Transform xml into html with XSLT

Question

I know that's easy, but I tried and XSLT is really non-understandable for me... That is why I would like to have an example over my code so I could examine it on real-case example and probably understand it.

So I have app that has GUI tree view and generates me XML of this sort:

<TreeView>
  <Parent text="Installation">
    <child text="Startup" type="video" file="startup.wmv" icon="c://user/desktop/blahblah.jpg"/>
    <child text="Getting there" type="video" file="gettingthere.wmv" icontPath="something"/>
    <child text="Steps" type="document" file="steps.docx" icon="asd"/>
    <child text="Pictures" type="image" file="pics.jpg" icon="dasdasdas"/>
  </Parent>
  <Parent text="Usage">
    <child text="Tilbud pane" type="video" file="tilbud.mwv" icon="asdasd"/>
    <child text="Report pane" type="document" file="report.docx" icon="gfdhd"/>
  </Parent>
</TreeView>

Then I need to transform this XML into HTML so I can update it in my website.

So I don't need the <html> and <body> tags. I need just the ordering of this XML into a list, whereas there should be some space before the child elements.

A desired output, viewed by the user in the browser will be like that:

Installation
  Startup
  Getting there
  Steps
  Pictures
Usage
  Tilbud pane
  Report pane

But because I will not have the same values in the attributes - I have to order upon the elements.

This is what I have:

<?xml version = "1.0" encoding = "utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="/">
    <ul>
      <xsl:for-each select="Parent">
        <li>
        <xsl:value-of select="@text"/>
          <ul>
            <xsl:for-each select="Parent/child"/>
            <li>
            <xsl:value-of select="@text"/>
            </li>
            </xsl:for-each>
          </ul>
        </li>
      </xsl:for-each>
    </ul>
  </xsl:template>
</xsl:stylesheet>

and the desired html I have to get is smth like this:

<ul>
    <li>Parent
         <ul>
               <li>Child</li>
               <li>Child</li>
 .....
         </ul>
    </li>
   <li>Parent
         <ul>
               <li>Child</li>
               <li>Child</li>
 .....
         </ul>
   </li>
.....
</ul>

but apparently it doesn't want to give me this... It gives me just <ul/> after I run the transformation...

Similar to stackoverflow.com/questions/3839730/…

fred02138
– fred02138

2013-08-19 13:16:07 +00:00
Commented Aug 19, 2013 at 13:16 — fred02138
– fred02138, Commented Aug 19, 2013 at 13:16

Jens Teich · Accepted Answer · 2013-08-19 14:30:38Z

1

Your XSLT does not work because it searches Parent directly at / but there is TreeView in between.

<xsl:template match="/TreeView">
   ...

answered Aug 19, 2013 at 14:30

Jens Teich

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2 Comments

Syspect Over a year ago

You were right about that, but I still don't get the children in the output. I get just the Parents.

Jens Teich Over a year ago

<xsl:for-each select="child"/>

fred02138 · Accepted Answer · 2013-08-19 13:42:54Z

0

Here is a stylesheet that does the transform:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:output method="html" encoding="utf-8"/>

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

    <xsl:template match="Parent">
        <li>
        <xsl:value-of select="@text"/>
        <br/>
        <ul>
            <xsl:apply-templates select="child"/>
        </ul>
        </li>
    </xsl:template>

    <xsl:template match="child">
        <li><xsl:value-of select="@text"/></li>
    </xsl:template>

</xsl:stylesheet>

If you just want text output:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:output method="text" encoding="utf-8"/>

    <xsl:template match="Parent">
        <xsl:value-of select="@text"/>
        <xsl:text>&#xa;  </xsl:text>
        <xsl:apply-templates select="child"/>
    </xsl:template>

    <xsl:template match="child">
        <xsl:text>  </xsl:text>
        <xsl:value-of select="@text"/>
        <xsl:text>&#xa;  </xsl:text>
    </xsl:template>

</xsl:stylesheet>

edited Aug 19, 2013 at 13:42

answered Aug 19, 2013 at 13:37

fred02138

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1 Comment

Syspect Over a year ago

So the correct way of doing these things is: making a template for every thing that I want to match and then calling "apply-templates"?

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