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I got the following xml:

<categories>
    <category>
        <id>1</id>
        <name>aaa</name>
                <url>lord</url>
        <categories>
            <category>
                <id>2</id>
                <name>bbb</name>
                                     <url>grrr</url>
            </category>
            <category>
                <id>3</id>
                <name>ccc</name>
                                     <url>grrr</url>
            </category>
        </categories>
    </category>
</categories>

What I need is to generate a html like:

<ul>
 <li>
  <a href="url">aaa</a>
  <ul>
   <li><a href="url">bbb</a></li>
   <li><a href="url">ccc</a></li>
  </ul> 
 </li>
<ul>

Any tip?

ps: I can have n category elements nested in the categories root, and n nested category elements inside each category.

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2
  • Sounds like a job for XML StyleSheets! (dum-da-da-dum) Commented Nov 27, 2011 at 22:19
  • i didnt, im not really sure how can i do that. Commented Nov 27, 2011 at 22:32

3 Answers 3

Reset to default
3

Using XSLT, something like this should work:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxsl="urn:schemas-microsoft-com:xslt"
    exclude-result-prefixes="msxsl">
  <xsl:output method="html" indent="yes"/>

  <xsl:template match="categories">
    <ul>
      <xsl:apply-templates select="category"/>
    </ul>
  </xsl:template>

  <xsl:template match="category">
    <li>
      <a>
        <xsl:attribute name="href"><xsl:value-of select="url"/></xsl:attribute>
        <xsl:value-of select="name"/>
      </a>
      <xsl:apply-templates select="categories"/>
    </li>
  </xsl:template>
</xsl:stylesheet>

Which yields:

<ul>
  <li><a href="lord">aaa</a><ul>
      <li><a href="grrr">bbb</a></li>
      <li><a href="grrr">ccc</a></li>
    </ul>
  </li>
</ul>
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1 Comment

your code works fine. I changed my original post to put the url problem, I know its not something that i should put after, but the xls syntax its not clear for me, sorry and thx again.
3

Use XSLT - this answer does not need to be longer

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3 Comments

I think it needs to be a bit longer. You should explain how this would be done with XSLT, and ideally give an example or link to a useful resource which gives tutorials in XLST.
Conciseness is better. Any good book in XSLT has that canonical example. Thanks. The tendency is to provide too much help. Someday I may need that level of help though.
You should have answered 'Go buy a book on XSLT'.
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The preferred method is probably XSLT, but I'm quite partial to LINQ-to-XML because I can do it completely in code (no reference or dependence on an external XSLT document). Actually I just enjoy writing LINQ-to-XML a lot more than XSLT.

Here's an example of projecting your category node into a li node

private XElement ConvertCategoryToListItem(XElement category)
{
    // new list item using the 'name' element
    var result = new XElement("li", category.Element("name").Value);

    // add any sub-categories to an ul element
    if (category.Element("categories") != null)
    {
        var nestedCategories = category.Element("categories").Elements("category");
        result.Add(new XElement("ul"), nestedCategories.Select(c => 
            ConvertCategoryToListItem(c)));
    }
    return result;
}

I'd then call this on all the categories in your original XML.

new XElement("ul", this.input.Elements("category").Select(c => 
    ConvertCategoryToListItem(c)));
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1 Comment

wow, thank you. very nice code. In my case scenario the user can change the html, so the xlst fits better! But I do prefer (like you) hard code everything when it applies.

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