Question

我有xml文档如下，

<footnote>
   <p type="Footnote Text">
      <link ref="http://www.apple.com/accessibility/iphone/hearing.html">
         <c type="Hyperlink">http:.apple.com/accessibility/iphone/hearing.html
         </c>
       </link>
     </s>
    </p>
</footnote>

我需要做的是从xslt我需要删除标记内url的子字符串。

示例：

像这样的原始xml文档，

<c type="Hyperlink">http:www.apple.com/accessibility/iphone/hearing.html
 </c>

我需要转换如下：

<c type="Hyperlink">www.apple.com </c>

我搜索了XSLT inbuild函数以删除字符串中的某些子字符串，但我找不到这样的函数。

你可以提出我怎么做的建议吗？

Answer 1

c元素可以这种方式转换：

<xsl:template match="c">
    <xsl:copy>
        <xsl:copy-of select="@*"/>
        <xsl:variable name="sa" select="substring-after(.,':')"/>
        <xsl:variable name="sb" select="substring-before($sa,'/')"/>
        <xsl:value-of select="$sb"/>
    </xsl:copy>
</xsl:template>

请注意，您的示例不是有效的XML文件，关闭标记时会出现问题。

XSLT：删除XML标记中的某些文本

1 个答案: