Question

我在尝试使用CDATA时读取换行符，但运气不佳。有什么我做错了吗？我一直在说：

This Picture was created by <br /> Type-Style back in 2007.作为输出，在allm处无法识别换行符

XML

<s xmlns:b="http://crownpublishing.com/imprint/crown-business/">
    <b:bio>
        <b:about>
        <![CDATA[This Picture was created by <br /> Type-Style back in 2007. ]]>
        </b:about>
    </b:bio>
</s>

XSL

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:b="http://crownpublishing.com/imprint/crown-business/" version="1.0">
<xsl:output cdata-section-elements="about"/>
<xsl:template match="/">
    <xsl:value-of select="s/b:bio/b:about"/>
</xsl:template>
</xsl:stylesheet>

Answer 1

此答案假定您无法更改输入文档。使用字符串替换将br字符串转换为实际换行符。使用XSLT 1.0，执行字符串替换的标准方法是编写递归模板。

XSLT样式表

从Mads的回答here中无耻地窃取了指定的模板，而Dave Pawson又从online here借用。

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:b="http://crownpublishing.com/imprint/crown-business/" version="1.0"
    exclude-result-prefixes="b">

    <xsl:output method="xml"  indent="yes"/>

    <xsl:template match="/">
    <result>
        <xsl:call-template name="replace-string">
          <xsl:with-param name="text" select="/s/b:bio/b:about"/>
          <xsl:with-param name="replace" select="'&lt;br /&gt;'" />
          <xsl:with-param name="with" select="'&#10;'"/>
        </xsl:call-template>
    </result>
    </xsl:template>

    <xsl:template name="replace-string">
    <xsl:param name="text"/>
    <xsl:param name="replace"/>
    <xsl:param name="with"/>
    <xsl:choose>
      <xsl:when test="contains($text,$replace)">
        <xsl:value-of select="substring-before($text,$replace)"/>
        <xsl:value-of select="$with"/>
        <xsl:call-template name="replace-string">
          <xsl:with-param name="text"
select="substring-after($text,$replace)"/>
          <xsl:with-param name="replace" select="$replace"/>
          <xsl:with-param name="with" select="$with"/>
        </xsl:call-template>
      </xsl:when>
      <xsl:otherwise>
        <xsl:value-of select="$text"/>
      </xsl:otherwise>
    </xsl:choose>
  </xsl:template>

</xsl:stylesheet>

XML输出

<?xml version="1.0" encoding="utf-8"?>
<result>
        This Picture was created by 
 Type-Style back in 2007. 
        </result>

尝试此解决方案https://momjian.us/main/writings/pgsql/features.pdf。

Answer 2

源文档中的文字换行符将继续输出到输出。尝试：

<s xmlns:b="http://crownpublishing.com/imprint/crown-business/">
    <b:bio>
        <b:about>
        <![CDATA[This Picture was created by
Type-Style back in 2007. ]]>
        </b:about>
    </b:bio>
</s>

<br />在XML中没有任何特殊含义。另一方面，如果你打算将XHTML作为输出产生，那么你所拥有的只在浏览器中有意义，并且它应该显示为换行符。

使用CDATA和XSL时读取XML文件中的中断

2 个答案: