how to include the node XML in my XSLT text output

问题:

I'm trying to convert an XML file into a flat, pipe-delimited file with XSLT (for bulk-loading into Postgres). I would like the last column in my output to be the actual XML of the node (for additional post-processing and debugging). For example:

<Library>
  <Book id="123">
    <Title>Python Does Everythig</Title>
    <Author>Smith</Author>
  </Book>

  <Book id="456">
    <Title>Postgres is Neat</Title>
    <Author>Wesson</Author>
  </Book>
</Library>

Should generate

Python Does Everything|Smith|<Book id="123"><Title>Python Does Everythig</Title>Author>Smith</Author></Book>
Postgres is Neat|Wesson|<Book id="456"><Title>Postgres is Neat</Title><Author>Wesson</Author></Book>

My current XSL is

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:strip-space elements="*" />
  <xsl:output method="text" omit-xml-declaration="yes" indent="no" /> 
  <xsl:template match="//Book">
    <xsl:value-of select="Title" />
    <xsl:text>|</xsl:text>
    <xsl:value-of select="Author" />

    <!-- put in the newline -->
    <xsl:text>&#xa;</xsl:text>
  </xsl:template>
</xsl:stylesheet>    

回答1:

I am not sure if this is a recommended solution, but you could try setting the output method to xml, and then just using the xsl:copy-of function.

So, the following XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
  <xsl:strip-space elements="*" /> 
  <xsl:output method="xml" omit-xml-declaration="yes" indent="no" />  
  <xsl:template match="//Book"> 
    <xsl:value-of select="Title" /> 
    <xsl:text>|</xsl:text> 
    <xsl:value-of select="Author" /> 
    <xsl:text>|</xsl:text>  
    <xsl:copy-of select="." />
    <!-- put in the newline --> 
    <xsl:text>&#xa;</xsl:text> 
  </xsl:template> 
</xsl:stylesheet>  

When applied to your sample XML, generates the following output

Python Does Everythig|Smith|<Book id="123"><Title>Python Does Everythig</Title><Author>Smith</Author></Book>
Postgres is Neat|Wesson|<Book id="456"><Title>Postgres is Neat</Title><Author>Wesson</Author></Book>

回答2:

Try that :

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output method="text"/>
    <xsl:strip-space elements="*"/>
    <xsl:template match="/">
        <xsl:apply-templates select="//Book"/>
    </xsl:template>
    <xsl:template match="Book">
        <xsl:value-of select="Title" />
        <xsl:text>|</xsl:text>
        <xsl:value-of select="Author" />
        <xsl:text>|</xsl:text>
        <xsl:apply-templates select="." mode="outputTags"/>
    </xsl:template>
    <xsl:template match="*"  mode="outputTags">
        <xsl:text>&lt;</xsl:text>
        <xsl:value-of select="local-name()"/>
        <xsl:apply-templates select="@*"/>
        <xsl:text>></xsl:text>
        <xsl:apply-templates mode="outputTags"/>
        <xsl:text>&lt;/</xsl:text>
        <xsl:value-of select="local-name()"/>
        <xsl:text>></xsl:text>
        <xsl:if test="self::Book">
            <xsl:text>&#x0A;</xsl:text>
        </xsl:if>
    </xsl:template>
    <xsl:template match="@*">
        <xsl:text> </xsl:text>
        <xsl:value-of select="local-name()"/>
        <xsl:text>="</xsl:text>
        <xsl:value-of select="."/>
        <xsl:text>"</xsl:text>
    </xsl:template>
</xsl:stylesheet>

It produces the following result from your input file :

Python Does Everythig|Smith|<Book id="123"><Title>Python Does Everythig</Title><Author>Smith</Author></Book>
Postgres is Neat|Wesson|<Book id="456"><Title>Postgres is Neat</Title><Author>Wesson</Author></Book>