格式化XSL转换中的URI

Question

I am using nginx and I am trying to do an directory listing using the xslt transformation

the xslt I got from here https://gist.github.com/wilhelmy/5a59b8eea26974a468c9

This is the line that prints out the file

But the problem is current() doesn't do any URI encoding and filenames with % cause problems.

From my limited knowledge I am using xslt v1 so I am missing out on some of the XQuery ? function line encode uri.

But I found http://www.getsymphony.com/download/xslt-utilities/view/55460/

which has a library ? template for encoding uri's

but I am unsure how to change

to include

if I try adding the above in the href=" i get errors.

Thanks

Answer 1

Found my answer

              <xsl:element name="a">
                 <xsl:attribute name="href">
                   <xsl:call-template name="url-encode">
                     <xsl:with-param name="str" select="current()" />
                   </xsl:call-template>
                 </xsl:attribute>
                 <xsl:value-of select="." />
               </xsl:element>