xsl：copy-of排除父项并替换

Question

Good afternoon! It's been a while since I've done XSLT and am a bit stumped. Trying to achieve the following:

When applied to the following xml..

<tag>
  content
  <br />
</tag>

would produce this result:

modified content
<br />

In other words, I'm trying to grab/output all contents on the main tag, including HTML but with a modified content item. Just selecting the element content is dead simple with a <xsl:copy-of select="./node()"/> and so is replacing the "content" with something else, but I can't seem to figure out how to combine the two. I'm also using XSLT 1.0 as well.

Answer 1

Use

<xsl:template match="@* | node()">
  <xsl:copy>
    <xsl:apply-templates select="@* | node()"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="tag">
  <xsl:apply-templates/>
</xsl:template>

<xsl:template match="tag/text()[contains(., 'content')]">
  <xsl:value-of select="concat('modified ', .)"/>
</xsl:template>

Of course if any first child text node of tag elements need to be manipulated then you could change the third template to say

<xsl:template match="tag/text()[1]">
  <xsl:value-of select="concat('modified ', .)"/>
</xsl:template>