How do i remove all linebreaks?

2019-01-23 15:02发布

I have something like this:

<node TEXT="   txt A   "/>
<node TEXT="

       txt X

"/>
<node>
   <html>
      <p>
        txt Y
      </p>
   </html>
</node>
<node TEXT="txt B"/>

and i want to use XSLT to get this:

txt A
txt X
txt Y
txt B

I want to strip all useless whitespaces and linebreaks of @TEXT's and CDATA's. The only XML-input that is giving structure to the output are the <node>-tags.

2条回答
聊天终结者
2楼-- · 2019-01-23 15:28

The following transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output method="text"/>
<xsl:template match="*">
  <xsl:apply-templates select="@TEXT | node()"/>
</xsl:template>

<xsl:template match="node/@TEXT | text()">
  <xsl:if test="normalize-space(.)">
    <xsl:value-of select=
     "concat(normalize-space(.), '&#xA;')"/>
  </xsl:if>

  <xsl:apply-templates />
</xsl:template>
</xsl:stylesheet>

when applied against this XML document

<t>
    <node TEXT="   txt A   "/>
    <node TEXT="       txt X"/>
    <node>
        <html>
            <p>        txt Y      </p>
        </html>
    </node>
    <node TEXT="txt B"/>
</t>

produces the wanted result:

txt A
txt X
txt Y
txt B

Do note the use of the standard XPath function normalize-space(), which strips off all leading and trailing spaces and replaces every sequence of other spaces with just one space.

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仙女界的扛把子
3楼-- · 2019-01-23 15:28

You probably want

 <xsl:strip-space elements="node"/>

explained here. And this article has a lot more details.

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