XML / XSL question

Hi all,

Just going through some basic XSL tutorials and I'm curious as to how I would loop through all (if any) node elements .

For example,

<catalog>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>1.90</price>
<year>1985</year>
</cd>

<cd>
<title>Apple Records</title>
<artist>Funny Monkeys</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1995</year>
</cd>
</catalog>

is my XML data and basically I'm looking for something like
<xsl:for-each select="/cd/*">
<xsl:value-of select="$1"/>
</xsl:for-each>

where $1 is the current element that is being processed.
I would have thought this was a basic task but can't seem to find it anywhere on the web, I've messed around with different possibilities but with no luck.

Any help would be appreciated on this, thanks (i'm sure it's a minor, niggly little issue somewhere)

Ginger

try here: http://www.w3schools.com/xsl/xsl_for_each.asp

Thanks but I've run through those tutorials, that specifies what elements are to be displayed, in the for-each loop.

Basically, I want to loop through all elements in a node set (if any), without knowing what could appear. Ah, I'll just mess around more, hope to stumble on the solution.

smcelhinney

Couple of things.

To filter for specific nodes, use Xpath. Xpath looks like a directory filter. Eg. to show all CD's for the year 1985 use

<xsl:for-each select="//cd/[year='1985']">
<xsl:value-of select="title" />
</xsl:for-each>

Do you want to actually display ALL the node info? i.e. in this case

<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>1.90</price>
<year>1985</year>

...

hmmm, I think then you need to be stricter, and use either schemas or DTD's. The only way I can think of doing this is programming. In Java you could use something like JDOM to do this.

 SAXBuilder parser = new SAXBuilder();
 Document doc = parser.build(urlToXMLFile);
// there are File implementations, I just cant be bothered checking what they are
Iterator i = doc.getRootElement().getChildren("cd").iterator();
while(i.hasNext()){
   Element e = (Element) i.next();
   out.println("<" + e.getName() + ">" + e.getText() + "</" + e.getName() + ">");
}

In the example above, you should obviously do some NodeType checking.

HTH.

Hi, yeah that Java example is that type of operation I was thinking of, except done through the style sheets.

Just thought there would have been an XPath expression to gather all of them and somehow, the stylesheet would hold the current element (whatever it may be) somewhere.

If it can't be done through bare XSL, fair enough, just thought something could have been worked out.

Thanks for your reply, I'm looking at the Java classes for XSL processing, look useful.

GreeBo

<xsl:template match="catalog">
<xsl:for-each select=".">
<xsl:value-of select="."/>
</xsl:for-each>
</xsl:template>

be careful using "//" in xpath, its *very* inefficient as its scans through every node in the document.
Use "/catalog/cd" instead.