Display contents of XML-file in HTML using jQuery stays empty

Question

I am trying to simply display contents of a XML file in an HTML list. For some reason the result remains empty and I don't know why. I did check out many of the questions and answers but still did not succeed.

Here is the code at JSbin: http://jsbin.com/arufep/edit

The goal is to have a list of clickable url's like this:

<li><a href="/dir/xyz.htm">description</a></li>

Thanks for looking into it.

SeanCannon · Accepted Answer · 2011-07-19 01:29:56Z

1

You're attempting to make a cross-domain ajax call to http://www.halo-photographs.com/test-code/panorama-list1.xml which is not allowed by the browser's Same Origin Policy.

You can look into JQuery's crossDomain:true ajax parameter to accomplish this but I believe it will not work with an XML response.

answered Jul 19, 2011 at 1:29

SeanCannon

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3 Comments

snahl Over a year ago

That XML file is located on that same domain which is my server.

SeanCannon Over a year ago

Try giving it a relative url path instead of an absolute path

snahl Over a year ago

Your are correct to use a relative path. But still there is no result, the alert-box in jsbin says [object Object] and the same page downloaded from jsbin reveals the content of <h2> only. So my assumption is that something in JavaScript code is still not as it should be to display the contents of that XML as an HTML list <li>.

2 revs, 2 users 97% · Accepted Answer · 2013-08-11 10:21:19Z

Below is the PHP5-page to used to insert data originating from an XML-file without client-side scripting.

Make sure your XSLT outputs what you intend to have displayed.

Comment: I must admit my initial question referred to jQuery, but I ended up with this rather simple and effective solution instead.

<?php
  $path_xml = "/...path-to.../list.xml";
  $path_style = "/...path-to.../list.xsl";  
  $xml_obj = new DomDocument;
  $xsl_obj = new DomDocument;

  if (!$xml_obj->load($path_xml)) {
      echo "Error! Unable to open " . $path_xml . "!\n";
      exit;
  }
  if (!$xsl_obj->load($path_style)) {
      echo "Error! Unable to open " . $path_style . "!\n";
      exit;
  }
  $xslt_parse = new xsltprocessor;
  $xslt_parse->importStyleSheet($xsl_obj);
?>
<!DOCTYPE...>
<head>
</head>
<body>

<p><!-- display XSLT-styled content of XML-data here: -->
    <?php echo $xslt_parse->transformToXML($xml_obj);?>
</p>
</body> 
</html>

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