arranging xml elements due to parent-child

ozane · Mar 7, 2005

hi,
i have a well formed xml file. wonna view it on browser via xsl file.
but the structure should be like in xml. i mean the tabs in browser should be set according to xml automatically. as an example:
XML
<a>
<b/>
<c>
<d/>
</c>
</a>

Browser view
a: value
b:value
c:value
d:value

is there a way or script to do it automatically??

JontyMC · Mar 8, 2005

Why not try:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="[URL unfurl="true"]http://www.w3.org/1999/XSL/Transform">[/URL]
  <xsl:template match="/">
    <html>
      <head>
        <title>XML view</title>
        <style type="text/css">p{margin: 0;}</style>
      </head>
      <body>
        <xsl:for-each select="node()">
          <xsl:call-template name="displayNodes">
            <xsl:with-param name="nodes" select="."/>
            <xsl:with-param name="indent" select="0"/>
          </xsl:call-template>
        </xsl:for-each>
      </body>
    </html>
  </xsl:template>
  <xsl:template name="displayNodes">
    <xsl:param name="nodes"/>
    <xsl:param name="indent"/>
    <xsl:if test="local-name($nodes)!=''">
      <p style="margin-left: {$indent}em">
        <xsl:value-of select="local-name($nodes)"/>
        <xsl:text>:</xsl:text>
        <xsl:value-of select="$nodes/text()"/>
      </p>
    </xsl:if>
    <xsl:if test="$nodes/node()">
      <xsl:for-each select="$nodes/node()">
        <xsl:call-template name="displayNodes">
          <xsl:with-param name="nodes" select="."/>
          <xsl:with-param name="indent" select="$indent + 1"/>
        </xsl:call-template>
      </xsl:for-each>
    </xsl:if>
  </xsl:template>
</xsl:stylesheet>

ozane · Mar 8, 2005

ooppss

thanks. will try.

but it will be hard to adapt my xsl to yours

JontyMC · Mar 8, 2005

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arranging xml elements due to parent-child

ozane

Technical User

JontyMC

Programmer

ozane

Technical User

JontyMC

Programmer

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