Node1>1234</Node1>
<Node2>abc<Node2>
<Node3>gfdf</Node3>
...
<Node10000>more text</Node10000>
</Data1>
<Data2>
...
</Data2>
</MyXml>
I want to transform this XML into another XML that looks exactly the same, but has a certain string concatinated to a certain node, say Node766. I am using an XSLT of course and wondering how I can tell it to copy everyhing as-is except for Node766, where I have to do something before outputing it.
我想将此 XML 转换为另一个看起来完全相同的 XML，但将某个字符串连接到某个节点，例如Node766。我当然正在使用 XSLT，想知道如何告诉它按原样复制除Node766之外的所有内容，我必须在输出它之前做一些事情。
回答by Mads Hansen
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<!--Identity template,
provides default behavior that copies all content into the output -->
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<!--More specific template for Node766 that provides custom behavior -->
<xsl:template match="Node766">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
<!--Do something special for Node766, like add a certain string-->
<xsl:text> add some text </xsl:text>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
回答by Don Roby
Start with an identity transform, and include a template match for your exception.
从身份转换开始，并为您的异常包含模板匹配。
